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# Solved Examples - The Metric System of Measurements and Units

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Solve all questions.
Use at least two methods as applicable.
State the measurement.
Show all work.

NOTE: Unless specified otherwise:
(1.) Use only the tables provided for you.
(2.) Please do not approximate intermediate calculations.

Metric to Metric Conversions
Prefix Symbol Multiplication Factor
yocto y $10^{-24}$
zepto z $10^{-21}$
atto a $10^{-18}$
femto f $10^{-15}$
pico p $10^{-12}$
nano n $10^{-9}$
micro $\mu$ $10^{-6}$
milli m $10^{-3}$
centi c $10^{-2}$
deci d $10^{-1}$
deka da $10^1$
hecto h $10^2$
kilo K $10^3$
mega M $10^6$
giga G $10^9$
tera T $10^{12}$
peta P $10^{15}$
exa E $10^{18}$
zetta Z $10^{21}$
yotta Y $10^{24}$

(1.) Convert $16.27$ meters to kilometers

Measurement is Length

$\underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 16.27\:m\:\:to\:\:km \\[3ex] = 16.27\:m * \dfrac{.....km}{.....m} \\[5ex] = 16.27\:m * \dfrac{1\:km}{10^3\:m} \\[5ex] = 16.27\:m * \dfrac{1\:km}{1000\:m} \\[5ex] = \dfrac{16.27}{1000}\:km \\[5ex] = 0.01627\:km \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] 1\:km = 10^3\:m \\[3ex] 1\:km = 1000\:m \\[3ex] Let\:\:p = length\:\:of\:\:16.27\:m\:\:in\:\:km \\[3ex]$
$km$ $m$
$1$ $1000$
$p$ $16.27$

$\dfrac{p}{1} = \dfrac{16.27}{1000} \\[5ex] p = 0.01627\:km \\[3ex] \therefore 16.27\:m = 0.01627\:km$
(2.) Convert $16.27$ kilometers to meters

Measurement is Length

$\underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 16.27\:km\:\:to\:\:m \\[3ex] = 16.27\:km * \dfrac{.....m}{.....km} \\[5ex] = 16.27\:km * \dfrac{10^3\:m}{1\:km} \\[5ex] = 16.27\:km * \dfrac{1000\:m}{1\:km} \\[5ex] = 16.27 * 1000 = 16270\:m \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] 1\:km = 10^3\:m \\[3ex] 1\:km = 1000\:m \\[3ex] Let\:\:p = length\:\:of\:\:16.27\:km\:\:in\:\:m \\[3ex]$
$km$ $m$
$1$ $1000$
$16.27$ $p$

$\dfrac{p}{1000} = \dfrac{16.27}{1} \\[5ex] \dfrac{p}{1000} = 16.27 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:1000 \\[3ex] 1000 * \dfrac{p}{1000} = 1000(16.27) \\[5ex] p = 16270\:m \\[3ex] \therefore 16.27\:km = 16270\:m$
(3.) Convert $25$ dekagrams to decigrams

Measurement is Mass

$\underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 25\:dag\:\:to\:\:dg \\[3ex] = 25\:dag * \dfrac{.....g}{.....dag} * \dfrac{.....dg}{.....g} \\[5ex] = 25\:dag * \dfrac{10\:g}{1\:dag} * \dfrac{1\:dg}{10^{-1}\:g} \\[5ex] = 25\:dag * \dfrac{10\:g}{1\:dag} * \dfrac{1\:dg}{0.1\:g} \\[5ex] = \dfrac{25 * 10}{0.1}\:dg \\[5ex] = \dfrac{250}{0.1}\:dg \\[5ex] = 2500\:dg \\[3ex] \underline{Third\:\:Method:\:\:Fast\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] 1\:dag = 10\:g \\[3ex] 1\:dg = 10^{-1}\:g \\[3ex] 1\:dg = 0.1\:g \\[3ex]$
$dag$ $g$ $dg$
$1$ $10$
$k$ $0.1$ $1$
$25$ $p$

$Let\:\:k = length\:\:of\:\:1\:dg\:\:in\:\:dag \\[3ex] Let\:\:p = length\:\:of\:\:25\:dag\:\:in\:\:dg \\[3ex] First: \\[3ex] \dfrac{k}{1} = \dfrac{0.1}{10} \\[5ex] k = 0.01\:dag \\[3ex] Next: \\[3ex] \dfrac{p}{1} = \dfrac{25}{k} \\[5ex] p = \dfrac{25}{0.01} \\[5ex] p = 2500\:dg \\[3ex] \therefore 25\:dag = 2500\:dg$
(4.) Convert $25$ decigrams to dekagrams

Measurement is Mass

$\underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 25\:dg\:\:to\:\:dag \\[3ex] = 25\:dg * \dfrac{.....g}{.....dg} * \dfrac{.....dag}{.....g} \\[5ex] = 25\:dag * \dfrac{10^{-1}\:g}{1\:dg} * \dfrac{1\:dag}{10\:g} \\[5ex] = 25\:dag * \dfrac{0.1\:g}{1\:dg} * \dfrac{1\:dag}{10\:g} \\[5ex] = \dfrac{25 * 0.1}{10}\:dg \\[5ex] = \dfrac{2.5}{10}\:dag \\[5ex] = 0.25\:dag \\[3ex] \underline{Third\:\:Method:\:\:Fast\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex] 1\:dg = 10^{-1}\:g \\[3ex] 1\:dg = 0.1\:g \\[3ex] 1\:dag = 10\:g \\[3ex]$
$dg$ $g$ $dag$
$1$ $0.1$ $a$
$10$ $1$
$25$ $c$

$Let\:\:a = length\:\:of\:\:1\:dg\:\:in\:\:dag \\[3ex] Let\:\:p = length\:\:of\:\:25\:dg\:\:in\:\:dag \\[3ex] First: \\[3ex] \dfrac{a}{1} = \dfrac{0.1}{10} \\[5ex] a = 0.01\:dag \\[3ex] Next: \\[3ex] \dfrac{c}{a} = \dfrac{25}{1} \\[5ex] \dfrac{c}{0.01} = 25 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\: 0.01 \\[3ex] 0.01 * \dfrac{c}{0.01} = 0.01(25) \\[5ex] c = 0.25\:dag \\[3ex] \therefore 25\:dg = 0.25\:dag$
(5.)

$Let\:\: \angle OPK = x \\[3ex] \angle OPR = \angle ORP = y ...base\:\:\angle s \:\:of\:\:isosceles\:\: \triangle \\[3ex] \underline{\triangle OPR} \\[3ex] y + y + 108 = 180 ... sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 2y = 180 - 108 \\[3ex] 2y = 72 \\[3ex] y = \dfrac{72}{2} \\[5ex] y = 36 \\[3ex] \angle POR = 2 * \angle PKR ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 108 = 2 * \angle PKR \\[3ex] \angle PKR = \dfrac{108}{2} \\[5ex] \angle PKR = 54 \\[3ex] \underline{\triangle PKR} \\[3ex] \angle KPR + \angle KRP + \angle PKR = 180 ... sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] \angle KPR = \angle OPK + \angle OPR \\[3ex] \angle KPR = x + 36 \\[3ex] \angle KRP = 20 + 36 = 56 \\[3ex] \therefore x + 36 + 56 + 54 = 180 \\[3ex] x + 146 = 180 \\[3ex] x = 180 - 146 \\[3ex] x = 34^\circ$
(6.)

$x = 2y ...\angle s \:\:in\:\:the\:\:same\:\:segment \\[3ex] p + 60 = 180 ...\angle s \:\:in\:\:a\:\:straight\:\:line \\[3ex] p = 180 - 60 \\[3ex] p = 120 \\[3ex] 120 + 2y + 2x = 180 ... sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 120 + x + 2x = 180 \\[3ex] 3x = 180 - 120 \\[3ex] 3x = 60 \\[3ex] x = \dfrac{60}{3} \\[5ex] x = 20 \\[3ex] x = 2y \rightarrow y = \dfrac{x}{2} \\[5ex] y = \dfrac{20}{2} \\[5ex] y = 10^\circ$
(7.) CSEC The scale on a map is $1:25000$
(i) Anderlin and Jersey are $31.8\:cm$ apart on the map.
Determine, in $km$, the actual distance between Anderlin and Jersey.

(ii) The actual distance between Clifton and James Town is $2.75\:km$
How many units apart are they on the map?

$(i) \\[3ex] Scale = 1:25000 \\[3ex] 31.8\:cm \rightarrow ? \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex]$
$Scale$ $Actual$
$1$ $25000$
$31.8$ $y$

$\dfrac{1}{31.8} = \dfrac{25000}{y} \\[5ex] Cross\:\:Multiply \\[3ex] 1(y) = 31.8(25000) \\[3ex] y = 795000\:cm \\[3ex] \therefore 31.8\:cm = 795000\:cm \\[3ex]$ We need to convert this distance to $km$

$\underline{First\:\:Method - Unity\:\:Fraction\:\:Method} \\[3ex] 795000\:cm\:\:to\:\:km \\[3ex] = 795000\:cm * \dfrac{0.01\:m}{1\:cm} * \dfrac{1\:km}{1000\:m} \\[5ex] = \dfrac{7950}{1000}\:km \\[5ex] = 7.95\:km \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex]$
$cm$ $m$
$1$ $0.01$
$795000$ $x$

$\dfrac{1}{795000} = \dfrac{0.01}{x} \\[5ex] Cross\:\:Multiply \\[3ex] 1(x) = 795000(0.01) \\[3ex] x = 7950\:m \\[3ex]$
$m$ $km$
$1000$ $1$
$7950$ $p$

$\dfrac{1000}{7950} = \dfrac{1}{p} \\[5ex] Cross\:\:Multiply \\[3ex] 1000p = 7950(1) \\[3ex] 1000p = 7950 \\[3ex] p = \dfrac{7950}{1000} \\[5ex] p = 7.95\:km \\[3ex] (ii) \\[3ex] \underline{First\:\:Method - Unity\:\:Fraction\:\:Method} \\[3ex] 2.75\:km\:\:to\:\:cm \\[3ex] = 2.75\:km * \dfrac{1000\:m}{1\:km} * \dfrac{1\:cm}{0.01\:m} \\[5ex] = \dfrac{2750}{0.01}\:km \\[5ex] = 275000\:cm \\[3ex] \underline{Second\:\:Method:\:\:Proportional\:\:Reasoning\:\:Method} \\[3ex]$
$Scale$ $Actual$
$1$ $25000$
$d$ $275000$

$\dfrac{1}{d} = \dfrac{25000}{275000} \\[5ex] \dfrac{25000}{275000} = \dfrac{5}{55} = \dfrac{1}{11} \\[5ex] \dfrac{1}{d} = \dfrac{1}{11} \\[5ex] Cross\:\:Multiply \\[3ex] d = 1(11) \\[3ex] d = 11\:cm \\[3ex]$ Clifton and James Town are $11\:cm$ apart on the map.
(8.)

$Obtuse\angle O = 2(65) ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] Obtuse\angle O = 130 \\[3ex] Reflex\angle ROP + Obtuse\angle ROP = 360 ...\angle s\:\:around\:\:a\:\:point \\[3ex] Reflex\angle ROP = x \\[3ex] Obtuse\angle ROP = 130 \\[3ex] x + 130 = 360 \\[3ex] x = 360 - 130 \\[3ex] x = 230^\circ$
(9.)

$\angle SQR = 79^\circ \\[3ex] \angle QRS = x \\[3ex] Reflex\:\:P\hat{O}S = 252^\circ \\[3ex] Reflex\:\:P\hat{O}S + Obtuse\:\:P\hat{O}S = 360 ...\angle s\:\:around\:\:a\:\:point \\[3ex] Obtuse\:\:P\hat{O}S = 360 - 252 \\[3ex] Obtuse\:\:P\hat{O}S = 108 \\[3ex] \underline{\triangle POS} \\[3ex] \angle SPO = \angle PSO = y...base\:\:\angle s\:\:of\:\:isosceles\:\:\triangle \\[3ex] \angle SPO + \angle POS + \angle PSO = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] x + 108 + x = 180 \\[3ex] 2x + 180 - 108 \\[3ex] 2x = 72 \\[3ex] x = \dfrac{72}{2} \\[5ex] x = 36 \\[3ex] Obtuse\:\:P\hat{O}S = 2 * \angle PQS ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 108 = 2 * \angle PQS \\[3ex] \angle PQS = \dfrac{108}{2} \\[5ex] \angle PQS = 54 \\[3ex] \underline{\triangle PQS} \\[3ex] \angle QPS = \angle QSP = p...base\:\:\angle s\:\:of\:\:isosceles\:\:\triangle \\[3ex] \angle QPS + \angle QSP + \angle PQS = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] p + p + 54 = 180 \\[3ex] 2p + 54 = 180 \\[3ex] 2p = 180 - 54 \\[3ex] 2p = 126 \\[3ex] p = \dfrac{126}{2} \\[5ex] p = 63 \\[3ex] \angle QSR = 63 ...\angle \:\:between\:\:tangent\:\:and\:\:chord=\angle\:\:in\:\:alternate\:\:segment \\[3ex] \\[3ex] OR \\[3ex] \angle PSQ = \angle PSO + \angle OSQ ...as\:\:shown \\[3ex] 63 = 36 + \angle OSQ \\[3ex] \angle OSQ = 63 - 36 \\[3ex] \angle OSQ = 27 \\[3ex] \angle OSR = 90^\circ ... radius \perp tangent\:\:at\:\:point\:\:of\:\:contact \\[3ex] \angle OSR = \angle OSQ + \angle QSR ...as\:\:shown \\[3ex] 90 = 27 + \angle QSR \\[3ex] \angle QSR = 90 - 27 \\[3ex] \angle QSR = 63 \\[3ex] \underline{\triangle QSR} \\[3ex] \angle QRS + \angle QRS + \angle SQR = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 63 + x + 79 = 180 \\[3ex] 142 + x = 180 \\[3ex] x = 180 - 142 \\[3ex] x = 38^\circ$
(10.)

$\angle OPT = 90^\circ ... radius \perp tangent\:\:at\:\:point\:\:of\:\:contact \\[3ex] \underline{\triangle QPT} \\[3ex] 30 + x + 90 + 2x + x = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 4x + 120 = 180 \\[3ex] 4x = 180 - 120 \\[3ex] 4x = 60 \\[3ex] x = \dfrac{60}{4} \\[5ex] x = 15 \\[3ex] \angle PTO = 2x \\[3ex] \angle PTO = 2(15) \\[3ex] \angle PTO = 30^\circ$
(11.)

$\underline{\triangle PSQ} \\[3ex] \angle PSQ = 50^\circ ...\angle \:\:between\:\:tangent\:\:and\:\:chord=\angle\:\:in\:\:alternate\:\:segment \\[3ex] \angle PSQ = \angle PQS = 50 ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \angle PSQ + \angle PQS + \angle SPQ = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 50 + 50 + \angle SPQ = 180 \\[3ex] 100 + \angle SPQ = 180 \\[3ex] \angle SPQ = 180 - 100 \\[3ex] \angle SPQ = 80 \\[3ex] \underline{Cyclic\:\:Quadrilateral\:\:SPQR} \\[3ex] \angle P + \angle R = 180 ...sum\:\:of\:\:interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:Quad \\[3ex] 80 + \angle R = 180 \\[3ex] \angle R = 180 - 80 \\[3ex] \angle R = \angle QRS = 100^\circ$
(12.)

$\angle OCA = \angle OAC = 48^\circ ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \angle ACB = 90^\circ ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \underline{\triangle ABC} \\[3ex] \angle CAB + \angle ABC + \angle ACB = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] \angle CAB = \angle OCA = 48 ...as\:\:shown \\[3ex] \rightarrow 48 + \angle ABC + 90 = 180 \\[3ex] 138 + \angle ABC = 180 \\[3ex] \angle ABC = 180 - 138 \\[3ex] \angle ABC = 42^\circ$
(13.)

$(i) \\[3ex] \angle HJK = 90^\circ ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \angle HJK = \angle HJL + \angle LJK...as\:\:shown \\[3ex] 90 = 20 + \angle LJK \\[3ex] \angle LJK = 90 - 20 \\[3ex] \angle LJK = 70 \\[3ex] \angle LHK = \angle LJK ...\angle s \:\:in\:\:a\:\:straight\:\:line \\[3ex] \rightarrow \angle LHK = 70 \\[3ex] \angle HLK = 90^\circ ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \underline{\triangle HKL} \\[3ex] \angle LHK + \angle HLK + \angle HKL = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 70 + 90 + \angle HKL = 180 \\[3ex] 160 + \angle HKL = 180 \\[3ex] \angle HKL = 180 - 160 \\[3ex] \angle HKL = 20^\circ \\[3ex] (ii) \\[3ex] \angle OKJ = \angle OJK ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \rightarrow \angle OJK = 50 \\[3ex] \underline{\triangle JOK} \\[3ex] \angle OKJ + \angle OJK + \angle JOK = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 50 + 50 + \angle JOK = 180 \\[3ex] 100 + \angle JOK = 180 \\[3ex] \angle JOK = 180 - 100 \\[3ex] \angle JOK = 80^\circ \\[3ex] (iii) \\[3ex] \underline{\triangle JHK} \\[3ex] \angle HJK + \angle JKH + \angle JHK = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 90 + 50 + \angle JHK = 180 \\[3ex] 140 + \angle JHK = 180 \\[3ex] \angle JHK = 180 - 140 \\[3ex] \angle JHK = 40^\circ$
(14.)

$\angle TSP = \angle TQP ... \angle s \:\:in\:\:the\:\:same\:\:segment \\[3ex] \angle TQP = 31 + 58 ...exterior\:\: \angle \:\:of\:\:a\:\: \triangle \\[3ex] \angle TQP = 89 \\[3ex] \therefore \angle TSP = 89^\circ$
(15.)

$x = y + 17...eqn.(1) ...\angle \:\:between\:\:tangent\:\:and\:\:chord=\angle\:\:in\:\:alternate\:\:segment \\[3ex] 2x - 43 = y...eqn.(2) ...\angle \:\:between\:\:tangent\:\:and\:\:chord=\angle\:\:in\:\:alternate\:\:segment \\[3ex] Substitute\:\:eqn.(1)\:\:into\:\:eqn.(2) \\[3ex] 2(y + 17) - 43 = y \\[3ex] 2y + 34 - 43 = y \\[3ex] 2y - y = 43 - 34 \\[3ex] y = 9^\circ$
(16.)

$\angle ORP = 90^\circ ... radius \perp tangent\:\:at\:\:point\:\:of\:\:contact \\[3ex] \angle ORP = \angle ORQ + \angle QRP \\[3ex] 90 = \angle ORQ + 34 \\[3ex] \angle ORQ = 90 - 34 \\[3ex] \angle ORQ = 56 \\[3ex] \angle ORQ = \angle OQR = 56 ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \underline{\triangle ORQ} \\[3ex] \angle ORG + \angle OQR + x = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 56 + 56 + x = 180 \\[3ex] 112 + x = 180 \\[3ex] x = 180 - 112 \\[3ex] x = 68^\circ$
(17.)

We can solve (i) in at least two ways

$(i) \\[3ex] \underline{First\:\:Method} \\[3ex] \angle AOD = 2 * \angle ACD ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 114 = 2 * \angle ACD \\[3ex] \angle ACD = \dfrac{114}{2} \\[5ex] \angle ACD = 57^\circ \\[3ex] \underline{Second\:\:Method} \\[3ex] \angle ODG = 90 ... radius \perp tangent\:\:at\:\:point\:\:of\:\:contact \\[3ex] \angle ODA = \angle OAD = y ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \underline{\triangle OAD} \\[3ex] \angle OAD + \angle ODA + \angle AOD = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] y + y + 114 = 180 \\[3ex] 2y + 114 = 180 \\[3ex] 2y = 180 - 114 \\[3ex] 2y = 66 \\[3ex] y = \dfrac{66}{2} \\[5ex] y = 33 \\[3ex] \angle ADE + \angle ODA + \angle ODG = 180 ...\angle s\:\:in\:\:a\:\:straight\:\:line \\[3ex] \angle ADE + 33 + 90 = 180 \\[3ex] \angle ADE + 123 = 180 \\[3ex] \angle ADE = 180 - 123 \\[3ex] \angle ADE = 57 \\[3ex] \angle ADE = \angle ACD ...\angle \:\:between\:\:tangent\:\:and\:\:chord = \angle\:\:in\:\:alternate\:\:segment \\[3ex] \rightarrow \angle ACD = 57^\circ \\[3ex] (ii) \\[3ex] \angle EAD = \angle ACD ...\angle \:\:between\:\:tangent\:\:and\:\:chord = \angle\:\:in\:\:alternate\:\:segment \\[3ex] \rightarrow \angle EAD = 57^\circ \\[3ex] \underline{\triangle AED} \\[3ex] \angle ADE + \angle EAD + \angle AED = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 57 + 57 + \angle AED = 180 \\[3ex] 114 + \angle AED = 180 \\[3ex] \angle AED = 180 - 114 \\[3ex] \angle AED = 66^\circ \\[3ex] (iii) \\[3ex] \angle OAD = \angle OAC + \angle CAD ...as\:\:shown \\[3ex] \angle CDG = \angle CAD ...\angle \:\:between\:\:tangent\:\:and\:\:chord=\angle\:\:in\:\:alternate\:\:segment \\[3ex] \therefore \angle CAD = 18 \\[3ex] \rightarrow 33 = \angle OAC + 18 \\[3ex] \angle OAC = 33 - 18 \\[3ex] \angle OAC = 15^\circ \\[3ex] (iv) \\[3ex] \underline{Quadrilateral\:\:ABCD} \\[3ex] \angle B + \angle D = 180 ...sum\:\:of\:\:interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:Quad \\[3ex] \angle D = \angle ODA + \angle ODC ...as\:\:shown \\[3ex] \angle ODG = \angle ODC + \angle CDG ...as\:\:shown \\[3ex] 90 = \angle ODC + 18 \\[3ex] \angle ODC = 90 - 18 \\[3ex] \angle ODC = 72 \\[3ex] \rightarrow \angle D = 33 + 72 \\[3ex] \angle D = 105 \\[3ex] \rightarrow \angle B + 105 = 180 \\[3ex] \angle B = 180 - 105 \\[3ex] \angle B = 75^\circ$
(18.)

$\angle MQP = 90 ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \angle MPQ = \angle MNQ ...\angle s\:\:in\:\:same\:\:segment \\[3ex] \rightarrow \angle MPQ = 42 \\[3ex] \underline{\triangle QMP} \\[3ex] \angle QMP + \angle MPQ + \angle MQP = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] \angle QMP + 42 + 90 = 180 \\[3ex] \angle QMP + 132 = 180 \\[3ex] \angle QMP = 180 - 132 \\[3ex] \angle QMP = 48^\circ$
(19.)

$(i) \\[3ex] \angle OQR = \angle ORQ = 32 ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \underline{\triangle OQR} \\[3ex] \angle OQR + \angle ORQ + \angle QOR = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 32 + 32 + \angle QOR = 180 \\[3ex] 64 + \angle QOR = 180 \\[3ex] \angle QOR = 180 - 64 \\[3ex] \angle QOR = 116 \\[3ex] \angle QOR = 2 * \angle QPR ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 116 = 2 * \angle QPR \\[3ex] \angle QPR = \dfrac{116}{2} \\[5ex] \angle QPR = 58^\circ \\[3ex] (ii) \\[3ex] \underline{Quadrilateral\:\:TPRQ} \\[3ex] \angle P + \angle Q = 180 ...sum\:\:of\:\:interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:Quad \\[3ex] \angle P = \angle TPQ + \angle QPR ...as\:\:shown \\[3ex] \angle P = 15 + 58 \\[3ex] \angle P = 73 \\[3ex] \angle Q = \angle TQO + \angle OQR ...as\:\:shown \\[3ex] \angle Q = \angle TQO + 32 \\[3ex] \rightarrow 73 + \angle TQO + 32 = 180 \\[3ex] \angle TQO + 105 = 180 \\[3ex] \angle TQO = 180 - 105 \\[3ex] \angle TQO = 75^\circ$
(20.)

$\underline{Quadrilateral\:\:TUVW} \\[3ex] \angle U + \angle W = 180 ...sum\:\:of\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:Quad \\[3ex] 3x + 20 + 88 = 180 \\[3ex] 3x + 108 = 180 \\[3ex] 3x = 180 - 108 \\[3ex] 3x = 72 \\[3ex] x = \dfrac{72}{3} \\[5ex] x = 24^\circ$

(21.)

$a) \\[3ex] \angle VZW = 51^\circ ...\angle \:\:between\:\:tangent\:\:and\:\:chord = \angle\:\:in\:\:alternate\:\:segment \\[3ex] b) \\[3ex] \angle VWZ = 78^\circ ...\angle \:\:between\:\:tangent\:\:and\:\:chord = \angle\:\:in\:\:alternate\:\:segment \\[3ex] \underline{Cyclic\:\:Quadrilateral\:\:WXYZ} \\[3ex] \angle XYZ = 78^\circ ...exterior\:\: \angle \:\:of\:\:a\:\:cyclic\:\:Quad = interior\:\:opposite\:\: \angle$
(22.)

$\angle SRQ = \angle SPQ ...\angle \:\:between\:\:tangent\:\:and\:\:chord = \angle\:\:in\:\:alternate\:\:segment \\[3ex] \rightarrow \angle SPQ = 50 \\[3ex] \angle SPQ = \angle SQP ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \rightarrow \angle SQP = 50 \\[3ex] \underline{\triangle SRQ} \\[3ex] \angle RSQ = \angle SPQ + \angle SQP ...exterior\:\: \angle \:\:of\:\:a\:\: \triangle \\[3ex] \angle RSQ = 50 + 50 \\[3ex] \angle RSQ = 100 \\[3ex] \angle SRQ + \angle RSQ + \angle SQR = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] \angle SRQ + 100 + 50 = 180 \\[3ex] \angle SRQ + 150 = 180 \\[3ex] \angle SRQ = 180 - 150 \\[3ex] \angle SRQ = 30^\circ$
(23.)

$\angle TAC = 30 ...\angle \:\:between\:\:tangent\:\:and\:\:chord = \angle\:\:in\:\:alternate\:\:segment \\[3ex] \angle ATC = 90 ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \underline{\triangle TKC} \\[3ex] \angle TCK = 90 + 30 ...exterior\:\: \angle \:\:of\:\:a\:\: \triangle \\[3ex] \angle TCK = 120 \\[3ex] \angle TKC + \angle TCK + \angle CTK = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] \angle TKC + 120 + 30 = 180 \\[3ex] \angle TKC + 150 = 180 \\[3ex] \angle TKC = 180 - 150 \\[3ex] \angle TKC = 30^\circ$
(24.)

$(i) \\[3ex] \angle BOA = 2 * \angle ACB ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 130 = 2 * \angle ACB \\[3ex] \angle ACB = \dfrac{130}{2} \\[5ex] \angle ACB = 65^\circ \\[3ex] (ii) \\[3ex] \angle CBD = \angle CAD ...\angle s\:\:in\:\:same\:\:segment \\[3ex] \therefore \angle CBD = 30^\circ \\[3ex] (iii) \\[3ex] \angle ADB = \angle ACB ...\angle s\:\:in\:\:same\:\:segment \\[3ex] \rightarrow \angle ADB = 65 \\[3ex] \underline{\triangle AED} \\[3ex] \angle DAE + \angle AED + \angle EDA = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] \angle EDA = \angle ADB = 65 ...as\:\:shown \\[3ex] 30 + \angle AED + 65 = 180 \\[3ex] \angle AED + 95 = 180 \\[3ex] \angle AED = 180 - 95 \\[3ex] \angle AED = 85^\circ$