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# Question about liminf of the sum of two sequences

1. Oct 22, 2009

### AxiomOfChoice

I know that for any two real sequences x_n and y_n, we have

$$\liminf_{n\to \infty} x_n + \liminf_{n\to \infty} y_n \leq \liminf_{n\to \infty} (x_n + y_n).$$

I also know that, if one of the sequences converges, the inequality becomes equality. My question is this: If I've managed to show that

$$\liminf_{n\to \infty} x_n + \liminf_{n\to \infty} y_n = \liminf_{n\to \infty}(x_n + y_n),$$

can I conclude that one, or both, of the sequences converge? A simple yes/no would suffice, but (of course) I'd prefer a short proof or counterexample. Thanks!

2. Oct 22, 2009

What if

$$x_n = y_n = (-1)^n$$

3. Oct 22, 2009

### AxiomOfChoice

Lame! I was hoping both of the sequences had to converge! And what a simple counterexample to prove me wrong! Thanks, though

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# request - solution/schematic/hex

#### arassel

Joined Nov 1, 2019
15
Hello,

If this is posted in the wrong section please correct me.

Would like a solution to a little project i'm working on :

I need a circuit ( using a 12F629 ) that will get powered with 12v (from a car battery) and will light up 3 led strips consecutively with a delay of 0.3-0.4 seconds (and stay powered as long as the circuit has power).

Any feedback would be appreciated ..

#### ericgibbs

Joined Jan 29, 2010
9,044
hi arassel,
We would need details of the LED strips. current rating, operating voltage etc...
Have you programmed the PIC, if so which programming language.?

E

#### BobTPH

Joined Jun 5, 2013
2,103
He said he wanted Hex, so no, he wants us to do everything including the programming.

In other words he is asking us to thousands of dollars worth of consulting work for free.

Bob

#### arassel

Joined Nov 1, 2019
15
Thanks for the reply, first i'd like to make you aware i know little to nothing about what i'm asking for, my electronic knowledge is scarce .. i'm some kind of scared about talking about it, not knowing exactly what i'm talking about .

please take into consideration i barely know what i'm doing ..

- ok so the 3 led strips are smd leds strips in series of 3 with a resistor , do i have to count them and get the current rating ?
- operating voltage is 12-14v, i'm trying to switch my taillights from bulb to led with a personal design. (also for changing from bulb to led what do i have to take into consideration?)

I've been looking alot for a solution to my idea as simple as possible .. i've attached pictures with what i've found and tried to build so far. any other simpler solution would be apreciated !

so far i've done this:

- i have the pic and the programmer, i used the hex i found and the programmer didnt give me any errors
- made the assembly as good as i could (please correct me if not)
- its not working

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#### arassel

Joined Nov 1, 2019
15
In other words he is asking us to thousands of dollars worth of consulting work for free.

Bob
My apologies, not my intent, i've seen so many other shared projects completely detailed with hex and schematics far more complicated than my 3 led strips yocto-project, figured i wasnt asking for much, yet alone for thousands of dollars of consulting.

#### BobTPH

Joined Jun 5, 2013
2,103
I never would have guessed from your post that you already had the schematic and the hex code, since that was what you wete asking for.

You say it is not working. So what does it do?

Bob

#### arassel

Joined Nov 1, 2019
15
You say it is not working. So what does it do?

Bob
when powering the circuit, nothing lights up, i will get a measuring tool today to see weather i'm not using the correct mosfets ... my thought is that the hex is not good .. but considering the most complex assembly i've ever done so was was a bike led blinker 15 years ago, i very much doubt my efforts in pulling this off.

first of all i dont know if the schematic for the soldering i did (last attachment) matches the schematic i found on the web (first attach) .. i dont even have refference to where i got the schematic and hex from, lost that trying to learn ho to program the pic .

Last edited:

#### BobTPH

Joined Jun 5, 2013
2,103
Make a copy of the schematic. Check each connection in the circuit you built and cross it off when it is verified. Make sure each set of connections (node) does not have any extra connections. Pay close attention to pinouts.

Unfortunately, you do not know if the hex is correct, no easy way to help you with that.

Verify that the strips work by using a wire to short across the source anddrain — that should light the up. Verify that the MOSFETs are working by putting 5V on the gate, that should also light the strips.

Bob

#### jpanhalt

Joined Jan 18, 2008
8,311
It's been awhile since I worked with the 12F629. Here's a (edit: raw ) disassembled version of that hex file:
Code:
    processor 12F629
#include <P12F629.INC>
__config 0x3194
;   _CPD_OFF & _CP_OFF & _BODEN_OFF & _MCLRE_OFF & _PWRTE_OFF & _WDT_OFF
;   & _INTRC_OSC_NOCLKOUT

; RAM-Variable
LRAM_0x5B equ 0x5B
LRAM_0x5C equ 0x5C
LRAM_0x5D equ 0x5D

; Program

Org 0x0000

;   Reset-Vector

Org 0x0003

MOVLW 0x80           ;   b'10000000'  d'128'
;   Interrupt-Vector
BSF STATUS,RP0       ; !!Bank Register-Bank(0/1)-Select
MOVWF TMR0           ; !!Bank!! TMR0 - OPTION_REG
CLRF INTCON
MOVLW 0x07           ;   b'00000111'  d'007'
BCF STATUS,RP0       ; !!Bank Register-Bank(0/1)-Select
MOVWF CMCON          ; !!Bank!! CMCON - VRCON
CLRF GPIO            ; !!Bank!! GPIO - TRISIO
MOVLW 0x38           ;   b'00111000'  d'056'  "8"
BSF STATUS,RP0       ; !!Bank Register-Bank(0/1)-Select
MOVWF GPIO           ; !!Bank!! GPIO - TRISIO
RETURN
BCF STATUS,RP0       ; !!Bank Register-Bank(0/1)-Select
BSF GPIO,0           ; !!Bank!! GPIO - TRISIO
MOVLW 0x06           ;   b'00000110'  d'006'
MOVWF LRAM_0x5B
MOVLW 0x13           ;   b'00010011'  d'019'
MOVWF LRAM_0x5C
MOVWF LRAM_0x5D
DECFSZ LRAM_0x5D,F
DECFSZ LRAM_0x5C,F
DECFSZ LRAM_0x5B,F
NOP
NOP
BSF GPIO,1           ; !!Bank!! GPIO - TRISIO
MOVLW 0x06           ;   b'00000110'  d'006'
MOVWF LRAM_0x5B
MOVLW 0x13           ;   b'00010011'  d'019'
MOVWF LRAM_0x5C
MOVWF LRAM_0x5D
DECFSZ LRAM_0x5D,F
DECFSZ LRAM_0x5C,F
DECFSZ LRAM_0x5B,F
NOP
NOP
BSF GPIO,2           ; !!Bank!! GPIO - TRISIO

End
I am taking a look at the code, but it will take awhile.

Last edited by a moderator:

Joined Apr 16, 2011
381
Hex file works OK on my home-made test board, except that 3 test LEDs light up at approximately 1 second intervals, not 0.3 to 0.4 seconds. .... but that is a good thing for testing - easier to work with....

PIC12F629 has a calibration word as the last instruction in memory. If that is missing, it can prevent the PIC from running.

I like BobTPH's idea of testing LED driver with no PIC fitted - proves the LED hardware.

Additionally you might wait 5 seconds after PIC power-up and check with multi-meter or oscilloscope that all three output pins are sitting at +5V (or probably a little less).

#### arassel

Joined Nov 1, 2019
15
Thank you all for your reply's ... Had a revelation trying to explain myself the schematic i done .. i totally messed up the mosfets drain-gate-source on my schematic .. no way it could work .. redoing it all and come back with feedback.

#### jpanhalt

Joined Jan 18, 2008
8,311
Here's slightly "corrected" code FWIW. Note, when in Bank1, a write to TMR0 actually writes to OPTION_REG, and similarly, a write to GPIO writes to TRISIO. I madethose changes and added asterisks to the code that was commented out.

Code:
    processor 12F629
#include <P12F629.INC>
__config 0x3194
;   _CPD_OFF & _CP_OFF & _BODEN_OFF & _MCLRE_OFF & _PWRTE_OFF & _WDT_OFF
;   & _INTRC_OSC_NOCLKOUT

; RAM-Variable
LRAM_0x5B equ 0x5B
LRAM_0x5C equ 0x5C
LRAM_0x5D equ 0x5D

; Program

Org 0x0000

;   Reset-Vector

Org 0x0003

MOVLW 0x80           ;
;   Interrupt-Vector
BSF STATUS,RP0       ;|B1
;   MOVWF TMR0*
MOVWF OPTION_REG     ;TMR0 = OPTION_REG when in |B1
CLRF INTCON          ;
MOVLW 0x07           ;
BCF STATUS,RP0       ;|B0
MOVWF CMCON          ;turn off comparators
CLRF GPIO            ;
MOVLW 0x38           ; b'00111000'
BSF STATUS,RP0       ;|B1
;   MOVWF GPI0*
MOVWF TRISIO         ;GPIO = TRISIO when in |B1, sets GPIO 0,1,2 to output
RETURN
BCF STATUS,RP0       ;|B0
BSF GPIO,0           ;
MOVLW 0x06           ;b'00000110'
MOVWF LRAM_0x5B
MOVLW 0x13           ;b'00010011'
MOVWF LRAM_0x5C
MOVWF LRAM_0x5D
DECFSZ LRAM_0x5D,F
DECFSZ LRAM_0x5C,F
DECFSZ LRAM_0x5B,F
NOP
NOP
BSF GPIO,1           ;
MOVLW 0x06           ;b'00000110'
MOVWF LRAM_0x5B
MOVLW 0x13           ;b'00010011'
MOVWF LRAM_0x5C
MOVWF LRAM_0x5D
DECFSZ LRAM_0x5D,F
DECFSZ LRAM_0x5C,F
DECFSZ LRAM_0x5B,F
NOP
NOP
BSF GPIO,2           ;

End
The code should work and compiles fine, except the delays have not been checked.

#### jpanhalt

Joined Jan 18, 2008
8,311
I just simulated the above program at 4 MHz using MPLAB SIM. First, the led attached to GPIO,0 turns on. After a delay of 1 second (1.00000x seconds), the led attached to GPIO,1 turns on, then after another delay of 1 second, the led attached to GPIO,2 turns on. It then idles in that state.

Those are in accord with @hexreader 's observations and suggest that the TS's expectations for delay are wrong. If he wants shorter intervals, that is easily fixed.

#### arassel

Joined Nov 1, 2019
15
Thanks alot for the help guys, i'm grateful for the effort in your reply's .. will try to redo the assembly after work ..

I just simulated the above program at 4 MHz using MPLAB SIM. First, the led attached to GPIO,0 turns on. After a delay of 1 second (1.00000x seconds), the led attached to GPIO,1 turns on, then after another delay of 1 second, the led attached to GPIO,2 turns on. It then idles in that state.

Those are in accord with @hexreader 's observations and suggest that the TS's expectations for delay are wrong. If he wants shorter intervals, that is easily fixed.
.. meanwhile

- the led strips i want to use have 21 smd leds eatch .. can the pic handle the amps ?

- what do i need to use to compile the code ? that gives me the hex file i need to program in the pic right ? (again, never done this before please dont stone me ) what do i need to change to make the delay at 300 or 400 ms instead of 1 second ? what i want is when circuit is powered 1st strip to light up, 0.3 seconds later 2nd strip lights up, 0.3 seconds after 3rd strip lights up and all stay on from the moment they are turned on by the pic.

- is there any delay from the moment of the circuit getting power to do the light up ?

#### jpanhalt

Joined Jan 18, 2008
8,311
.. meanwhile
- the led strips i want to use have 21 smd leds eatch .. can the pic handle the amps ?
The PIC cannot run that many LED's DIRECTLY, but that is the purpose of the mosfets. Be sure you have logic level mosfets, and there is no reasonable limit to what you can run.

- what do i need to use to compile the code ? that gives me the hex file i need to program in the pic right ? (again, never done this before please dont stone me ) what do i need to change to make the delay at 300 or 400 ms instead of 1 second ?
You do not need to compile the code. What you have works. The code I posted simply converted that hex file to Assembly language. There are a few "quirks" in the way such disassembly works. I changed them to make it more readable. They were not errors. If you want to compile it, you will need to download a compiler from Microchip. They are generally free for compiling Assembly. I use MPLAB 8.92, which includs the compiler. If you really want the delay to be different, I or many others here can do that for you and post the code. It is really no big deal. However, the original author of that code carefully planned it to be 1 second. Anyone who does that can easily post the new hex code.

what i want is when circuit is powered 1st strip to light up, 0.3 seconds later 2nd strip lights up, 0.3 seconds after 3rd strip lights up and all stay on from the moment they are turned on by the pic.
That is what the current code does, except instead of 0.3 seconds, it is 1 second. 0.3 seconds is really quite brief, but if that is what you need, just say so.

- is there any delay from the moment of the circuit getting power to do the light up ?
As I recall, there is a short delay of about 18 microseconds, which is not perceptible by humans. Your switch may bounce that long. It can be shortened a little but not made "zero."

#### arassel

Joined Nov 1, 2019
15
@ jpanhalt thanks again, really appreciate you taking the time for me ... i may have to ask if you can change the code, make 2 different hex files ?

these are the mosfets .. i bent the legs to get less routing on the circuit board i drawn after the schematic, tried different routes, but i dont have much space so i went for the leg bending one .

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#### jpanhalt

Joined Jan 18, 2008
8,311
I cannot make out clearly the whole ID on the mosfets. Please tell . Generally, the pins are GDS in that order, and the metal tab is electrically connected to the middle pin.

I'll work on shortening the delays.

John

#### bertus

Joined Apr 5, 2008
20,171
Hello,

How are you driving the mosfet?
A full schematic would give us the information needed.
If the gate voltage is to low, the drain current will be to low.

Bertus

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#### arassel

Joined Nov 1, 2019
15
I cannot make out clearly the whole ID on the mosfets. Please tell . Generally, the pins are GDS in that order, and the metal tab is electrically connected to the middle pin.
this is the data sheet of the mosfet.

I'll work on shortening the delays.

John
is it possible to make another hex file from the one i got but modify it to be with 0.5 s delay but to keep the strips light up only for just 0.3 seconds ? so the operation order would be :

power to circuit - 1 led lights up only for 0.3s - 2nd led strip lights up 0.2s after the 1st strip for another 0.3s - 3rd led strip lights up with 0.2s from the 2nd for 0.3s.

hope i made myself clear on that

#### arassel

Joined Nov 1, 2019
15
Hello,

How are you driving the mosfet?
A full schematic would give us the information needed.
If the gate voltage is to low, the drain current will be to low.
the only schematic i have is the one i found on the internets here, posted it earlyer, attached it again now

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text:
## Directory Services Overview

Beyond directory services, what are some other uses for LDAP? What type of structure does LDAP use for data? You do not need to memorize the abbreviations used in LDAP, but you should be aware of them. This video reviews active directories, which use both LDAP and Kerberos for authentication.

#### Video Transcript

This video presents a brief overview of directory services and related concepts. This information is very helpful in understanding the larger role that many of these services play in modern computer networks.

At its core, a directory service is simply a piece of software that can store and retrieve information. The information stored could take many forms, including user accounts, groups, system information, resources, security settings, and more. By providing a centralized location to store that information, it becomes much simpler to share that information across a large enterprise network.

We've already worked with one example of a directory service in this course. The modern Domain Name Service is, at its core, a directory service. It stores a directory of domain names matched to IP addresses, as well as additional information about a domain such as the location of the mail server. It can then provide that information to the systems on a large network, providing a level of consistency that is difficult to achieve if each computer has to store a local version of the same information.

Directory services for computers have also been around for quite a while. This page gives a brief timeline of events, starting in the late 1980s and culminating with the release of Microsoft's Active Directory Domain Services in 1999. Let's look at each of these items in detail.

First, the X.500 set of standards was released in 1988, and was one of the first projects that aimed to provide a consistent directory of user account information. It was actually developed as a directory for the X.400 set of standards for email, which at that time was the dominant way that email was handled on the internet. Those standards were developed using the OSI networking protocols. However, with the rise of the TCP/IP set of internet protocols, these systems were quickly made obsolete. X.400 was replaced by the SMTP standards used today to send email between MTAs, while X.500 forms the basis of many later protocols. What is most notable about X.500 is that it actually defined several underlying protocols, including a Directory Access Protocol (DAP), as well as a Directory System Protocol (DSP) and others. Today, we primarily use protocols based on the original DAP defined by X.500.

Soon after the release of the X.500 standards, work began on an implementation of the DAP protocol using the TCP/IP internet protocols. It was eventually named the Lightweight Directory Access Protocol, or LDAP. LDAP is the underlying protocol used by many of todays directory services, including Microsoft Active Directory.

However, it is very important to understand that LDAP is not a complete implementation of the original X.500 DAP. Instead, it implements a core set of features, and then adds some additional features not present in the original protocol, while several parts of the original protocol are not implemented. The maintainers of the X.500 standard are very quick to point out this difference, but in practice most systems fully support LDAP as it stands today.

LDAP also has uses beyond just a directory service for user accounts on workstations. For example, LDAP can be used to provide a user lookup service to an email client, or it can be linked with a website to provide user authentication from a central repository of users.

LDAP, along with many other directory services, defines a tree structure for the data. In this way, the data can be organized in a hierarchical manner, closely resembling the real relationships between people, departments, and organizations in an enterprise. For example, in this diagram, you can clearly see nodes in the tree for the overarching organization, the organizational unit, and the individual user.

LDAP uses several short abbreviations for different fields and types of data, and at times they can be a bit confusing to new system administrators. This slide lists a few of the most common ones, as well as a good way to remember them. For example, an  ou  in LDAP is an organizational unit, which could be a department, office, or some other collection of users and systems.

You can see these abbreviations clearly in this sample LDAP entry. Here, we can see the distinguished name includes the canonical name, as well as the domain components identifying its location in the tree. We can also see the common name and surname for this user, as well as the distinguished name for his supervisor. As you work with LDAP throughout this module, you'll see data presented in this format several times.

Switching gears a bit, in the early 1990s, there were many different software programs developed to help share files and users between systems. One of the first was Samba. Samba is an open source project that originally reverse engineered the SMB (hence the name) and CIFS file sharing protocols used by Windows. By doing so, they were able to enable file sharing between Linux based systems, including Macs, and Windows. Eventually, Microsoft worked with the developers of the project to allow them to fully implement the protocol, and in more recent versions, a computer running Samba can even act as part of an Active Directory domain. We'll work with Samba a bit later this semester when we deal with application servers.

Around the same time, Novell released their Directory Services software. This was one of the first directory services targeted at large organizations, and for many years it was the dominant software in that area. It was built on top of the IPX/SPX internet protocols, so it required computers to run special software in order to use the directory service.

Much like the LDAP example earlier, NDS also supported a tree structure for storing information in the directory. It also had some novel features for assigning permissions across different users and ous, which made it a very powerful tool. Many older system administrators have probably worked with Novell in the past. I remember that both my high school, as well as some departments at K-State, used NDS until they switched to Microsoft Active Directory in the mid 2000s.

Before we discuss Active Directory, we should first discuss Windows Workgroups and Homegroups. Most home networks don't need a full directory service, but it is very helpful to have a way to share files easily between a few computers connected to the same network. Prior to Windows 7, Windows used a "Workgroup" to share these files.

Computers with the same Workgroup name would be able to discover each other across a local network, and any shared files or resources would be available. However, user accounts were not shared, so you wanted to access a resource shared from a particular computer, you'd have to know a username and password for that system as well. For smaller networks, this usually wasn't a problem at all. However, as the network grows larger, it becomes more and more difficult to maintain consistent user accounts across all systems.

That all changed in 1999 with the introduction of Microsoft Active Directory. It coincided with the release of Windows 2000, one of the first major Windows releases directly targeted at large enterprises. Active Directory uses LDAP and Kerberos to handle authentication, and provides not only a central repository of user accounts and systems, but it can also be used to share security configurations across those systems as well. It was really ideal for large enterprise networks, and gave administrators the level of control they desired. Active Directory quickly became the dominant directory service, and most enterprises today use Active Directory to manage user accounts for their Windows systems. In fact, both K-State and K-State Computer Science primarily use Active Directory to manage their user accounts across a variety of systems and platforms.

With a shared domain, the user account information is stored on the central domain controller, instead of the individual workstations. In this way, any user who has a valid account on the domain and permissions to access a resource can do so, regardless of which system they are connected to. This is what allows you to use the same user account to log on to computers across campus, as well as online resources such as webmail and Canvas. One of the major activities in this lab will be to install and configure an Active Directory Domain, so you can get first-hand experience of how this works.

Lastly, no discussion of directory services would be complete without discussing Kerberos. Kerberos is named after Cerberus, the three-headed dog from ancient Greek mythology. It allows a client and a server to mutually authenticate each other's identity through a third party, hence the three-headed mascot.

Kerberos was originally developed at MIT in the 1980s, and fully released in 1993 as RFC 1510. It is one of the primary ways that directory services today handle authentication, and it is used by many other types of security software on the internet today.

The Kerberos protocol is quite complex, and it uses many different cryptographic primitives to ensure the communication is properly secured and authenticated. However, this diagram gives a quick overview of the protocol and how it works.

1. First, the client will communicate with the authentication server.
2. The server responds with two messages, labelled A and B in this diagram. Message A is encrypted by the server using the client's key. When the client receives that message, it attempts to decrypt it. If it is successful, it can use the contents inside the message to prove its identity.
3. Then, it can send two new messages, C and D, to the ticket granting server, which in practice is usually the authentication server, but it doesn't have to be. Message C includes the contents of message B along with information about what the client would like to access, and message D is the decrypted contents of the original message A. The ticket granting server then looks at those messages and confirms the user's identity.
4. If it is correct, it will then send two more messages, E and F, to the client. Message E is the ticket that can be given to the server containing the desired resource, and message F contains a key that can be used to communicate with that server.
5. So, the client finally can contact the destination server, sending along message E, the ticket from earlier, as well as a new message G that is protected with the key contained in message F.
6. Finally, the server decrypts message G and uses it to create and send back message H, which can be confirmed by the client to make sure the server is using the same key.

If everything works correctly, both the client and server know they can trust each other, and they have established keys that allow them to communicate privately and share the requested information. Kerberos is a very powerful protocol, and if you'd like to learn more, I highly recommend reading the linked resources below this video. Kerberos is also discussed in several of the cybersecurity courses here at K-State.

That's all the background information you'll need for this module. Next, you'll work on installing Microsoft Active Directory on Windows and OpenLDAP on Ubuntu, as well as configuring clients to use those systems for authentication.

Source: Russell Feldhausen, https://cis527.russfeld.me/4-directory-services/02-directory-services-overview/


text:
# Connected But Not Path-Connected?

Can you think of any spaces that are connected but not path connected apart from the Topologist's Sine Curve?

-

Here are a whole bunch from $\pi$-Base, a searchable version of Steen and Seebach's Counterexamples in Topology. You can visit the search result to learn more about any of these spaces.

An Altered Long Line

A Pseudo-Arc

Cantor's Leaky Tent

Closed Topologist's Sine Curve

Countable Complement Extension Topology

Countable Complement Topology

Double Pointed Countable Complement Topology

Finite Complement Topology on a Countable Space

Gustin's Sequence Space

Indiscrete Irrational Extension of $\mathbb{R}$

Indiscrete Rational Extension of $\mathbb{R}$

Irrational Slope Topology

Lexicographic Ordering on the Unit Square

Nested Angles

One Point Compactification of the Rationals

Pointed Irrational Extension of $\mathbb{R}$

Pointed Rational Extension of $\mathbb{R}$

Relatively Prime Integer Topology

Roy's Lattice Space

Smirnov's Deleted Sequence Topology

The Extended Long Line

The Infinite Broom

The Infinite Cage

Topologist's Sine Curve

-
Spacebook has been phased out in favor of $\pi$-Base, so I've updated the answer to point there. – Austin Mohr Jul 12 '14 at 18:46

Another standard example is the extended long line. Counterexamples in Topology will have more, but my copy isn't to hand right now.

-
Nice example!___ – George Lowther Mar 26 '12 at 20:53

An example of a connected space that is not path-connected is the deleted comb space: $$(\{0\} \times \{0,1\}) \cup (K \times [0,1]) \cup ([0,1] \times \{0\})$$ where $K = \{ \frac{1}{n} \mid n \in \mathbb{N} \}$

Taken from here.

-

The canonical example is the extended long line. You can think of the regular line $[0,\infty)$ as the product of $[0,1)$ and $\omega$ in the dictionary order topology—effectively, a countable number of copies of $[0, \infty)$ pasted end-to-end.

The long line is the same way, except that instead of a countable number of copies you use an uncountable number of copies: take $[0, 1)\times\omega_1$ in the dictionary order topology, where $\omega_1$ is the smallest uncountable ordinal. Then to get the extended long line, you add one more point $p$ onto the far end. It's clearly connected, but it isn't path-connected because the path from any finite point, say $(1/2, 1)$, is too far from $p$ for the path between them to be the image of $[0,1]$.

The book Counterexamples in Topology by Seebach and Steen is good for answering questions like this.

-
The long line is path connected. All locally path connected and connected spaces are path connected. – George Lowther Mar 26 '12 at 20:47
@George: The space I described is not locally path-connected in a neighborhood of the extra point $p$. – MJD Mar 26 '12 at 20:49
I think I misunderstood what you meant. Shouldn't this be called the extended long line (or ray)? I don't think the long line has an extra point added at the end. – George Lowther Mar 26 '12 at 20:56
@George: That is my fault. I will edit the reply. Thanks for pointing out that I misspoke. – MJD Mar 26 '12 at 20:58
Ah, ok, I agree that this is a valid example now. – George Lowther Mar 26 '12 at 21:02


text:
### Instantons in complex geometry: talks and abstracts

#### 14-18 March 2011, Laboratory of Algebraic Geometry,Higher School of Economics, Moscow

Home | Venue | Schedule | Program | Poster | PDF

#### Vladimir Baranovsky (Irvine) Uhlenbeck compactification in arbitrary dimension

We will outline a definition of a moduli functor which compactifies the moduli of principal G-bundles on a smooth projective variety X of any dimension (for G a reductive group). Our work is based on the technique of intersection bundles due to Deligne and the semistable reduction for principal bundles due to Balaji.

#### Alexander Braverman (Brown Univ.) Towards categorical approach to instanton counting

We present a list of (partially proven) conjectures about the structure of derived categories of constructible sheaves on various instanton moduli spaces and explain how these conjectures can be applied to questions of "instanton counting" (in particular, to a version of Alday-Gaiotto-Tachikawa conjecture).

#### Vasile Brinzanescu (IMAR, Bucharest) Vector bundles on non-Kahler Calabi-Yau type 3-folds

We study moduli of relatively semi-stable vector bundles on non-Kahler principal elliptic bundles, which are Calabi-Yau type 3-folds. The main technical tools used are the twisted Fourier-Mukai transform and a spectral cover construction (joint work with A. Halanay and G. Trautmann).

#### Ugo Bruzzo (SISSA, Trieste) Uhlenbeck-Donaldson Compactification for Framed Sheaves

We introduce the moduli spaces of framed sheaves. A reason for being interested in these objects is that under suitable conditions they are nice moduli spaces (quasi-projective, smooth and fine) and provide desingularizations of moduli spaces of framed instantons. We study a "partial compactification" for these moduli spaces, called the Uhlenbeck-Donaldson compactification, and relate it to the moduli space of "ideal" framed instantons.

#### Laura Costa (Univ. de Barselona) Linear Monads on Instanton bundles on hyperquadrics

Monads appear in a wide variety of context within Algebraic Geometry. We will focus our attention on linear monads as a tool for constructing indecomposable vector bundles on hyperquadrics and on the particular case of instanton bundles on them. This is a join work with R.M. Miro-Roig.

#### Michael Finkelberg (HSE) Parabolic bundles on P^2, Quiver varieties, and Quantization

The moduli space of certain parabolic torsion free sheaves on P^2 plays an important role in the geometric representation theory (under the name of Affine Laumon space). The singularities of its affinization contain the "limit singularities" of Schubert varieties of the affine Grassmannian of SL_n. It can be realized as a quiver variety, and quantized via the quantum Hamiltonian reduction. The quantization is described in terms of affine Yangians.

#### Sergei Gorchinskiy (Steklov Institute) Polar homology

The talk is based on a joint work with A.Rosly. We discuss new complexes that compute cohomology groups of locally free sheaves on smooth algebraic varieties in characteristic zero. These complexes are algebro-geometric analogs of singular complexes for real smooth manifolds and have arised from E.Witten's program of holomorphic Chern-Simons theory.

#### Laurent Gruson (Univ. de Versailles) About the mathematical instantons with c_2 = 4

In his thesis (1996) Frederic Han gave a description of a dense open set of mathematical instantons for c_2 = 4,5 . He assumed that the second order jumping lines are in codimension 3, but there was a gap in the proof he gave of this fact. I would like to discuss the state of affairs around this in the case c_2 = 4.

#### Norbert Hoffmann (Freie Univ. Berlin) Rational families of instanton bundles on P^{2n+1}

I'll discuss the rationality problem for two kinds of moduli spaces for instanton bundles on P^{2n+1}, namely for special instantons in the sense of Spindler-Trautmann and for symplectic 't Hooft instantons in the sense of Ottaviani. The latter is joint work with Costa, Miro-Roig and Schmitt.

#### Marcos Jardim (UNICAMP) Moduli spaces of framed instanton bundles on CP^3

I will present the ADHM construction of framed instanton sheaves on CP^3, which yields a parametrization of the (fine) moduli space of framed instanton bundles in terms of matrices satisfying some quadratic equations. Using twistor theory, one can show that such moduli space coincides with the space of twistor section of the moduli space of framed bundles on CP^2. We also introduce a the notion of trisymplectic structures on a complex manifolds, and the notion of trisymplectic reduction. We show that the moduli space of framed instanton bundles can also be described in terms of a trisymplectic reduction, and conclude that it is a smooth trisymplectic manifold of the expected dimension. Joint work with Misha Verbitsky and with Amar Henni and Renato Vidal Martins.

#### Alexander Kuznetsov (Steklov Institute) Instanton bundles on Fano threefolds

I will talk about instanton bundles on Fano threefolds of index 2. In particular, jumping lines and monadic representations will be discussed.

#### Adrian Langer (University of Warsaw) Moduli spaces of framed perverse instantons on P^3

I will talk about moduli spaces of framed perverse instantons on P^3. As an open subset they contain the moduli space of framed instantons studied by I. Frenkel and M. Jardim. I will explain the connection with the moduli space of pairs on the blow up of P^3 along a line. I will also study the map between the Gieseker and Donaldson-Uhlenbeck partial compactifications of the moduli space of instantons on P^3. Finally I will construct a few counterexamples to earlier conjectures and results concerning these moduli spaces.

#### Eyal Markman (University of Massachussets) Morrison's movable cone conjecture for projective irreducible holomorphic symplectic manifolds.

We prove a version of the conjecture in the title as a consequence of the Global Torelli Theorem for irreducible holomorphic symplectic manifolds X. Let Bir(X) be the group of birational automorphisms of X. As consequence it is shown that for each non-zero integer d there are only finitely many Bir(X)-orbits of complete linear systems, which contain a reduced and irreducible divisor of Beauville-Bogomolov degree d. A variant hold for degree zero as well.

#### Rosa M. Miro-Roig (Univ. de Barcelona) Monday, Tuesday or Wednesday COHOMOLOGICAL CHARACTERIZATION OF VECTOR BUNDLES

In my talk, I will address the problem of giving a cohomological characterization of vector bundles on algebraic varieties. This is a longstanding problem in Algebraic Geometry which has its roots in an old paper by Horrocks where he gave a cohomological characterization of line bundles on projective spaces P^n.

In my talk, I will give a cohomological characterization of the bundle of p-differential forms on multiprojective spaces P^{n_1}\times ... \times P^{n_s} and a cohomological characterization of Steiner bundles on algebraic varieties. As a main tool I will use a generalized version of Beilinson's spectral sequence.

This is joint work with Costa and Soares

#### Ruxandra Moraru (Univ. Waterloo) Generalized holomorphic vector bundles and Poisson modules on Hopf surfaces

Hopf surfaces are interesting examples of compact complex surfaces that do not admit Kaehler metrics, but nonetheless admit generalized Kaehler structures. In this talk, I will describe the geometry of generalized holomorphic vector bundles on these surfaces, thus providing new non-trivial examples of generalised holomorphic vector bundles and their moduli. Time permitting, I will also describe how these bundles relate to Poisson modules on Hopf surfaces.

#### Christian Okonek (Univ. Zurich) Abelian Yang-Mills theory on a Real torus, and theta divisors of Klein surfaces

In joint work with Andrei Teleman we determine certain natural theta line bundles of Klein surfaces as elements in the Grothendieck cohomology group which classifies Real line bundles in the sense of Atiyah. These line bundles are important since they control the orientability of certain moduli spaces in Real gauge theory to a large extend.

#### Dmitri Panov (King's College London) Hyperbolic geometry and fundamental groups of complex 3-folds

We give an alternative short proof of Taubes' theorem stating that compact complex 3-folds can have arbitrary finitely presented fundamental group. This is related to a question of Gromov: Is it true that every compact manifold is homeomorphic to a quotient of the hyperbolic space H^n by an isometric (non-free) action of a discreet group? This talk is based on a joint work with Anton Petrunin.

#### Prabhakar Rao (Univ. of Missouri) ACM vector bundles on hypersurfaces

An Arithmetically Cohen Macaulay vector bundle on a hypersurface X of projective space is a bundle E for which $H^i(X,E(a))=0$ for any integer a and any i between 1 and dim(X)-1. We will discuss existence, constructions and applications of ACM bundles on smooth hypersurfaces.

#### Alexander Schmitt (Freie Univ. Berlin) Quiver Bundles

In this talk, I will first review the "classical" formalism of quiver representations, with a view on sheaves and instantons on projective manifolds. Then, I will give a survey on results on quiver bundles on projective manifolds and their moduli spaces. At the end, I will present joint work with Garcia-Prada and Heinloth on the motive of the moduli space of Higgs bundles of rank four and odd degree on a compact Riemann surface.

#### Andrei Teleman (Univ. de Provence) Holomorphic bundles and holomorphic curves on class VII surfaces. The classification problem for class VII surfaces.

The classification of complex surfaces is not finished yet. The most important gap in the Kodaira-Enriques classification table concerns the Kodaira class VII, e.g. the class of surfaces $X$ having $kod(X)=-\infty$, $b_1(X)=1$. These surfaces are interesting from a differential topological point of view, because they are non-simply connected 4-manifolds with definite intersection form. The main conjecture which (if true) would complete the classification of class VII surfaces, states that any minimal class VII surface with $b_2>0$ contains $b_2$ holomorphic curves. We explain a new approach, based on ideas from Donaldson theory, which gives existence of holomorphic curves on class VII surfaces with small $b_2$. In particular, for $b_2=1$ we obtain a proof of the conjecture, and for $b_2=2$ we prove the existence of a cycle of curves.

#### Nadezhda Timofeeva (Yaroslavl University) On Giseker-Maruyama moduli for a surface: new interpretation of classical scheme

Main (containing locally free sheaves) components of GM moduli scheme will be re alized as moduli scheme of locally free scheaves on some special class of schemes.

#### Gunther Trautmann (Univ. Kaiserslautern) Replacing the boundary by treelike vector bundles

Given a moduli space of semistable sheaves, one can try to replace its boundary of non-locally free sheaves by a variety of classes of locally free sheaves in order to eventually better understand the degenerations and the singularities of the sheaves in the boundary. This will be done for Gieseker-Maruyama moduli spaces M_S(2; P) of semistable sheaves of rank 2 with Hilbert polynomial P on a projective surface S. The replacement in this situation resembles the bubbling in Yang-Mills theory for vector bundles on 4-manifolds. A second type of replacement by vector bundles will be discussed for 1-dimensional sheaves in a Simpson moduli space M of semistable sheaves on the projective plane.

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# Section9.1Morley's miracle

Probably you have heard of the impossibility of trisecting an angle. (Hold on for a quick rant about the importance of understanding your hypotheses…) What's actually true is that you can't trisect a generic angle if you accept the restriction of using the old-fashioned tools of Euclidean geometry: the compass and straight-edge. There are a lot of constructions that can't be done using just a straight-edge and compass — angle trisection, duplication of a cube 1 , squaring a circle, constructing a regular heptagon, et cetera.

If you allow yourself to use a ruler — i.e. a straight-edge with marks on it (indeed you really only need two marks a unit distance apart) then angle trisection can be done via what is known as a neusis construction. Nevertheless, because of the central place of Euclid's Elements in mathematical training throughout the centuries, and thereby, a very strong predilection towards that which is possible via compass and straight-edge alone, it is perhaps not surprising that a perfectly beautiful result that involved trisecting angles went undiscovered until 1899, when Frank Morley stated his Trisector Theorem. There is much more to this result than we will state here — so much more that the name “Morley's Miracle” that has been given to the Trisector theorem is truly justified — but even the simple, initial part of this beautiful theory is arguably miraculous! To learn more about Morley's theorem and its extension see  [8].

So, let's state the theorem!

Start with an arbitrary triangle ${\triangle}ABC$. Trisect each of its angles to obtain a diagram something like that in <<Unresolved xref, reference "fig_morley_setup"; check spelling or use "provisional" attribute>>.

The six angle trisectors that we've just drawn intersect one another in quite a few points.

##### Exercise9.1.1

You could literally count the number of intersection points between the angle trisectors on the diagram, but you should also be able to count them (perhaps we should say “double-count them”) combinatorially. Give it a try!

Among the points of intersection of the angle trisectors there are three that we will single out — the intersections of adjacent trisectors. In <<Unresolved xref, reference "fig_morley_1st_triangle"; check spelling or use "provisional" attribute>> the intersection of adjacent trisectors are indicated, additionally, we have connected them together to form a small triangle in the center of our original triangle.

Are you ready for the miraculous part? Okay, here goes!

In other words, that little blue triangle in <<Unresolved xref, reference "fig_morley_1st_triangle"; check spelling or use "provisional" attribute>> that kind of looks like it might be equilateral actually does have all three sides equal to one another. Furthermore, it doesn't matter what triangle we start with, if we do the construction above we'll get a perfect $60^\circ - 60^\circ - 60^\circ$ triangle in the middle!

Sources differ, but it is not clear whether Morley ever proved his theorem. The first valid proof (according to R. K. Guy in  [8] was published in 1909 by M. Satyanarayana  [15]. There are now many other proofs known, for instance the cut-the-knot website (http://www.cut-the-knot.org/) exposits no less than nine different proofs. The proof by Satyanarayana used trigonometry. The proof we'll look at here is arguably the shortest ever produced and it is due to John Conway. It is definitely a “Book proof”!

Let us suppose that an arbitrary triangle ${\triangle}ABC$ is given. We want to show that the triangle whose vertices are the intersections of the adjacent trisectors is equilateral — this triangle will be referred to as the Morley triangle. Let's also denote by $A$, $B$ and $C$ the measures of the angles of ${\triangle}ABC$. (This is what is generally known as an “abuse of notation” — we are intentionally confounding the vertices ($A$, $B$ and $C$) of the triangle with the measure of the angles at those vertices.) It turns out that it is fairly hard to reason from our knowledge of what the angles $A$, $B$ and $C$ are to deduce that the Morley triangle is equilateral. How does the following plan sound: suppose we construct a triangle, that definitely does have an equilateral Morley triangle, whose angles also happen to be $A$, $B$ and $C$. Such a triangle would be similar 2  to the original triangle ${\triangle}ABC$ — if we follow the similarity transform from the constructed triangle back to ${\triangle}ABC$ we will see that their Morley triangles must coincide; thus if one is equilateral so is the other!

One of the features of Conway's proof that leads to its great succinctness and beauty is his introduction of some very nice notation. Since we are dealing with angle trisectors, let $a$, $b$ and $c$ be angles such that $3a=A$, $3b=B$ and $3c=C$. Furthermore, let a superscript star denote the angle that is $\pi/3$ (or $60^\circ$ if you prefer) greater than a given angle. So, for example, \begin{equation*} a^\star = a + \pi/3 \end{equation*} and \begin{equation*} a^{\star\star} = a + 2\pi/3. \end{equation*}

Now, notice that the sum $a+b+c$ must be $\pi/3$. This is an immediate consequence of $A+B+C=\pi$ which is true for any triangle in the plane. It follows that by distributing two stars amongst the three numbers $a$, $b$ and $c$ we will come up with three quantities which sum to $\pi$. In other words, there are Euclidean triangles having the following triples as their vertex angles:

 $(a, b, c^{\star\star})$ $(a, b^\star, c^\star)$ $(a, b^{\star\star}, c)$ $(a^\star, b^\star, c)$ $(a^{\star\star}, b, c)$ $(a^\star, b, c^\star)$
##### Exercise9.1.3

What would a triangle whose vertex angles are $(0^\star, 0^\star, 0^\star)$ be?

In a nutshell, Conway's proof consists of starting with an equilateral triangle of unit side length, adding appropriately scaled versions of the six triangles above and ending up with a figure (having an equilateral Morley triangle) similar to ${\triangle}ABC$. The generic picture is given in <<Unresolved xref, reference "fig_morley_conway_puzzle"; check spelling or use "provisional" attribute>>. Before we can really count this argument as a proof, we need to say a bit more about what the phrase “appropriately scaled” means. In order to appropriately scale the triangles (the small acute ones) that appear green in <<Unresolved xref, reference "fig_morley_conway_puzzle"; check spelling or use "provisional" attribute>> we have a relatively easy job — just scale them so that the side opposite the trisected angle has length one; that way they will join perfectly with the central equilateral triangle.

The triangles (these are the larger obtuse ones) that appear purple in <<Unresolved xref, reference "fig_morley_conway_puzzle"; check spelling or use "provisional" attribute>> are a bit more puzzling. Ostensibly, we have two different jobs to accomplish — we must scale them so that both of the edges that they will share with green triangles have the correct lengths. How do we know that this won't require two different scaling factors? Conway also developed an elegant argument that handles this question as well. Consider the purple triangle at the bottom of the diagram in <<Unresolved xref, reference "fig_morley_conway_puzzle"; check spelling or use "provisional" attribute>> — it has vertex angles $(a,b,c^{\star\star})$. It is possible to construct triangles similar (via reflections) to the adjacent green triangles $(a, b^\star, c^\star)$ and $(a^\star, b, c^\star)$ inside of triangle $(a,b,c^{\star\star})$. To do this just construct two lines that go through the top vertex (where the angle $c^{\star\star}$ is) that cut the opposite edge at the angle $c^\star$ in the two possible senses — these two lines will coincide if it should happen that $c^\star$ is precisely $\pi/2$ but generally there will be two and it is evident that the two line segments formed have the same length. We scale the purple triangle so that this common length will be 1. See <<Unresolved xref, reference "fig_morley_conway_puzzle_scaling"; check spelling or use "provisional" attribute>>.

##### Exercise9.1.4

If it should happen that $c^\star = \pi/2$, what can we say about $C$?

Of course the other two obtuse triangles can be handled in a similar way.

# Subsection9.1.1Exercises

1. What value should we get if we sum all of the angles that appear around one of the interior vertices in the finished diagram? Verify that all three have the correct sum. \ifx\SetFigFont\undefined\gdef\SetFigFont#1#2#3#4#5{ \reset@font\fontsize{#1}{#2pt} \fontfamily{#3}\fontseries{#4}\fontshape{#5} \selectfont}\fi

2. In this section we talked about similarity. Two figures in the plane are similar if it is possible to turn one into the other by a sequence of mappings: a translation, a rotation and a scaling. Geometric similarity is an equivalence relation. To fix our notation, let $T(x,y)$ represent a generic translation, $R(x,y)$ a rotation and $S(x,y)$ a scaling — thus a generic similarity is a function from $\Reals^2$ to $\Reals^2$ that can be written in the form $S(R(T(x,y)))$. Discuss the three properties of an equivalence relation (reflexivity, symmetry and transitivity) in terms of geometric similarity.


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# Increasing resources overtime for a browser-based game [duplicate]

I've been interested in developing a browser game ala OGame or Ikariam etc. Basically, a menu is displayed with your resources and, overtime, they increase depending on how advanced are you in the game.

Now then, the "resources" information for all players is stored in a MySQL database.

I've considered two methods to keep track of a player's resources:

## Update database every second

Every second, something like

UPDATE Players SET Gold = Gold + GoldProduction

This is one option. However, performing a query every second for a potentially large table of players is not exactly appealing - but I might be underestimating my server. Am I?

## Calculate it when needed

Since you get gold every second, technically the amount of gold any player has should be

(currentTime - creationTime) * goldProduction

What if the player spends gold? Well, gold spending should affect a variable goldModifications:

goldModifications + ((currentTime - creationTime) * goldProduction)

So if you spend 100 gold, goldModifications would be -100.

But wait. Players can upgrade their gold production. Hence, goldProduction isn't constant and this formula is flawed...

... Maybe when you upgrade your gold production, goldModifications will be modified by your current amount of gold (so if you had 1500, we would do goldModifications += 1500), then creationTime will be overwritten to the current time.

So I have two methods. One is incredibly simple (just update the database), but I am unsure if it is too much overhead or not. The other is very complex as there could be any number of factors affecting gold production, but certainly requires less processing power.

What should I do? What do those popular games actually do?

• The other method is not very complex, you simply need to keep track of the last time and the last amount of gold and update that when production rate changes and or when a player makes a transaction. Updating every second might be unneeded overhead but updating whenever the player makes a transaction is just like that ides of using a goldModification (which is uneeded). Just make sure that changes in production rate require a transaction or use a cron job that handles these. – wolfdawn Aug 28 '14 at 9:46

tl;dr

The resources should only be controlled by the Server (securely) as a decision making entity; The actual work could be offloaded to Clients and later verified lazily by the Server.

Long story:

How often you update the resources is up to you, your hardware and desired gameplay experience. I don't see why it requires you to update every second? What about every 10 seconds? But if it does, I am sure it is entirely possible with the right hardware. Remember, data that changes frequently could be stored in game memory and only updated to database for storage / backup purposes (with constraint validation).

When a user logs in (assuming production continues when offline), one could simply check the last date the user was online, how much X resource she had and use that to calculate her current amount of resource X. When she plays, let her handle the money, again checking if she is cheating by verifying constraints online. When the client makes a transaction, you could check that the current amount of resource X corresponds with the constraints (what the clients claims matches what the server thinks), calculating the resource state and leaving a tiny buffer for synchronization.

It all depends on the complexity of the game. In general I think you are doing three things that don't necessarily mix well (if taken to extreme) but could work fine depending on your server HW.

1. Updating game state frequently.
2. Stored all data on a server DB.
3. Run game as HTTP POST/GET requests, not a concurrent game server.

Again, I don't see why updating every second should be a huge concern but some benchmarks and tests you could easily perform would alleviate or affirm your suspicions. Then you could lower the rate of updates and/or employ a different strategy (offloading some of the work to client-side [Angular]) or something like that.

I am also trying to make a online game in PHP and encounter also this problem. My solution was actualy pretty simple. In Assuming you are using different city's or planets, you will need a database table for the city/planet. In this database you store the current resources and the resources per hour. And also a timestamp.

For upgrading resources facility (gold mine for example) i have a database that has all the building ques with a begin time and end time.

i then have made a page that i include on every page in the game. It first checks if there are any building ques for the selected city that should be finished since last time database was entered. If that gold mine was finished at let say 12.00 and the timestamp on your city was 11.00 and current time is 13.00. What the page does, it calculates all the resources for the time 11.00 to 12.00. Then it will update the gold per hour in the city table. And then it recalculates the resources from 12.00 to 13.00.

when all the above is done, it puts the timestamp on current. But as you have all the resources per hour from the second run, you can store it in an array and let a local timer run to update the resources on screen. But since you include the above mentioned page on every page, when you go to the next page it will acces the database to see what the current resources should be. So the local data is only for viewing.

I hope all is clear. Else just ask


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# Selina solutions for Class 10 Mathematics chapter 22 - Height and Distances

## Chapter 22: Height and Distances

Ex. 22AEx. 22BEx. 22C

#### Chapter 22: Height and Distances Exercise 22A solutions [Page 0]

The height of a tree is √3 times the length of its shadow. Find the angle of elevation of the sun.

The angle of elevation of the top of a tower from a point on the ground and at a distance of 160 m from its foot, is found to be 60o. Find the height of the tower.

A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is 2.4 m from the wall and the ladder is making an angle of 68o with the ground. Find the height, upto which the ladder reaches.

Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30o and 38o respectively. Find the distance between them, if the height of the tower is 50 m.

A kite is attached to a string. Find the length of the string, when the height of the kite is 60 m and the string makes an angle 30o with the ground

A boy, 1.6 m tall, is 20 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45o, (ii) 60o. Find the height of the tower in each case.

The upper part of a tree, broken over by the wind, makes an angle of 45o with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. What was the height of the tree before it was broken?

The angle of elevation of the top of an unfinished tower at a point distance 80 m from its base is 30°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60°?

At a particular time, when the sun’s altitude is 30o, the length of the shadow of a vertical tower is 45 m. Calculate (i) the length of the tower. (ii) the length of the shadow of the same tower, when the sun’s altitude is (a) 45° (b) 60°

Two vertical poles are on either side of a road. A 30 m long ladder is placed between the two poles. When the ladder rests against one pole, it makes angle 32o24′ with the pole and when it is turned to rest against another pole, it makes angle 32o24′ with the road. Calculate the width of the road.

Two climbers are at points A and B on a vertical cliff face. To an observer C, 40m from the foot of the cliff, on the level ground, A is at an elevation of 48o and B of 57°. What is the distance between the climbers?

A man stands 9 m away from a flag-pole. He observes that angle of elevation of the top of the pole is 28o and the angle of depression of the bottom of the pole is 13o. Calculate the height of the pole.

From the top of a cliff 92 m high, the angle of depression of a buoy is 20o. Calculate, to the nearest metre, the distance of the buoy from the foot of the cliff.

#### Chapter 22: Height and Distances Exercise 22B solutions [Page 0]

In the figure, given below, it is given that AB is perpandiculer to BD  and is of length X metres. DC = 30m .∠ ADB = 30º and  ∠ ACB = 45°. Without using tables, find X

Find the height of a tree when it is found that on walking away from it 20 m, in a horizontal line through its base, the elevation of its top changes from 60o to 30o.

Find the height of a building, when it is found that on walking towards it 40 m in a horizontal line through its base the angular elevation of its top changes from 30o to 45o.

From the top of a light house 100 m high, the angles of depression of two ships are observed as 48o and 36o respectively. Find the distance between the two ships(in the nearest metre) if: (i) the ships are on the same side of the light house. (ii) the ships are on the opposite sides of the light house.

Two pillars of equal heights stand on either side of a roadway, which is 150 m wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are 60o and 30o ; find the height of the pillars and the position of the point.

From the figure , given below , calculate the lenth of CD .

The angle of elevation of the top of a tower is observed to be 60°. At a point, 30 m vertically above the first point of observation, the elevation is found to be 45°. Find: (i) the height of the tower, (ii) its horizontal distance from the points of observation.

From the top of a cliff, 60 metres high, the angles of depression of the top and bottom of a tower are observed to be 30o and 60o. Find the height of the tower.

A man on a cliff observes a boat, at an angle of depression 30°, which is sailing towards the shore to the point immediately beneath him. Three minutes later, the angle of depression of the boat is found to be 60°. Assuming that the boat sails at a uniform speed, determine:

(i) how much more time it will take to reach the shore.
(ii) the speed of the boat in metre per second, if the height of the cliff is 500 m.

A man in a boat rowing away from a lighthouse 150 m high, takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60o to 45o. Find the speed of the boat.

A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60o. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find: (i) the height of the tree, correct to 2 decimal places, (ii) the width of the river.

The horizontal distance between two towers is 75 m and the angular depression of the top of the first tower as seen from the top of the second, which is 160 m high, is 45°. Find the height of the first tower.

The length of the shadow of a tower standing on level plane is found to be 2y metres longer when the sun’s altitude is 30o than when it was 45°. Prove that the height of the tower is y(√3 + 1) metres.

An aeroplane flying horizontally 1 km above the ground and going away from the observer is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°; find the uniform speed of the aeroplane in km per hour.

From the top of a hill, the angles of depression of two consecutive kilometer stones, due east, are found to be 30° and 45° respectively. Find the distances of the two stones from the foot of the hill.

#### Chapter 22: Height and Distances Exercise 22C solutions [Page 0]

In the following diagram , AB is a floor- board : PQRS is a cubical box with each edge = 1m and ∠B = 60 °. Calculate the lenght of the board AB.

Calculate BC .

The radius of a circle is given as 15 cm and chord AB subtends an angle of 131o at the centre C of the circle. Using trigonometry, calculate:

(i) the length of AB;

(ii) the distance of AB from the centre C.

At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is 5/12. On walking 192 metres towards the tower, the tangent of the angle is found to be 3/4. Find the height of the tower.

A vertical tower stand on a horizontel plane and is surmounted by a vertical flagstaff of height h metre. At a point on the plane, the angle of elevation of the bottom of the flagstaff is ∝  and at the  top of the flagstaff is β . prove that the height of the tower is

"h   tan  ∝ "/"tan  β - tan  ∝"

With reference to the given figure, a man stands on the ground at point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m. The man’s eye s 2 m above the ground. He observes the angle of elevation of C, the top of the pole, as xo , where tan xo = 2/5.

Calculate:

(i) the distance AB in metres;

(ii) angle of elevation of the top of the pole when he is standing 15 metres from the pole. Give your answer to the nearest degree.

The angles of elevation of the top of a tower from two points on the ground at distances a and b metres from the base of the tower and in the same line are complementary. Prove that the height of the tower is √ab metre.

From a window A , 10 m above the ground the angle of elevation of the top C of a tower is X °, Where tan x ° = 5/2  and the angle of depression of the foot D of the tower is Y° , Where tan y ° = 1/4.  calcu8late the height CD of the tower in metres .

A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower?

A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 50 m away from the bank, he finds the angle of elevation to be 30°. Calculate:

(i) the width of the river;

(ii) the height of the tree.

A 20 m high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole is 60° and the angle of elevation of the top of the pole, as seen from the foot of the tower is 30°. Find:

(i) the height of the tower ;

(ii) the horizontal distance between the pole and the tower.

A vertical pole and a vertical tower are on the same level ground in such a way that from the top of the pole, the angle of elevation of the top of the tower is 60o and the angle of depression of the bottom of the tower is 30o. Find: (i) the height of the tower, if the height of the pole is 20 m; (ii) the height of the pole, if the height of the tower is 75 m.

From a point, 36 m above the surface of a lake, the angle of elevation of a bird is observed to be 30o and the angle of depression of its image in the water of the lake is observed to be 60o. Find the actual height of the bird above the surface of the lake.

A man observes the angle of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60 m, the angle of elevation changes to 60°. Find the height of the building correct to the nearest metre.

As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships, on the same side of a light house in a horizontal line with its base, are 30° and 40° respectively. Find the distance between the two ships. Give your answer corrected to the nearest metre.

In the given figure, from the top of a building AB = 60 m hight, the angle of depression of the top and bottom of a vertical lamp post CD are observed to be 30 ° and 60 ° respectively. find :

1) the horizental  distance between AB and CD ;

2) the height of the lamp post.

An aeroplane, at an altitude of 250 m, observes the angles of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number.

The horizontal dist6ance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the top of the second i9s 30 ° and 24 ° respectively. find the height of the two towers. give your answers correct to 3 significant figures.

## Chapter 22: Height and Distances

Ex. 22AEx. 22BEx. 22C

## Selina solutions for Class 10 Mathematics chapter 22 - Height and Distances

Selina solutions for Class 10 Maths chapter 22 (Height and Distances) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CISCE Selina ICSE Concise Mathematics for Class 10 (2018-2019) solutions in a manner that help students grasp basic concepts better and faster.

Further, we at shaalaa.com are providing such solutions so that students can prepare for written exams. Selina textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

Concepts covered in Class 10 Mathematics chapter 22 Height and Distances are Trigonometric Ratios of Complementary Angles, Trigonometric Identities, Heights and Distances - Solving 2-D Problems Involving Angles of Elevation and Depression Using Trigonometric Tables, Trigonometry Problems and Solutions.

Using Selina Class 10 solutions Height and Distances exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Selina Solutions are important questions that can be asked in the final exam. Maximum students of CISCE Class 10 prefer Selina Textbook Solutions to score more in exam.

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