5.04 Angles of elevation and depression

An angle of elevation is the angle from the lower object to the higher one, while an angle of depression is the angle from the higher object to the lower one. Both angles are measured with respect to the horizontal plane of the reference object.

The angle of elevation from point A to B is the angle between the horizontal line at A and the line connecting the two points.

The angle of depression from point B to A is the angle between the horizontal line at B and the line connecting the two points.

Notice that the angle of elevation between two points will always be equal to the angle of depression between those two points, since they are alternate angles on parallel lines (since all horizontal planes will be parallel).

Combining the angles of elevation or depression between two objects with trigonometry can help us to solve problems involving missing lengths or angles.

When given the angle of elevation or depression between two objects, we will always be able to model their relative position using a right-angled triangle. Using trigonometry, if we are given any side length of this triangle then we can solve for the other side lengths in the triangle.

Alternatively, there are three distances between two objects: horizontal distance, vertical distance and direct distance. These will represent the adjacent, opposite and hypotenuse sides respectively, and if any two are given then we can find the angle of elevation and depression.

Examples

Example 1

Find the angle of depression from point B to point D. Use x as the angle of depression and round your answer to two decimal places.

Worked Solution

Create a strategy

Identify where the angle of depression is and apply the appropriate trigonometric ratio.

Apply the idea

The angle of depression is the angle \angle CBD, with adjacent side BC=9 and hypotenuse BD=19. We can let x be the unknown angle.

\displaystyle \cos x	\displaystyle =	\displaystyle \dfrac{\text{Adjacent}}{\text{Hypotenuse}}	Use the cosine ratio
\displaystyle \cos x	\displaystyle =	\displaystyle \dfrac{9}{19}	Subsitute the values
\displaystyle x	\displaystyle =	\displaystyle \cos^{-1}\left(\dfrac{9}{19}\right)	Apply inverse cosine on both sides
	\displaystyle \approx	\displaystyle 61.73\degree	Evaluate using a calculator

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Example 2

A fighter jet, flying at an altitude of 4000 m is approaching a target. At a particular time the pilot measures the angle of depression to the target to be 13\degree. After a minute, the pilot measures the angle of depression again and finds it to be 16\degree.

a

Find the distance AC, to the nearest metre.

Worked Solution

Create a strategy

Use the trigonometric ratio of tangent.

Apply the idea

With respect to the given angle 13 \degree, the opposite side is 4000, and the adjacent side is AC, so we can use the tangent ratio.

\displaystyle \tan \theta	\displaystyle =	\displaystyle \frac{\text{Opposite }}{\text{Adjacent}}	Use the tangent ratio
\displaystyle \tan {13\degree}	\displaystyle =	\displaystyle \frac{4000}{AC}	Substitute the values
\displaystyle AC \times \tan {13\degree}	\displaystyle =	\displaystyle 4000	Multiply both sides by AC
\displaystyle AC	\displaystyle =	\displaystyle \frac{4000}{\tan {13\degree}}	Divide both sides by \tan {13\degree}
	\displaystyle \approx	\displaystyle 17\,326 \text{ m}	Evaluate using a calculator

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b

Find the distance BC, to the nearest metre.

Worked Solution

Create a strategy

Use the same method that we used to find AC in part (a).

Apply the idea

With respect to the given angle 16 \degree, the opposite side is 4000, and the adjacent side is BC, so we can use the tangent ratio.

\displaystyle \tan \theta	\displaystyle =	\displaystyle \frac{\text{Opposite }}{\text{Adjacent}}	Use the tangent ratio
\displaystyle \tan {16\degree}	\displaystyle =	\displaystyle \frac{4000}{BC}	Substitute the values
\displaystyle BC \times \tan {16\degree}	\displaystyle =	\displaystyle 4000	Multiply both sides by BC
\displaystyle BC	\displaystyle =	\displaystyle \frac{4000}{\tan {16\degree}}	Divide both sides by \tan {16\degree}
	\displaystyle \approx	\displaystyle 13\,950 \text{ m}	Evaluate using a calculator

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c

Now find the distance covered by the jet in one minute to the nearest metre.

Worked Solution

Create a strategy

Find the difference between the length of AC and BC.

Apply the idea

The distance covered by the jet in one minute is given by the length of AB.

\displaystyle AB	\displaystyle =	\displaystyle 17\,326-13\,950	Subtract BC from AC
	\displaystyle =	\displaystyle 3376 \text{ m}	Simplify the difference

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