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# Solved Examples on Angles of Elevation and Depression

Verify your answers with these Calculators as applicable.

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 a wrong answer.

For SAT Students
Any question labeled SAT-C is a question that allows a calculator.
Any question labeled SAT-NC is a question that does not allow a calculator.

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

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE and Malta Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions.
Draw diagrams as applicable.
Show all work.

(1.) ACT Skyline Tours is offering hot-air balloon tours.
The tables below give information about the balloon, the equipment, and the tours offered.
 Hot-air-balloon information Volume of balloon Maximum capacity of basket Weight of balloon Weight of basket Weight of burner $80,000$ cubic feet $8$ people $200$ pounds $150$ pounds $50$ pounds

 Tour information Tour Ticket price Duration, in minutes Maximum altitude, in feet A B C $\$100\$125$ $\$2004560905006001,000$Jarrod is looking up at a hot-air balloon. The balloon is currently at the maximum altitude duirng Tour C The angle of elevation from the horizon is$37^\circ$, as shown in the figure below. Which of the following expressions is closest to the distance,$d$feet, from Jarrod to the basket?$ F.\;\; \dfrac{1,000}{\sin 37^\circ} \\[5ex] G.\;\; \dfrac{1,000}{\cos 37^\circ} \\[5ex] H.\;\; 1,000 \sin 37^\circ \\[3ex] J.\;\; 1,000 \cos 37^\circ \\[3ex] K.\;\; 1,000 \tan 37^\circ \\[3ex]  SOH:CAH:TOA \\[3ex] \sin 37 = \dfrac{opp}{hyp} = \dfrac{1000}{d} \\[5ex] d * \sin 37 = 1000 \\[3ex] d = \dfrac{1,000}{\sin 37^\circ} $(2.) ACT Josh is standing in a pool and looking up at his friend Olivia. Olivia is lying on her stomach on the diving board looking at Josh. The horizontal and vertical distances, in meters, between Josh and Olivia are given in the diagram below. What is the measure of the angle of elevation,$\theta$, of Josh's line of sight? (Note: Drawing is NOT to scale.)$ F.\;\; Arcsin\left(\dfrac{3}{8}\right) \\[5ex] G.\;\; Arccos\left(\dfrac{3}{8}\right) \\[5ex] H.\;\; Arctan\left(\dfrac{3}{8}\right) \\[5ex] J.\;\; Arccot\left(\dfrac{3}{8}\right) \\[5ex] K.\;\; Arccsc\left(\dfrac{3}{8}\right) \\[5ex]  SOHCAHTOA \\[3ex] \tan \theta = \dfrac{3}{8} \\[5ex] \theta = \tan^{-1}\left(\dfrac{3}{8}\right) \\[5ex] \theta = Arctan\left(\dfrac{3}{8}\right) $(3.) NZQA One spider makes a large spider web by fixing its web between a vertical tree at point T and two points on the ground at W and S Point B is at the base of the tree, below T Points B, W, and S are all on the same horizontal level (ground level) Angle SWT = Angle WBT = 90° SW = 120 cm Angle TWB = 12° WT = 80 cm (i) Show that the height of the spider web, BT, is 16.63 cm Show your working clearly. (ii) Find the angle of elevation of T above S, angle TSB Show your working clearly.$ (i) \\[3ex] \underline{\triangle TWB} \\[3ex] \sin 12 = \dfrac{|BT|}{80}...SOHCAHTOA \\[5ex] 80\sin 12 = |BT| \\[3ex] |BT| = 80\sin 12 \\[3ex] |BT| = 80(0.2079116908) \\[3ex] |BT| = 16.63293527 \\[3ex] |BT| \approx 16.63\;cm \\[3ex] (ii) \\[3ex] \underline{\triangle STW} \\[3ex] |ST|^2 = 120^2 + 80^2 ...Pythagorean\;\;theorem \\[3ex] |ST|^2 = 14400 + 6400 \\[3ex] |ST|^2 = 20800 \\[3ex] |ST| = \sqrt{20800} \\[3ex] |ST| = 144.222051 \\[3ex] \underline{\triangle STB} \\[3ex] \sin \angle TSB = \dfrac{|BT|}{ST} ...SOHCAHTOA \\[5ex] \sin \angle TSB = \dfrac{16.63293527}{144.222051} \\[5ex] \sin \angle TSB = 0.1153286557 \\[3ex] \angle TSB = \sin^{-1} (0.1153286557) \\[3ex] \angle TSB = 6.622581764^\circ \\[3ex] \angle TSB \approx 7^\circ $(4.) ACT Josiah stands on level ground 15 ft from the base of a cliff. The angle of elevation from where Josiah is standing to the top of the cliff is 50°, as shown below. Which of the following values is closest to the height, in feet, of the cliff? (Note:$\sin 50^\circ = 0.8;\;\;\cos 50^\circ = 0.6;\;\;\tan 50^\circ = 1.2$)$ A.\;\; 12 \\[3ex] B.\;\; 13 \\[3ex] C.\;\; 18 \\[3ex] D.\;\; 25 \\[3ex] E.\;\; 60 \\[3ex]  \tan 50 = \dfrac{height}{15} ...SOHCAHTOA \\[5ex] 15\tan 50 = height \\[3ex] height = 15 * \tan 50 \\[3ex] height = 15(1.2) \\[3ex] height = 18\;ft $(5.) curriculum.gov.mt PQ is a vertical pole fixed on horizontal ground. It is attached to the ground by two vents PR and PS such that$R\hat{Q}S = 109^\circ$and$Q\hat{P}S = 28^\circ$Calculate: (a) the angle of elevation of P from R (b) the distance RQ (c) the distance QS (d) the distance RS$ (a) \\[3ex] \angle\;\;of\;\;elevation\;\;of\;\;P\;\;from\;\;R = P\hat{R}Q \\[3ex] \underline{\triangle RPQ} \\[3ex] \sin P\hat{R}Q = \dfrac{|PQ|}{|PQ|}...SOHCAHTOA \\[5ex] \sin P\hat{R}Q = \dfrac{11}{14.5} \\[5ex] \sin P\hat{R}Q = 0.7586206897 \\[3ex] P\hat{R}Q = \sin^{-1}(0.7586206897) \\[3ex] P\hat{R}Q = 49.34275136^\circ \\[3ex] (b) \\[3ex] \underline{\triangle RPQ} \\[3ex] |PR|^2 = |RQ|^2 + |PQ|^2 ...Pythagorean\;\;theorem \\[3ex] 14.5^2 = |RQ|^2 + 11^2 \\[3ex] |RQ|^2 + 11^2 = 14.5^2 \\[3ex] |RQ|^2 + 121 = 210.25 \\[3ex] |RQ|^2 = 210.25 - 121 \\[3ex] |RQ|^2 = 89.25 \\[3ex] |RQ| = \sqrt{89.25} \\[3ex] |RQ| = 9.447221814\;m \\[3ex] (c) \\[3ex] \underline{\triangle PQS} \\[3ex] \tan Q\hat{P}S = \dfrac{|QS|}{|PQ|}...SOHCAHTOA \\[5ex] \tan 28 = \dfrac{|QS|}{11} \\[5ex] 11\tan 28 = |QS| \\[3ex] |QS| = 11\tan 28 \\[3ex] |QS| = 11(0.5317094317) \\[3ex] |QS| = 5.848803748\;m \\[3ex] $Construction: Draw the length from point R to point S$ \underline{\triangle RQS} \\[3ex] |RS|^2 = |RQ|^2 + |QS|^2 - 2 * |RQ| * |QS| * \cos R\hat{Q}S...Cosine\;\;Law \\[3ex] |RS|^2 = 9.447221814^2 + 5.848803748^2 - 2(9.447221814)(5.848803748) * \cos 109 \\[3ex] |RS|^2 = 89.25 + 34.20850528 - 110.5098927 * -0.3255681545 \\[3ex] |RS|^2 = 123.4585053 + 35.97850182 \\[3ex] |RS|^2 = 159.4370071 \\[3ex] |RS| = \sqrt{159.4370071} \\[3ex] |RS| = 12.62683678\;m $(6.) ACT The figure below shows a flying kite. At a certain moment, the kite string forms an angle of elevation of 75° from point A on the ground. At the same moment, the angle of elevation of the kite at point B, 240 ft from A on level ground, is 45°. What is the length, in feet, of the string?$ A.\;\; 60\sqrt{3} \\[3ex] B.\;\; 80\sqrt{6} \\[3ex] C.\;\; 144 \\[3ex] D.\;\; 180 \\[3ex] E.\;\; 240 \\[3ex] $Let the length of string = string$ \underline{\triangle AKB} \\[3ex] \angle KAB + \angle KBA + \angle AKB = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;\triangle AKB \\[3ex] 75 + 45 + \angle AKB = 180 \\[3ex] 120 + \angle AKB = 180 \\[3ex] \angle AKB = 180 - 120 \\[3ex] \angle AKB = 60^\circ \\[3ex] \dfrac{string}{\sin 45^\circ} = \dfrac{240}{\sin 60^\circ} ...Sine\;\;Law \\[5ex] \sin 45 = \dfrac{\sqrt{2}}{2} ...Special\;\;Angles \\[5ex] \sin 60 = \dfrac{\sqrt{3}}{2} ... Special\;\;Angles \\[3ex] \implies \\[3ex] string \div \sin 45 = 240 \div \sin 60 \\[3ex] string \div \dfrac{\sqrt{2}}{2} = 240 \div \dfrac{\sqrt{3}}{2} \\[5ex] string * \dfrac{2}{\sqrt{2}} = 240 * \dfrac{2}{\sqrt{3}} \\[5ex] \dfrac{\sqrt{2}}{2} * string * \dfrac{2}{\sqrt{2}} = \dfrac{\sqrt{2}}{2} * 240 * \dfrac{2}{\sqrt{3}} \\[5ex] string = \dfrac{240\sqrt{2}}{\sqrt{3}} \\[5ex] = \dfrac{240\sqrt{2}}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\[5ex] = \dfrac{240\sqrt{6}}{3} \\[5ex] = 80\sqrt{6} $(7.) (8.) HSC A tower BT has height h metres. From point A, the angle of elevation to the top of the tower is 26° as shown. Which of the following is the correct expression for the length of AB?$ A.\;\; h\tan 26^\circ \\[3ex] B.\;\; h\cot 26^\circ \\[3ex] C.\;\; h\sin 26^\circ \\[3ex] D.\;\; h\;cosec\; 26^\circ \\[3ex]  \tan 26^\circ = \dfrac{h}{|AB|}...SOHCAHTOA \\[5ex] |AB|\tan 26 = h \\[3ex] |AB| = \dfrac{h}{\tan 26} \\[5ex] |AB| = h * \dfrac{1}{\tan 26} \\[5ex] |AB| = h * \cot 26 \\[3ex] |AB| = h\cot 26^\circ $(9.) (10.) CSEC The diagram below, not drawn to scale, shows two ships, R and S at anchor on a lake of calm water. FT is a vertical tower. FSR is a straight line and RS = 150 m. The angles of elevation of T, the top of a tower, from R and S, are 22° and 40° respectively. F is the foot of the tower. Calculate, giving your answer to 1 decimal place where appropriate (i) the measure of ∠RTS (ii) the length of ST (iii) the height of the tower, FT.$ \angle RST + \angle FST = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle RST + 40 = 180 \\[3ex] \angle RST = 180 - 40 \\[3ex] \angle RST = 140^\circ \\[3ex] (i) \\[3ex] \underline{\triangle TRS} \\[3ex] \angle RTS + \angle RST + \angle TRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle RTS + 140 + 22 = 180 \\[3ex] \angle RTS + 162 = 180 \\[3ex] \angle RTS = 180 - 162 \\[3ex] \angle RTS = 18^\circ \\[3ex] (ii) \\[3ex] \dfrac{|ST|}{\sin \angle TRS} = \dfrac{|RS|}{\sin \angle RTS}...Sine\;\;Law \\[5ex] \dfrac{|ST|}{\sin 22^\circ} = \dfrac{150}{\sin 18^\circ} \\[5ex] |ST| = \dfrac{150\sin 22^\circ}{\sin 18^\circ} \\[5ex] |ST| = \dfrac{150(0.3746065934)}{0.3090169944} \\[5ex] |ST| = 181.8378602 \\[3ex] |ST| \approx 181.8\;m \\[3ex] (iii) \\[3ex] \underline{\triangle TSF} \\[3ex] \sin \angle TSF = \dfrac{|FT|}{ST|}...SOHCAHTOA \\[5ex] \sin 40^\circ = \dfrac{|FT|}{181.8378602} \\[5ex] |FT| = 181.8378602 \sin 40 \\[3ex] |FT| = 181.8378602(0.6427876097) \\[3ex] |FT| = 116.8831235 \\[3ex] |FT| \approx 116.9\;m $(11.) CSEC A person at the top of a lighthouse, TB, sees two ships, S1 and S2, approaching the coast as illustrated in the diagram below. The angles of depression are 20° and 12° respectively. The ships are 110 m apart. (i) Complete the diagram below by inserting the angles of depression and the distance between the ships. (ii) Determine, to the nearest metre, (a) the distance, TS2, between the top of the lighthouse and Ship 2 (b) the height of the lighthouse, TB. (i) The angles of depression and the distance between the ships are shown in the diagram below:$ (ii) \\[3ex] \angle TS_1B = 20^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle TS_1S_2 + \angle TS_1B = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle TS_1S_2 + 20 = 180 \\[3ex] \angle TS_1S_2 = 180 - 20 \\[3ex] \angle TS_1S_2 = 160^\circ \\[3ex] \underline{\triangle TS_2S_1} \\[3ex] (a) \\[3ex] \dfrac{|TS_2|}{\sin\angle TS_1S_2} = \dfrac{|S_1S_2|}{\sin\angle S_1TS_2}...Sine\;\;Law \\[5ex] \dfrac{|TS_2|}{\sin 160^\circ} = \dfrac{110}{\sin 8^\circ} \\[5ex] |TS_2| = \dfrac{110\sin 160^\circ}{\sin 8^\circ} \\[5ex] |TS_2| = \dfrac{110(0.3420201433)}{0.139173101} \\[5ex] |TS_2| = 270.3267766\;m \\[3ex] |TS_2| \approx 270\;m \\[3ex] (b) \\[3ex] \angle TS_2B = 12^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \underline{\triangle TS_2B} \\[3ex] \sin \angle TS_2B = \dfrac{|TB|}{|TS_2|}...SOHCAHTOA \\[5ex] \sin 12^\circ = \dfrac{|TB|}{270.3267766} \\[5ex] |TB| = 270.3267766\sin 12 \\[3ex] |TB| = 270.3267766(0.2079116908) \\[3ex] |TB| = 56.2040972 \\[3ex] |TB| \approx 56\;m \$
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