Trigonometry (SOHCAHTOA): O-Level / IGCSE Maths (0580)

Syllabus C6.2, E6.2 · Strand 6 Trigonometry

Questions
3
Total marks
13
Tier mix
1 Core · 2 Extended

This topic is about working with right-angled triangles: given some sides and angles, you use trigonometry to find the ones you do not yet know. The three ratios are sine, cosine and tangent, and they connect an acute angle to two of the triangle’s sides. Relative to a chosen angle θ\theta, the sides are named the opposite, the adjacent, and the hypotenuse (the longest side, always facing the right angle). The mnemonic SOHCAHTOA records the three relationships compactly: sinθ=opphyp\sin\theta = \dfrac{\text{opp}}{\text{hyp}}, cosθ=adjhyp\cos\theta = \dfrac{\text{adj}}{\text{hyp}}, and tanθ=oppadj\tan\theta = \dfrac{\text{opp}}{\text{adj}}.

The method has two main directions. To find an unknown side, label the sides relative to the known angle, pick the ratio that uses the side you want and the side you already have, then rearrange and evaluate, for example x=12sin35x = 12\sin 35^\circ. To find an unknown angle, form the correct ratio from two known sides and apply the inverse function, such as θ=tan1 ⁣(oppadj)\theta = \tan^{-1}\!\left(\dfrac{\text{opp}}{\text{adj}}\right). Make sure your calculator is in degree mode, and keep enough accuracy in intermediate steps before rounding the final answer.

A common application is angles of elevation and depression: the angle measured up from the horizontal to an object above you (elevation), or down from the horizontal to an object below you (depression). These problems are solved by sketching the right-angled triangle hidden in the situation and choosing the appropriate ratio. The worked examples below are original, written from the syllabus objective to show these exam-style techniques, with each worked solution laid out step by step.

Question 1

Structured Core 4 marks

A surveyor is checking the height of a vertical communication tower TFTF that stands on level horizontal ground.

She stands at a point PP on the ground, 28 m28\text{ m} from the foot FF of the tower. From PP, the angle of elevation of the top TT of the tower is 3737^\circ.

Because the tower TFTF is vertical and PP and FF are on the same horizontal level, triangle PFTPFT is right-angled at FF.

(a) Calculate the height of the tower, TFTF. Give your answer correct to 1 decimal place. [2]

(b) Calculate the distance PTPT from the surveyor to the top of the tower. Give your answer correct to the nearest metre. [2]

Question 2

Structured Extended 6 marks

A vertical lighthouse stands on a horizontal sea wall. The lamp room window WW is 48 m48\text{ m} vertically above sea level. A lighthouse keeper at WW watches two fishing boats, AA and BB, which are on the sea on the same side of the lighthouse, in line with its base FF.

Boat AA is at a horizontal distance of 95 m95\text{ m} from the foot of the lighthouse FF. Boat BB is closer in, at a horizontal distance of 60 m60\text{ m} from FF.

(All distances are measured in a vertical plane. You may assume the line FWFW is vertical and the sea surface through FF, BB, AA is horizontal.)

(a) Calculate the angle of depression of boat AA from the window WW. Give your answer correct to 11 decimal place. [3]

(b) Calculate the angle of depression of boat BB from the window WW. Give your answer correct to 11 decimal place. [2]

(c) Hence find how many degrees larger the angle of depression of BB is than that of AA. [1]

Question 3

Multiple choice Extended 3 marks

A surveyor stands on level ground at point PP, which is 24 m24\text{ m} from the base BB of a vertical flagpole. From PP the angle of elevation to the top TT of the flagpole is 3737^\circ. Triangle PBTPBT is right-angled at BB. Calculate the height BTBT of the flagpole, giving your answer correct to 33 significant figures.