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Year 10AC9M10M03

Angles of Elevation and Depression

Solve practical problems applying Pythagoras' theorem and trigonometry of right-angled triangles, including problems involving direction and angles of elevation and depression.

Almost every practical trigonometry question is the same question wearing different clothes. Draw the triangle, label what you know, choose the ratio. The words change; the method does not.

Builds onYr 9 · The Basics

Method

The three ratios
sin⁡θ=OHcos⁡θ=AHtan⁡θ=OA\sin\theta = \frac{O}{H} \qquad \cos\theta = \frac{A}{H} \qquad \tan\theta = \frac{O}{A}
SOH CAH TOA. Label the sides from the angle you are using, not from the page.
Elevation looks up, depression looks down
Both are measured from the horizontal, never from the vertical. This is the most common setup error.
They are equal
elevation=depression\text{elevation} = \text{depression}
The angle of depression from the top equals the angle of elevation from the bottom. Alternate angles on parallel horizontals.
Add the eye height last
If the observer's height is given, the triangle sits on their eye line. Add it back at the end, or you lose the final mark.

Worked example

e.g. From a point 40 m from the base of a tower, the angle of elevation to the top is 32∘\text{From a point } 40 \text{ m from the base of a tower, the angle of elevation to the top is } 32^\circ
  1. Draw the triangle and mark the right angle. The ground and the tower meet at 90∘90^\circ.
    Triangle drawn to scale32°?40 mThe angle of elevation is measured from the horizontal ground
  2. Write down what you have and what you want, from the angle given.A=40,O=?,θ=32∘A = 40,\quad O = ?,\quad \theta = 32^\circ
  3. Choose the ratio that uses only those two sides.tan⁡θ=OA\tan\theta = \frac{O}{A}
  4. Substitute.tan⁡32∘=h40\tan 32^\circ = \frac{h}{40}
  5. Solve. Multiply up before you reach for the calculator.h=40tan⁡32∘=24,997…h = 40\tan 32^\circ = 24{,}997\ldots
  6. Round sensibly and write the unit.h≈25,0 mh \approx 25{,}0 \text{ m}
    Triangle drawn to scale32°25,0 m40 mThe same triangle with the height now known
  7. If an eye height was given, add it now.tower=25,0+1,6=26,6 m\text{tower} = 25{,}0 + 1{,}6 = 26{,}6 \text{ m}

Practice

Draw the triangle every time, even when you think you can do it in your head.

1
A kite string is 60 m long at an angle of elevation of 48∘. Find the height.\text{A kite string is } 60 \text{ m long at an angle of elevation of } 48^\circ. \text{ Find the height.}
Answersin⁡48∘=h60∴h≈44,6 m\sin 48^\circ = \frac{h}{60} \quad\therefore\quad h \approx 44{,}6 \text{ m}
2
From a cliff 80 m high, the angle of depression to a boat is 25∘. How far is the boat from the cliff base?\text{From a cliff } 80 \text{ m high, the angle of depression to a boat is } 25^\circ. \text{ How far is the boat from the cliff base?}
Answertan⁡25∘=80d∴d≈171,6 m\tan 25^\circ = \frac{80}{d} \quad\therefore\quad d \approx 171{,}6 \text{ m}
3
A ramp rises 1,2 m over a horizontal distance of 7 m. Find the angle of elevation.\text{A ramp rises } 1{,}2 \text{ m over a horizontal distance of } 7 \text{ m. Find the angle of elevation.}
Answertan⁡θ=1,27∴θ≈9,7∘\tan\theta = \frac{1{,}2}{7} \quad\therefore\quad \theta \approx 9{,}7^\circ
4
A 1,7 m tall observer 30 m from a flagpole measures an elevation of 38∘. Find the pole height.\text{A } 1{,}7 \text{ m tall observer } 30 \text{ m from a flagpole measures an elevation of } 38^\circ. \text{ Find the pole height.}
Answer30tan⁡38∘=23,4∴23,4+1,7=25,1 m30\tan 38^\circ = 23{,}4 \quad\therefore\quad 23{,}4 + 1{,}7 = 25{,}1 \text{ m}
5
Why does the angle of depression from a tower equal the angle of elevation from the ground?\text{Why does the angle of depression from a tower equal the angle of elevation from the ground?}
AnswerThe two horizontals are parallel, so the angles are alternate angles.\text{The two horizontals are parallel, so the angles are alternate angles.}
6
A plane at 900 m sees the runway at a depression of 12∘. Find the ground distance.\text{A plane at } 900 \text{ m sees the runway at a depression of } 12^\circ. \text{ Find the ground distance.}
Answertan⁡12∘=900d∴d≈4234 m\tan 12^\circ = \frac{900}{d} \quad\therefore\quad d \approx 4234 \text{ m}
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