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Elevation, depression & bearings · Lesson 1 of 3

Angles of elevation and depression

Measure from the horizontal, spot that elevation and depression are equal, and handle eye height, two angles, ladders and shadows.

20 minYou should already know: Trigonometric ratios
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These questions are right-angled triangles described in words: towers, cliffs, trees, ladders and aeroplanes. Once the diagram is right, one trig ratio finishes the job.

Two angles from the horizontal

  • The angle of elevation is the angle you look up through, measured from the horizontal.
  • The angle of depression is the angle you look down through, again measured from the horizontal.
Elevation and depressionDrag the observer
40 m45 m42°42°
41.6°angle of elevation41.6°angle of depression40 ÷ 45tan θ = height ÷ distance
The angle of elevation is measured up from the horizontal to the line of sight. From the top, the angle of depression is measured down from the horizontal. The two horizontals are parallel, so these are alternate angles and always equal. Then tan θ = 40 ÷ 45.
elevationdepressionhorizontal
Elevation and depressionBoth from the horizontal, and equal: alternate angles

Eye height

If the observer’s height is given, the line of sight starts at their eyes, not their feet. Take the eye height off the height of the object first.

eH − ed
Eye heighttan θ = (H − e) ÷ d

Worked example · WAEC 2023

WAEC 2023 · Paper 1 · Q24

Mrs Kebeh stands at a distance of 110 m110\text{ m} away from a building of vertical height 58 m58\text{ m}. If Kebeh is 2 m2\text{ m} tall, find the angle of elevation of the top of the building from her eye.

  1. Draw the horizontal at eye level

    Draw a horizontal line from Mrs Kebeh’s eyes, 2 m above the ground, across to the building. The angle of elevation sits between this line and her line of sight to the top.

    Think first. Her eyes are 2 m up. How far above her eyes is the top of the building?

  2. Find the sides of the triangle

    The side opposite the angle is 58−2=56 m58 - 2 = 56\text{ m}. The adjacent side is the horizontal distance, 110 m110\text{ m}.

  3. Use tan

    tan⁡θ=56110=0.5091⇒θ≈27∘\tan\theta = \frac{56}{110} = 0.5091 \quad\Rightarrow\quad \theta \approx 27^\circ

    The answer is A. (Using 58 instead of 56 gives 27.8∘≈28∘27.8^\circ \approx 28^\circ, which is option B: the trap.)

More: one right-angled triangle

Two angles, one unknown distance

A common theory question gives two angles of elevation from two points in a line, a known distance apart. There are two right-angled triangles sharing the same height hh.

  1. Call the unknown distance from the nearer point xx, so the farther point is x+dx + d away.
  2. Write hh from each triangle: h=xtan⁡αh = x\tan\alpha and h=(x+d)tan⁡βh = (x + d)\tan\beta.
  3. Set them equal and solve for xx, then find hh.
βαABdxh
Two triangles, one heighth = x tan α = (x + d) tan β

The other common version has one observer and two angles to the top and bottom of something, such as a flagpole on a building or a tower seen from an aeroplane. Both triangles share the horizontal distance, and the object’s height is the difference between the two vertical sides.

More: two angles

Ladders and shadows

A ladder against a wall makes a right-angled triangle, with the ladder as the hypotenuse. If it is LL long at angle θ\theta to the ground, it reaches Lsin⁡θL\sin\theta up the wall and its foot is Lcos⁡θL\cos\theta from the wall. When a ladder slips, its length doesn’t change: find the new height or distance, then the new angle.

θLL sin θL cos θdashed: slipped, same length L
A ladder against a wallHeight L sin θ, distance out L cos θ

A shadow works the same way. The sun’s angle of elevation θ\theta satisfies tan⁡θ=\tan\theta = (height of the object) ÷ (length of its shadow), so the lower the sun, the longer the shadow.

More: ladders and shadows

Your turn

WAEC 2024 · Paper 1 · Q44

A cliff on the bank of a river is 87 m87\text{ m} high. A boat on the river is 22 m22\text{ m} away from the cliff. Calculate, correct to the nearest degree, the angle of depression of the boat from the top of the cliff.

Worked solution (try it first)
  1. The angle of depression equals the angle of elevation of the cliff top from the boat (alternate angles).
  2. The height 87 m is opposite the angle and 22 m is adjacent: tan⁡θ=8722≈3.955\tan\theta = \frac{87}{22} \approx 3.955.
  3. So θ≈75.8∘\theta \approx 75.8^\circ, which is 76∘76^\circ to the nearest degree, option A.

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