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Class 10 Trigonometric Ratios of Some Specific Angles

### topic to be covered

☼ Trigonometric Ratios of Some Specific Angles .

### Trigonometric Ratios of Some Specific Angles

From geometry, you are already familiar with the construction of angles of 30°, 45°, 60° and 90°.

In this section, we will find the values of the trigonometric ratios for these angles and, of course, for 0°.

Trigonometric Ratios of 45°
In Delta ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e., ∠ A = ∠ C = 45° (see Fig. 8.14).

So, BC = AB
Now, Suppose BC = AB = a.
Then by Pythagoras Theorem, AC^2 = AB^2 + BC^2 = a^2 + a^2 = 2a^2,

and, therefore AC = a sqrt2

sin45^0 = text(side opposite to angle 45°)/text(hypotenuse) = (BC)/(AC) = a/(a sqrt2) = 1/sqrt2

cos45^0 = text(side adjacent to angle 45°)/text(hypotenuse) = (AB)/(AC) = a/(a sqrt2) = 1/sqrt2

tan45^0 = text(side opposite to angle 45°)/text(side adjacent to angle 45°) = (BC)/(AB) = a/a = 1

also cosec45^0 = 1/(sin45^0) = sqrt2 , sec45^0 = 1/(cos45^0) = sqrt2 , cot45^0 = 1/(tan45^0) = 1.

Trigonometric Ratios of 30° and 60° :

Let us now calculate the trigonometric ratios of 30° and 60°. Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°, therefore,

Draw the perpendicular AD from A to the side BC (see Fig. 8.15).

Now Delta ABD ≅ Delta ACD
Therefore, BD = DC
and ∠ BAD = ∠ CAD (CPCT)
Now observe that:

Delta ABD is a right triangle, right- angled at D with ∠ BAD = 30° and ∠ ABD = 60°
(see Fig. 8.15).

As you know, for finding the trigonometric ratios, we need to know the lengths of the
sides of the triangle. So, let us suppose that AB = 2a.

Then, BD = 1/2 BC = a

and AD^2 = AB^2 – BD^2 = (2a)^2 – (a)^2 = 3a^2,

Therefore, AD = a sqrt3

Now, we have :

sin30^0 = (BD)/(AB) = a/(2a) = 1/2 , cos30^0 = (AD)/(AB) = ( a sqrt3)/(2a) = sqrt3/2

tan30^0 = (BD)/(AD) = a/(a sqrt3) = 1/sqrt3

Also cosec30^0 = 1/(sin30^0) = 2 , sec30^0 = 1/(cos30^0) = 2/sqrt3

cot 30^0 = 1/(sin30^0) = sqrt3

Similarly,  sin60^0 = (AD)/(AB) = (a sqrt3)/(2a) = sqrt3/2 , cos60^0 = 1/2 , tan 60^0 = sqrt3

cosec60^0 = 2/sqrt3 , sec60^0 = 2 and cot60^0 = 1/sqrt3

bbul"Trigonometric Ratios of 0° and 90°"

Let us see what happens to the trigonometric ratios of angle A, if it is made smaller and smaller in the right triangle ABC
(see Fig. 8.16), till it becomes zero.

As ∠ A gets smaller and smaller, the length of the side BC decreases.The point C gets closer to point B, and finally when ∠ A becomes very close to 0°, AC becomes almost the same as AB (see Fig. 8.17).

When ∠ A is very close to 0°, BC gets very close to 0 and so the value of
sinA = (BC)/(AC)  is very close to 0. Also, when ∠ A is very close to 0°, AC is nearly the
same as AB and so the value of cos A = (AB)/(AC) is very close to 1.

This helps us to see how we can define the values of sin A and cos A when A = 0°. We define : sin 0° = 0 and cos 0° = 1.
Using these, we have : tan0^0 = (sin 0^0)/(cos 0^0) = 0 , cot 0^0 = 1/(tan 0^0) which is not defined.

sec^0 = 1/( cos 0^0) = 1 and cosec 0^0 = 1/( sin 0^0) which is again not defined.

Now, let us see what happens to the trigonometric ratios of ∠ A, when it is made larger and larger in Delta ABC till it becomes 90°. As ∠ A gets larger and larger, ∠ C gets smaller and smaller.

Therefore, as in the case above, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when ∠ A is very close to 90°, ∠ C becomes very close to 0° and the side AC almost coincides with side BC (see Fig. 8.18).

When ∠ C is very close to 0°, ∠ A is very close to 90°, side AC is nearly the same as side BC, and so sin A is very close to 1. Also when ∠ A is very close to 90°, ∠ C is very close to 0°, and the side AB is nearly zero, so cos A is very close to 0.
So, we define : sin 90° = 1 and cos 90° = 0.

We shall now give the values of all the trigonometric ratios of 0°, 30°, 45°, 60°
and 90° in Table 8.1, for ready reference

bbul"Remark" : From the table above you can observe that as ∠ A increases from 0° to 90°, sin A increases from 0 to 1 and cos A decreases from 1 to 0. Let us illustrate the use of the values in the table above through some examples.

Q 3250101014

In Delta ABC, right-angled at B, AB = 5 cm and ∠ ACB = 30° (see Fig). Determine the lengths of the sides BC and AC.
Class 10 Chapter 8 Example 6
Solution:

To find the length of the side BC, we will choose the trigonometric ratio involving BC and the given side AB. Since BC is the side adjacent to angle C and AB is the side opposite to angle C, therefore

(AB)/(BC) = tanC

5/(BC) = tan30^0 =1/sqrt3

which gives BC = 5 sqrt3 cm

To find the length of the side AC, we consider

sin30^0 = (AB)/(AC)  (Why?)

1/2 = 5/(AC)

AC = 10 cm

Note that alternatively we could have used Pythagoras theorem to determine the third side in the example above,

AC = sqrt(AB^2+BC^2) = sqrt(5^2+(5 sqrt3)^2) cm = 10 cm
Q 3260101015

In Delta PQR, right -angled at Q (see Fig), PQ = 3 cm and PR = 6 cm.
Determine ∠ QPR and ∠ PRQ.
Class 10 Chapter 8 Example 7
Solution:

Given PQ = 3 cm and PR = 6 cm

Therefore, (PQ)/(PR) = sinR
or sinR = 3/6 = 1/2

so ∠ PRQ = 30°

and therefore, ∠ QPR = 60°.  (Why?)

You may note that if one of the sides and any other part (either an acute angle or any side) of a right triangle is known, the remaining sides and angles of the triangle can be determined.
Q 3270101016

If sin (A – B) = 1/2 , cos (A+B) = 1/2 , 0^0 < A + B ≤ 90°, A > B, find A and B
Class 10 Chapter 8 Example 8
Solution:

sin (A – B) = 1/2 , cos (A+B) = 1/2 , therefore A - B = 30^0  ......(1)

Also, since cos (A + B) = 1/2 therefore, A + B = 60° .........(2)

Solving (1) and (2), we get : A = 45° and B = 15°.


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Thelaunching mechanism of a toy gun consists of a spring of unknownspring constant, as shown in the figure below. If its spring iscompressed a distance of 0.080 m and the gun fired vertically asshown, the gun can launch a 17.0 g projectile to a maximum heightof18.0 m above the starting pointof the projectile.
(a) (2pts) Neglecting all resistive forces, determine the springconstant.
N/m

(b) (2 pts) Neglecting all resistive forces, determine the speed ofthe projectile as it moves through the equilibrium position of thespring (where x = 0), as shown in figure(b). Assume for this question that the springconstant, k, = 900 N/m.
m/s

format: plain


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# Easy Geometry & Mensuration Solved QuestionAptitude Discussion

 Q. Four equal sized maximum circular plates are cut off from a square paper sheet of area $784$ cm$^2$. The circumference of each plate is:
 ✖ A. 22 cm ✔ B. 44 cm ✖ C. 66 cm ✖ D. 88 cm

Solution:
Option(B) is correct

Consider the diagram given below:

Side of square paper

$=\sqrt{784}$ cm $=28$ cm

$=\dfrac{1}{4}\times 28$cm $=7$ cm

Circuference of each circular plate

$=2\times \dfrac{22}{7}\times 7$

$=44$ cm

## (2) Comment(s)

Poonam Pipaliya
()

Sir, I have a confusion.

How 28 is divided by 4?

28 is just a side value.

Then how only one side can be divided for four square?

Deepak
()

Because if you visualize, then side of the square will contain 2 circles (or 4 radius).

So in order to get the radius, it is divided by 4.


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Canadian Mathematical Society www.cms.math.ca
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Search: All articles in the CMB digital archive with keyword reductive group

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1. CMB 2012 (vol 57 pp. 97)

Levy, Jason
 Rationality and the Jordan-Gatti-Viniberghi decomposition We verify our earlier conjecture and use it to prove that the semisimple parts of the rational Jordan-Kac-Vinberg decompositions of a rational vector all lie in a single rational orbit. Keywords:reductive group, $G$-module, Jordan decomposition, orbit closure, rationalityCategories:20G15, 14L24
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Forums Forums Classes Demon Hunters Havoc 8.0.x [UPDATED 10/15/18] DV8 DH Havoc (Mythic+ and Raiding) Reply To: [UPDATED 10/15/18] DV8 DH Havoc (Mythic+ and Raiding)

DV8
Sergeant
Post count: 12

@DV8 Update to the macro sometime soon? or is there no need for a update?

I’ll check over it but it still seems to be working fine.

DV8, can you post the version with eye beams in the macro or send it to me please?

Hey sorry I’ve been away for an extended period of time. Eye Beam in the rotation?

Yep :). saw a post where you had sent one to another player.

Sure do you want me to PM it to you?

just cleared 5/9 mythic bod using this, seems to still be the best havoc macro.
going to run some +13’s later and i’ll update how it performs; which i imagine is still the same as last season.

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