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Solved Examples on Trigonometry (All Topics)

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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.
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NSC (1.1) Without using a calculator, write the following expressions in terms of $\sin 11^\circ$:
(1.1.1) $\sin 191^\circ$
(1.1.2) $\cos 22^\circ$

(1.2) Simplify $\cos (x - 180^\circ) + \sqrt{2}\sin (x + 45^\circ)$ to a single trigonometric ratio.

(1.3) Given: $\sin P + \sin Q = \dfrac{7}{5}$ and $\hat{P} + \hat{Q} = 90^\circ$

Without using a calculator, determine the value of $\sin 2P$

$ (1.1.1) \\[3ex] \sin 191 \\[3ex] = \sin(180 + 11)...Sum\;\;Formula \\[3ex] = \sin 180 \cos 11 + \cos 180 \sin 11 \\[3ex] = 0 * \cos 11 + -1 * \sin 11 \\[3ex] = 0 + -\sin 11 \\[3ex] = 0 - \sin 11 \\[3ex] = -\sin 11 \\[3ex] (1.1.2) \\[3ex] \cos 22 \\[3ex] = \cos (2 * 11)...Double-Angle\;\;Formula \\[3ex] = 1 - 2\sin^2(11) \\[3ex] (1.2) \\[3ex] \cos (x - 180^\circ) + \sqrt{2}\sin (x + 45^\circ) \\[3ex] \cos (x - 180) \\[3ex] = \cos x \cos 180 + \sin x \sin 180 ...Difference\;\;Formula \\[3ex] = \cos x * -1 + \sin x * 0 \\[3ex] = -\cos x + 0 \\[3ex] = -\cos x \\[3ex] \sin(x + 45) \\[3ex] = \sin x \cos 45 + \cos x \sin 45 ...Sum\;\;Formula \\[3ex] = \sin x \left(\dfrac{\sqrt{2}}{2}\right) + \cos x\left(\dfrac{\sqrt{2}}{2}\right) \\[5ex] = \dfrac{\sqrt{2}}{2} \sin x + \dfrac{\sqrt{2}}{2} \cos x \\[5ex] \sqrt{2} \sin (x + 45) \\[3ex] = \sqrt{2}\left(\dfrac{\sqrt{2}}{2} \sin x + \dfrac{\sqrt{2}}{2} \cos x\right) \\[5ex] = \sin x + \cos x \\[3ex] \implies \\[3ex] \cos (x - 180^\circ) + \sqrt{2}\sin (x + 45^\circ) \\[3ex] = -\cos x + \sin x + \cos x \\[3ex] = \sin x \\[3ex] (5.3) \\[3ex] \hat{P} + \hat{Q} = 90 \implies \hat{P} \;\;and\;\; \hat{Q} \;\;are\;\;complements \\[3ex] \implies \sin Q = \cos P ...cofunctions\;\;of\;\;complementary\;\;\angle s \\[3ex] \sin P + \sin Q = \dfrac{7}{5} \\[5ex] \implies \sin P + \cos P = \dfrac{7}{5} \\[5ex] Square\;\;both\;\;sides \\[3ex] (\sin P + \cos P)^2 = \left(\dfrac{7}{5}\right)^2 \\[5ex] (\sin P + \cos P)(\sin P + \cos P) = \dfrac{49}{25} \\[5ex] \sin^2P + \sin P \cos P + \cos P \sin P + \cos^2P = \dfrac{49}{25} \\[5ex] \sin^2P + 2\sin P \cos P + \cos^2P = \dfrac{49}{25} \\[5ex] \sin^2P + \cos^2P + 2\sin P \cos P = \dfrac{49}{25} \\[5ex] \sin^2P + \cos^2P = 1 ...Pythagorean\;\;Identity \\[3ex] \implies 1 + 2\sin P \cos P = \dfrac{49}{25} \\[5ex] 2\sin P \cos P = \dfrac{49}{25} - 1 \\[5ex] 2\sin P \cos P = \dfrac{49}{25} - \dfrac{25}{25} \\[5ex] 2\sin P \cos P = \dfrac{24}{25} $

(2.) JAMB A chord of a circle subtends an angle of $60^\circ$ at the centre of a circle of radius $14\;cm$
Find the length of the chord.

$ A.\;\; 7\;cm \\[3ex] B.\;\; 14\;cm \\[3ex] C.\;\; 21\;cm \\[3ex] D.\;\; 28\; cm \\[3ex] $

$ x^2 = 14^2 + 14^2 - 2(14)(14) \cos 60^\circ...Cosine\;\;Law \\[3ex] x^2 = 196 + 196 - 392\left(\dfrac{1}{2}\right) \\[5ex] x^2 = 392 - 196 \\[3ex] x^2 = 196 \\[3ex] x^2 = 196 \\[3ex] x = \sqrt{196} \\[3ex] x = 14\;cm $

(3.) ACT What is the area of $\triangle AOB$ in square coordinate units?

$ A.\;\; 6 \\[3ex] B.\;\; 6\sqrt{2} \\[3ex] C.\;\; 12 \\[3ex] D.\;\; 18 \\[3ex] E.\;\; 36 \\[3ex] $

$ \underline{\triangle OAB} \\[3ex] Area = \dfrac{1}{2} * OA * OB * \sin \angle AOB \\[5ex] Area = \dfrac{1}{2} * 6 * 6 * \sin 90 \\[5ex] Area = 3 * 6 * 1 \\[3ex] Area = 18 \;\;square\;\;units $

(4.) ACT What is the length of $\overline{AB}$ in coordinate units?

$ F.\;\; 2\sqrt{6} \\[3ex] G.\;\; 6\sqrt{2} \\[3ex] H.\;\; 6\sqrt{3} \\[3ex] J.\;\; 6 \\[3ex] K.\;\; 12 \\[3ex] $

$ \underline{\triangle OAB} \\[3ex] hyp^2 = leg^2 + leg^2 ... Pythagorean\;\;theorem \\[3ex] leg = 6 \\[3ex] leg = 6 \\[3ex] hyp^2 = 6^2 + 6^2 \\[3ex] hyp^2 = 36 + 36 \\[3ex] hyp^2 = 72 \\[3ex] hyp = \sqrt{72} \\[3ex] hyp = \sqrt{36 * 2} \\[3ex] hyp = \sqrt{36} * \sqrt{2} \\[3ex] hyp = 6 * \sqrt{2} \\[3ex] hyp = 6\sqrt{2} $

(5.) ACT Which of the following is an equation of $\overleftrightarrow{AB}$ ?

$ A.\;\; y = -x + 6 \\[3ex] B.\;\; y = x - 6 \\[3ex] C.\;\; y = x + 6 \\[3ex] D.\;\; y = -6x - 6 \\[3ex] E.\;\; y = 6x + 6 \\[3ex] $

$ A = Point\;\;1 = (0, 6) \\[3ex] x_1 = 0 \\[3ex] y_1 = 6 \\[3ex] B = Point\;\;2 = (6, 0) \\[3ex] x_2 = 6 \\[3ex] y_2 = 0 \\[3ex] Slope = m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{0 - 6}{6 - 0} \\[5ex] m = \dfrac{-6}{6} \\[5ex] m = -1 \\[3ex] b = y-intercept = 6 \\[3ex] Equation:\;\; y = mx + b \\[3ex] y = -1 * x + 6 \\[3ex] y = -x + 6 $

(6.) JAMB

In the diagram above, $\overline{XZ}$ is the diameter of the circle XYZW, with the centre O and radius $\dfrac{15}{2}\;cm$
If $|XY| = 12\;cm$, find the area of the triangle.

$ A.\;\; 45\;cm^2 \\[3ex] B.\;\; 27\;cm^2 \\[3ex] C.\;\; 75\;cm^2 \\[3ex] D.\;\; 54\;cm^2 \\[3ex] $

$ Radius = OX = OZ = \dfrac{15}{2} \\[5ex] Diameter = XZ = 2 * \dfrac{15}{2} = 15 \\[5ex] \underline{\triangle XYZ} \\[3ex] hyp^2 = leg^2 + leg^2 ...Pythagorean\;\;theorem \\[3ex] hyp = XZ = 15\;cm \\[3ex] leg = XY = 12\;cm \\[3ex] leg = YZ = ? \\[3ex] \implies \\[3ex] 15^2 = 12^2 + leg^2 \\[3ex] leg^2 = 15^2 - 12^2 \\[3ex] leg^2 = 225 - 144 \\[3ex] leg^2 = 81 \\[3ex] leg = \sqrt{81} \\[3ex] leg = 9\;cm \\[3ex] \underline{Circle\;XYZW} \\[3ex] \angle XYZ = 90^\circ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] Area\;\;of\;\;\triangle XYZ \\[3ex] = \dfrac{1}{2} * XY * YZ * \sin \angle XYZ \\[5ex] = \dfrac{1}{2} * 12 * 9 * \sin 90 \\[5ex] = 6 * 9 * 1 \\[3ex] = 54\;cm^2 $

(7.) CSEC (a)

In the diagram above, not drawn to scale, $O$ is the centre of the circle.
$\angle AOB = 130^\circ$, $\angle DAE = 30^\circ$, and $AEC$ and $BED$ are chords of the circle.
(a) Calculate the size of EACH of the following angles, giving reasons for EACH step of your answers.

$ (i)\:\: \angle ACB \\[3ex] (ii)\:\: \angle CBD \\[3ex] (iii)\:\: \angle AED \\[3ex] $ (b) Show that $\triangle BCE$ and $\triangle ADE$ are similar.

(c) Given that CE = 6 cm, EA = 9.1 cm and DE = 5 cm,
(i) calculate the length of EB
(ii) calculate correct to 1 decimal place the area of $\triangle AED$

The (a) part is Circle Theorems

$ (i) \\[3ex] \angle BOA = 2 * \angle ACB \\[3ex] ...\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 = 30^\circ ...\angle s\:\:in\:\:same\:\:segment\;\;are\;\;equal \\[3ex] (iii) \\[3ex] \angle ADB = \angle ACB = 65^\circ ...\angle s\:\:in\:\:same\:\:segment\;\;are\;\;equal \\[3ex] \underline{\triangle AED} \\[3ex] \angle DAE + \angle AED + \angle EDA = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle AED \\[3ex] \angle EDA = \angle ADB = 65 ...diagram \\[3ex] 30 + \angle AED + 65 = 180 \\[3ex] \angle AED + 95 = 180 \\[3ex] \angle AED = 180 - 95 \\[3ex] \angle AED = 85^\circ \\[3ex] (b) \\[3ex] \underline{\triangle BCE} \\[3ex] \angle BCE = \angle ACB = 65^\circ ...diagram \\[3ex] \angle CBE = \angle CBD = 30^\circ ...diagram \\[3ex] \underline{\triangle ADE} \\[3ex] \angle ADE = \angle BCE = 65^\circ ...\angle s\:\:in\:\:same\:\:segment\;\;are\;\;equal \\[3ex] \angle DAE = 30^\circ...given \\[3ex] Because: \\[3ex] \angle BCE = \angle ADE = 65^\circ \\[3ex] And: \\[3ex] \angle CBE = \angle DAE = 30^\circ \\[3ex] \therefore \triangle BCE \sim \triangle ADE ...AA\sim\;\;Postulate \\[3ex] (c) \\[3ex] (i) \\[3ex] \dfrac{|EB|}{|EA|} = \dfrac{|CE|}{|DE|} ...Scale\;\;Factors \\[5ex] \dfrac{|EB|}{9.1} = \dfrac{6}{5} \\[5ex] |EB| = \dfrac{9.1(6)}{5} \\[5ex] |EB| = \dfrac{54.6}{5} \\[5ex] |EB| = 10.92\;cm \\[3ex] (ii) \\[3ex] \underline{\triangle AED} \\[3ex] \angle AED + \angle ADE + \angle DAE = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;\triangle AED \\[3ex] \angle AED + 65 + 30 = 180 \\[3ex] \angle AED + 95 = 180 \\[3ex] \angle AED = 180 - 95 \\[3ex] \angle AED = 85^\circ \\[3ex] Area\;\;of\;\;\triangle AED \\[3ex] = \dfrac{1}{2} * |EA| * |DE| * \sin \angle AED \\[5ex] = 0.5 * 9.1 * 5 * \sin 85^\circ \\[5ex] = 22.75 * 0.9961946981 \\[3ex] = 22.66342938 \\[3ex] \approx 22.7\;cm^2 $

(8.) ACT Graphed in the standard $(x, y)$ coordinate plane below is line l and the circle with equation $(x - 2)^2 + y^2 = 1$
Line l passes through $O(0, 0)$ and is tangent to the circle at A, and B is the center of the circle.
What is the measure of $\angle AOB$?

$ F.\;\; 15^\circ \\[3ex] G.\;\; 22.5^\circ \\[3ex] H.\;\; 30^\circ \\[3ex] J.\;\; 45^\circ \\[3ex] K.\;\; 60^\circ \\[3ex] $

$ Based\;\;on\;\; (x - 2)^2 + y^2 = 1 \\[3ex] (x - 2)^2 + (y - 0)^2 = 1^2 \\[3ex] \implies \\[3ex] Circle\;\;center = (2, 0)...at\;\;B \\[3ex] \implies OB = 2 \\[3ex] Radius = AB = 1 \\[3ex] $

$ \underline{\triangle OAB} \\[3ex] \dfrac{\sin \angle AOB}{1} = \dfrac{\sin 90}{2} \\[5ex] \sin \angle AOB = \dfrac{1}{2} \\[5ex] \angle AOB = \sin^{-1}\left(\dfrac{1}{2}\right) \\[5ex] \angle AOB = 30^\circ $

(9.) NZQA C is the centre of the circle.
The radius of the circle is 28 mm.
Straight lines LMT and LNS are tangents to the circle at points M and N, respectively.
Angle MLN = 72°

Calculate the length LC
Justify your answer with clear geometrical reasoning.

$ \angle CML = 90^\circ ... radius\;CM \perp tangent\;LMT \;\;at\;\;point\;\;of\;\;contact\;M \\[3ex] \angle CLM = \dfrac{1}{2} * 72 \\[3ex] ... the\;\;line\;CL\;\;joining\;\;the\;\;external\;\;point\;\;from\;\;two\;\;tangents\;\;and\;\;the\;\;centre\;C\;\;bisects\;\;the\;\;\angle\;\;between\;\;the\;\;tangents \\[3ex] \angle CLM = 36^\circ \\[3ex] $

$ \underline{\triangle CLM} \\[3ex] \sin \angle CLM = \dfrac{|MC|}{|LC|} ... SOHCAHTOA \\[5ex] \sin 36 = \dfrac{28}{|LC|} \\[5ex] |LC| * \sin 36 = 28 \\[3ex] |LC| = \dfrac{28}{\sin 36} \\[5ex] |LC| = \dfrac{28}{0.5877852523} \\[5ex] |LC| = 47.63644527\;mm $

(10.) CSEC Given that $\sin\theta = \dfrac{\sqrt{3}}{2},\;\;\; 0^\circ \le \theta \le 90^\circ$

(i.) Express in fractional or surd form the value of $\cos\theta$

(ii.) Show that the area of triangle $CDE$ is $150\sqrt{3}$ square units, where $CD = 30$ units and $DE = 20$ units.

(iii.) Calculate the length of the side $EC$

$ (i.) \\[3ex] 0^\circ \le \theta \le 90^\circ \\[3ex] SOH:CAH:TOA \\[3ex] \sin \theta = \dfrac{\sqrt{3}}{2} = \dfrac{opp}{hyp} \\[5ex] opp = \sqrt{3} \\[3ex] hyp = 2 \\[3ex] hyp^2 = opp^2 + adj^2 ...Pythagorean\;\;theorem \\[3ex] 2^2 = (\sqrt{3})^2 + adj^2 \\[3ex] 4 = 3 + adj^2 \\[3ex] 4 - 3 = adj^2 \\[3ex] 1 = adj^2 \\[3ex] adj^2 = 1 \\[3ex] adj = \sqrt{1} \\[3ex] adj = 1 \\[3ex] \cos \theta \\[3ex] = \dfrac{adj}{hyp} \\[5ex] = \dfrac{1}{2} \\[5ex] (ii.) \\[3ex] \underline{\triangle CDE} \\[3ex] \cos \theta = \dfrac{1}{2} \\[5ex] \theta = \cos^{-1}\left(\dfrac{1}{2}\right) \\[5ex] \theta = 60^\circ \\[3ex] Area = \dfrac{1}{2} * |CD| * |DE| * \sin\theta \\[5ex] = \dfrac{1}{2} * 30 * 20 * \sin 60 \\[5ex] = 15 * 20 * \dfrac{\sqrt{3}}{2} \\[5ex] = 15 * 10 * \sqrt{3} \\[3ex] = 150\sqrt{3}\;square\;\;units \\[3ex] (iii.) \\[3ex] |EC|^2 = |CD|^2 + |DE|^2 - 2(|CD|)(|DE|)\cos\theta...Cosine\;\;Law \\[3ex] |EC|^2 = 30^2 + 20^2 - 2(30)(20) * \cos 60 \\[3ex] |EC|^2 = 900 + 400 - 1200 * \dfrac{1}{2} \\[5ex] |EC|^2 = 1300 - 600 \\[3ex] |EC|^2 = 700 \\[3ex] |EC| = \sqrt{700} \\[3ex] |EC| = 26.45751311 \\[3ex] |EC| \approx 26\;units $

(11.) curriculum.gov.mt

PRT is a triangle with points Q and S on PR and RT respectively such that SQ is parallel to TP.

(a) Show that triangles RTP and RSQ are similar.

(b) Calculate the size of SQ

(c) Given that the area of triangle RTP is 35 $cm^2$, calculate:
(i) the area of triangle RSQ
(ii) the area of quadrilateral QSTP

$ (a) \\[3ex] \angle RQS = \angle RPT...corresponding\;\;\angle s\;\;are\;\;equal \\[3ex] \angle RSQ = \angle RTP ...corresponding\;\;\angle s\;\;are\;\;equal \\[3ex] \angle QRS = \angle PRT...common\;\;\angle\;\;R \\[3ex] Any\;\;of\;\;the\;\;two\;\;can\;\;be\;\;used\;\;for\;\;the\;\;AA\sim\;\;Postulate \\[3ex] \therefore \triangle RTP \sim \triangle RSQ...AA\sim\;\;Postulate \\[3ex] (b) \\[3ex] \dfrac{|SQ|}{7.8} = \dfrac{7.2}{7.2 + 6} \\[5ex] \dfrac{|SQ|}{7.8} = \dfrac{7.2}{13.2} \\[5ex] |SQ| = \dfrac{7.8(7.2)}{13.2} \\[5ex] |SQ| = \dfrac{56.16}{13.2} \\[5ex] |SQ| = 4.254545455 \\[3ex] |SQ| \approx 4.3\;cm \\[3ex] (c) \\[3ex] (i) \\[3ex] Scale\;\;Factor = \dfrac{|RQ|}{|RP|} \\[5ex] Area\;\;Ratio = \dfrac{|RQ|^2}{|RP|^2} ...Area\;\;Ratio-Scale\;\;Factor\;\;Theorem \\[5ex] Also: \\[3ex] Area\;\;Ratio = \dfrac{Area\;\;of\;\;\triangle RSQ}{Area\;\;of\;\;\triangle RTP} \\[5ex] \implies \\[3ex] \dfrac{Area\;\;of\;\;\triangle RSQ}{Area\;\;of\;\;\triangle RTP} = \dfrac{|RQ|^2}{|RP|^2} \\[5ex] \dfrac{Area\;\;of\;\;\triangle RSQ}{35} = \dfrac{(7.2)^2}{(7.2 + 6)^2} \\[5ex] \dfrac{Area\;\;of\;\;\triangle RSQ}{35} = \dfrac{(7.2)^2}{(13.2)^2} \\[5ex] \dfrac{Area\;\;of\;\;\triangle RSQ}{35} = \dfrac{51.84}{174.24} \\[5ex] Area\;\;of\;\;\triangle RSQ \\[3ex] = \dfrac{35 * 51.84}{174.24} \\[5ex] = \dfrac{1814.4}{174.24} \\[5ex] = 10.41322314 \\[3ex] \approx 10.4\;cm^2 \\[3ex] (ii) \\[3ex] Area\;\;of\;\;Quadrilateral\;QSTP = Area\;\;of\;\;\triangle RTP - Area\;\;of\;\;\triangle RSQ ...diagram \\[3ex] = 35 - 10.41322314 \\[3ex] = 24.58677686 \\[3ex] \approx 24.6\;cm^2 $

(12.) CSEC The digram below, not drawn to scale, shows the sector of a circle with centre $O$
$\angle MON = 45^\circ$ and $ON = 15\;cm$

Calculate, giving your answer correct to 2 decimal places
(i.) the length of the minor arc MN
(ii.) the perimeter of the figure MON
(iii.) the area of the figure MON

$ (i.) \\[3ex] Length\;\;of\;\;arc\;MN = L \\[3ex] L = \dfrac{2\pi r\theta}{360} ...\theta \:\:in\:\:DEG \\[5ex] \pi = 3.14 \\[3ex] \theta = 45^\circ \\[3ex] r = 15\;cm \\[3ex] L = \dfrac{2 * 3.14 * 15 * 45}{360} \\[5ex] L = \dfrac{4239}{360} \\[5ex] L = 11.775 \\[3ex] L \approx 11.78\;cm \\[3ex] (ii.) \\[3ex] \underline{Figure\;\;MON} \\[3ex] Perimeter = r + r + L \\[3ex] Perimeter = 15 + 15 + 11.775 \\[3ex] Perimeter = 41.775 \\[3ex] Perimeter \approx 41.78\;cm \\[3ex] (iii.) \\[3ex] \underline{Figure\;\;MON} \\[3ex] Area\;\; of\;\; Figure\;\;MON = Area\;\;of\;\;sector\;OMN = A \\[3ex] A = \dfrac{\pi r^2\theta}{360} ...\theta \:\:in\:\:DEG \\[5ex] A = \dfrac{3.14 * 15^2 * 45}{360} \\[5ex] A = \dfrac{31792.5}{360} \\[5ex] A = 88.3125 \\[3ex] A \approx 88.31\;cm^2 $

(13.) NZQA In the diagram drawn below, A represents a point due south of the base of a vertical tower BT.
The point C is due east of B.
The points A, B, and C are all on horizontal ground.
Kenny, lying at the point A, measured the angle of elevation of the top of the tower, T, to be 32°.
The point C is on a bearing of 056° from A.
The distance AC is 86 metres.

(i) Show that the height, BT, of the tower is 30.05 metres.
Show your working clearly.

(ii) Calculate the angle of elevation from C to T.
Show your working clearly.

$ (i) \\[3ex] \underline{\triangle ABC} \\[3ex] \cos \angle BAC = \dfrac{|AB|}{|AC|} ...SOHCAHTOA \\[5ex] \cos 56^\circ = \dfrac{|AB|}{86} \\[5ex] |AB| = 86\cos 56^\circ \\[3ex] \underline{\triangle ATB} \\[3ex] \tan \angle TAB = \dfrac{|BT|}{|AB|} ...SOHCAHTOA \\[5ex] \tan 32^\circ = \dfrac{|BT|}{86\cos 56^\circ} \\[5ex] |BT| = 86\cos 56^\circ\tan 32^\circ \\[3ex] |BT| = 86(0.5591929035)(0.6248693519) \\[3ex] |BT| = 30.05033562 \\[3ex] |BT| \approx 30.05\;m \\[3ex] (ii) \\[3ex] \underline{\triangle ABC} \\[3ex] \sin \angle BAC = \dfrac{|BC|}{|AC|} ...SOHCAHTOA \\[5ex] \sin 56^\circ = \dfrac{|BC|}{86} \\[5ex] |BC| = 86\sin 56^\circ \\[3ex] |BC| = 86(0.8290375726) \\[3ex] |BC| = 71.29723124 \\[3ex] \angle\;\;of\;\;elevation\;\;from\;\;C\;\;to\;\;T \\[3ex] = \angle BCT \\[3ex] \underline{\triangle BCT} \\[3ex] \tan \angle BCT = \dfrac{|BT|}{|BC|} ...SOHCAHTOA \\[5ex] \tan \angle BCT = \dfrac{30.05033562}{71.29723124} \\[5ex] \tan \angle BCT = 0.4214796998 \\[3ex] \angle BCT = \tan^{-1}(0.4214796998) \\[3ex] \angle BCT = 22.85443544 \\[3ex] \angle BCT \approx 23^\circ $

(14.) CSEC The regular polygon EFGHIJ, shown below, has centre O.
Triange OEF is equilateral and EF = 5 cm.

(i) What is the name of the polygon shown above?
(ii) Calculate the perimeter of the polygon EFGHIJ
(iii) Determine the size of each interior angle of the polygon.
(iv) Show, by calculation, that the area of the polygon, to the nearest whole number, is 65 cm².

(i)
The regular polygon has 6 sides.
It is known as a regular hexagon.
A regular hexagon is a polygon with six equal sides and six equal angles.

$ (ii) \\[3ex] |EF| = |FG| = |GH| = |HI| = |IJ| = |JE| = 5\;cm \\[3ex] ...regular\;\;hexagon \\[3ex] Perimeter\;\;of\;\;polygon\;EFGHIJ \\[3ex] = |EF| + |FG| + |GH| + |HI| + |IJ| + |JE| \\[3ex] = 5 + 5 + 5 + 5 + 5 + 5 \\[3ex] = 5(6) \\[3ex] = 30\;cm \\[3ex] Sum\;\;of\;\;interior\;\;\angle s\;\;of\;\;a\;\;regular\;\;polygon \\[3ex] = 180(n - 2) \\[3ex] For\;\;a\;\;regular\;\;polygon:\;\; n = 6 \\[3ex] Sum = (6 - 2) * 180 \\[3ex] = 4(180) \\[3ex] = 720^\circ \\[3ex] (iii) \\[3ex] Each\;\;interior\;\;\angle = \dfrac{Sum}{n} \\[5ex] = \dfrac{720}{6} \\[5ex] = 120^\circ \\[3ex] $ (iv)

$ |OE| = |OF| = |EF| = 5\;cm...equilateral\;\triangle \\[3ex] \angle EOF = \angle OEF = \angle OFE = 60^\circ ...equilateral\;\triangle \\[3ex] Area\;\;of\;\;the\;\;equilateral\;\triangle OEF \\[3ex] = \dfrac{1}{2} * |OE| * |OF| * \sin \angle EOF \\[5ex] = \dfrac{1}{2} * 5 * 5 * \sin 60^\circ \\[5ex] = \dfrac{25}{2} * \dfrac{\sqrt{3}}{2} \\[5ex] = \dfrac{25\sqrt{3}}{4} \\[5ex] Area\;\;of\;\;the\;\;polygon \\[3ex] = 6 * Area\;\;of\;\;the\;\;equilateral\;\triangle OEF \\[3ex] = 6 * \dfrac{25\sqrt{3}}{4} \\[5ex] = 3 * \dfrac{25\sqrt{3}}{2} \\[5ex] = \dfrac{75\sqrt{3}}{2} \\[5ex] = \dfrac{75(1.732050808)}{2} \\[5ex] = \dfrac{129.9038106}{2} \\[5ex] = 64.95190528 \\[3ex] \approx 65\;cm^2 $

(20.) CSEC The diagram below shows a quadrilateral PQRS where PQ and SR are parallel, SQ = 8 cm, ∠SPQ = 90°, ∠SQR = 82° and ∠QSR = 30°.

Determine
(i) the length PS
(ii) the length PQ
(iii) the area of PQRS.

$ |PQ| \;\;\;||\;\;\; |SR| \\[3ex] \angle PQS = \angle QSR = 30^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \underline{\triangle SPQ} \\[3ex] (i) \\[3ex] \sin \angle PQS = \dfrac{|PS|}{|SQ|}...SOHCAHTOA \\[5ex] \sin 30^\circ = \dfrac{|PS|}{8} \\[5ex] |PS| = 8\sin 30 \\[3ex] |PS| = 8(0.5) \\[3ex] |PS| = 4\;cm \\[3ex] (ii) \\[3ex] \cos \angle PQS = \dfrac{|PQ|}{|SQ|}...SOHCAHTOA \\[5ex] \cos 30^\circ = \dfrac{|PQ|}{8} \\[5ex] |PQ| = 8\cos 30 \\[3ex] |PQ| = 8\left(\dfrac{\sqrt{3}}{2}\right) \\[5ex] |PQ| = 4\sqrt{3}\;cm \\[3ex] |PQ| = 6.92820323\;cm \\[3ex] $ We can determine the area of PQRS using at least two approaches.
Use any approach you prefer.

$ \underline{\triangle SPQ} \\[3ex] \angle PSQ + \angle SPQ + \angle PQS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle PSQ + 90 + 30 = 180 \\[3ex] \angle PSQ + 120 = 180 \\[3ex] \angle PSQ = 180 - 120 \\[3ex] \angle PSQ = 60^\circ \\[3ex] \angle PSR = \angle PSQ + \angle QSR \\[3ex] \angle PSR = 60 + 30 \\[3ex] \angle PSR = 90^\circ \\[3ex] $ This implies that PQRS is a trapezium where PS is the perperndicular height.
The parallel sides are PQ and SR
We have found the length of PQ
Let us find the length of SR

$ \underline{\triangle SQR} \\[3ex] \angle SRQ + \angle SQR + \angle QSR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle SRQ + 82 + 30 = 180 \\[3ex] \angle SRQ + 112 = 180 \\[3ex] \angle SRQ = 180 - 112 \\[3ex] \angle SRQ = 68^\circ \\[3ex] \dfrac{|SR|}{\sin \angle SQR} = \dfrac{|SQ|}{\sin \angle SRQ}...Sine\;\;Law \\[5ex] \dfrac{|SR|}{\sin 82} = \dfrac{8}{\sin 68}\\[5ex] |SR| = \dfrac{8\sin 82}{\sin 68}\\[5ex] |SR| = \dfrac{8(0.9902680687)}{0.9271838546} \\[5ex] |SR| = 8.544308134\;cm \\[5ex] \underline{First\;\;Approach:\;\;Area\;\;of\;\;a\;\;Trapezium} \\[3ex] Area\;\;of\;Trapezium\;\;PQRS \\[3ex] = \dfrac{1}{2} * (|PQ| + |SR|) * |PS| \\[5ex] = 0.5 * (6.92820323 + 8.544308134) * 4 \\[3ex] = 2(15.47251136) \\[3ex] = 30.94502273\;cm^2 \\[5ex] \underline{Second\;\;Approach:\;\;Areas\;\;of\;\;two\;\;Triangles} \\[3ex] Area\;\;of\;\;PQRS \\[3ex] = Area\;\;of\;\;\triangle SPQ + Area\;\;of\;\;\triangle SQR \\[3ex] = \dfrac{1}{2} * |PQ| * |PS| + \dfrac{1}{2} * |SQ| * |SR| * \sin \angle QSR \\[5ex] = 0.5(6.92820323)(4) + 0.5(8)(8.544308134)(\sin 30) \\[3ex] = 13.85640646 + 34.17723254(0.5) \\[3ex] = 13.85640646 + 17.08861627 \\[3ex] = 30.94502273\;cm^2 $

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(24.) CSEC The figure below, not drawn to scale, shows a circle with centre O.
The radius of the circle is 21 cm and angle HOK = 40°.

Determine
(i) the area of the minor sector HOK
(ii) the area of triangle HOK
(iii) the area of the shaded segment.

$ radius = r = 21\;cm \\[3ex] (i) \\[3ex] Area\;\;of\;\;minor\;\;sector\;HOK \\[3ex] = \dfrac{\angle HOK}{360} * \pi r^2 \\[5ex] = \dfrac{40}{360} * \dfrac{22}{7} * 21 * 21 \\[5ex] = 22 * 7 \\[3ex] = 154\;cm^2 \\[3ex] (ii) \\[3ex] |OH| = |OK| = 21\;cm ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle HOK \\[3ex] Area\;\;of\;\;\triangle HOK \\[3ex] = \dfrac{1}{2} * |OH| * |OK| * \sin \angle HOK \\[5ex] = \dfrac{1}{2} * 21 * 21 * \sin 40^\circ \\[5ex] = 0.5(21)(21)(0.6427876097) \\[3ex] = 141.7346679\;cm^2 \\[3ex] (iii) \\[3ex] Area\;\;of\;\;the\;\;shaded\;\;segment \\[3ex] = Area\;\;of\;\;minor\;\;sector\;HOK - Area\;\;of\;\;\triangle HOK...diagram \\[3ex] = 154 - 141.7346679 \\[3ex] = 12.2653321\;cm^2 $