Angles of elevation and depression

If a person stands and looks up at an object, the angle of elevation is the angle between the horizontal line of sight and the object.

Person with line of sight, object, angle of elevation and horizontal labelled

If a person stands and looks down at an object, the angle of depression is the angle between the horizontal line of sight and the object.

Person with object, angle of depression and horizontal labelled

Trigonometry can be used to solve problems that use an angle of elevation or depression.

Example

A man is 1.8 m tall.

He stands 50 m away from the base of a building. His angle of elevation to the top of the building is 70°.

Calculate an estimate of the height of the building. Give your answer to an appropriate degree of accuracy.

Triangle at 70degrees 50m from a skyscraper

Label the sides of the triangle o, a and h.

Next choose the correct ratio from s^o_h~c^a_h~t^o_a.

In the triangle the length a is known and the length o must be calculated.

Use \tan{x} = \frac{o}{a}

\tan{70} = \frac{z}{50}

Make z the subject by multiplying both sides by 50.

z = 50 \times \tan{70}

z = 137.4~\text{m}

Assume the man’s eye-level is (approximately) 1.7 m above the ground.

The building is roughly 1.7 + 137.4 = 139.1~m tall.

That is far too precise (being to the nearest 10 cm). An estimate to an appropriate degree of accuracy would be 140 m. (Note this is also about the same degree of accuracy as the 50 m and 70° that are used in the question).

Question

From the top of a 72 m high vertical cliff, a boat has an angle of depression of 32°. How far is the boat from the base of the cliff? Give your answer to an appropriate degree of accuracy. State any assumptions you have made in your calculation.

Triangle showing distance of yacht from coastline

In the triangle the length o is known and the length a must be calculated.

Use \tan{x} = \frac{o}{a}

\tan{32} = \frac{72}{y}

Rearrange the equation to make y the subject.

Multiply both sides by y.

y \times \tan{32} = 72

Divide both sides by \tan{32}.

y = \frac{72}{\tan{32}}

y = 115.2~\text{m}

The boat is 115.2 m from the base of the cliff.

115.2 m is far too precise (being to the nearest 10 cm). 110 m or 120 m are appropriate answers. (Note these are about the same degree of accuracy as the 72 m and 32° used in the question.)

There are various assumptions that you may have made.

For example:

Is the distance to be found to the middle of the boat or to the end of the boat which is nearest the cliff?

Was the angle of depression measured to deck level or sea level? (This affects whether 72 m or something a little less is used in the calculation).