Angles of Elevation and Depression

Worked example

1. 1

   $\tan 34^{\circ }=\frac{h}{85}$ — The height $h$ is opposite the angle of elevation and the 85 m along the ground is next to it, so use tan.

2. 2

   $h=85\times \tan 34^{\circ }$ — $h$ is the unknown on top of the fraction, so multiply both sides by 85.

3. 3

   $h=57.33\dots$ — Work it out with your calculator in degree mode.

4. 4

   $h=57.3\text{ m}$ — Round to 3 significant figures and give the unit.

Worked example

1. 1

   $21.3-6.5=14.8$ — The triangle only goes from the top of the short mast up to the top of the tall one, so use the difference in heights.

2. 2

   $\tan x=\frac{14.8}{60}$ — 14.8 is opposite the angle and the 60 m gap is next to it, so use tan.

3. 3

   $\tan x=0.2467\dots$ — Divide 14.8 by 60 to get the ratio.

4. 4

   $x={\tan }^{-1}(0.2467\dots )$ — You want the angle, not a side, so use inverse tan.

5. 5

   $x=13.86\dots =13.9^{\circ }$ — Round to 1 decimal place.

Worked example

1. 1

   $\tan 21^{\circ }=\frac{48}{d}$ — From the boat, the cliff top is at 21°. The cliff height 48 is opposite, and the distance $d$ along the sea is next to the angle, so use tan.

2. 2

   $d=48÷\tan 21^{\circ }=125.04$ — $d$ is the unknown on the bottom, so divide.

3. 3

   $\tan 26^{\circ }=\frac{\text{top}}{125.04}$ — The mast top is seen from the same boat, so the same distance $d$ is next to the angle. The height up to the mast top is opposite 26°.

4. 4

   $\text{top}=125.04\times \tan 26^{\circ }=60.99$ — The height is the unknown on top, so multiply. This is measured from the sea.

5. 5

   $60.99-48=13.0\text{ m}$ — The mast stands on the cliff, so take off the cliff height to leave the mast alone. Round to 3 significant figures.

Worked example

1. 1

   $\text{angle at the boat}=18^{\circ }$ — The angle of depression from the top equals the angle of elevation from the boat, because they are alternate angles. So 18° is inside the triangle, at the boat.

2. 2

   $\sin 18^{\circ }=\frac{h}{120}$ — The height $h$ is opposite the 18° angle and the 120 m line of sight is the hypotenuse, so use sin.

3. 3

   $h=120\times \sin 18^{\circ }=37.1\text{ m}$ — $h$ is the unknown on top, so multiply. Round to 3 significant figures.

4. 4

   $\cos 18^{\circ }=\frac{d}{120}$ — The distance $d$ along the ground is next to the 18° angle and 120 is the hypotenuse, so use cos.

5. 5

   $d=120\times \cos 18^{\circ }=114\text{ m}$ — Multiply again, since $d$ is on top. Round to 3 significant figures.

6. 6

   $\sqrt{37.08^2+114.12^2}=120$ — Check with Pythagoras: the two answers should make a triangle with hypotenuse 120.