Angles of Elevation and Depression
An angle of elevation is measured up from the horizontal to an object above, and an angle of depression is measured down from the horizontal to an object below. This NSW Year 9 Advanced maths subtopic uses both in right-angled triangle problems (NESA outcome MA5‑TRG‑C‑02).
You will find heights and distances for flagpoles, cliffs, towers, kites and balloons, and allow for the observer's eye height. Two-triangle problems include cars on opposite sides of a spire, a boat sailing towards a lighthouse, a flagpole on a roof and a tower seen from two points. You will also find when a descending balloon reaches a given angle. It is part of Year 9 Advanced Trigonometry.
Theory
An angle of elevation is measured up from the horizontal; an angle of depression is measured down from it. This Year 9 Advanced maths page covers why the two are equal, eye height, angles to the nearest minute, two-triangle problems and finding speeds from changing angles.
The line of sight is the straight line from an observer's eye to an object. The angle of elevation is the angle between the horizontal and the line of sight when looking up. The angle of depression is the angle between the horizontal and the line of sight when looking down.
Both angles are measured from the horizontal, never from the vertical. Draw a dashed horizontal line at the observer before you mark an angle of depression.
The angle of depression from \(A\) to \(B\) equals the angle of elevation from \(B\) to \(A\). The horizontals at \(A\) and \(B\) are parallel, so the two angles are alternate angles. This lets you put the angle inside the right-angled triangle at ground level.
If the angle is measured from an observer's eyes or from a window, the right-angled triangle starts at that height. Find the part above (or below) it, then add or subtract the starting height.
When the angle is unknown, use an inverse ratio: \(\theta\) \(= \tan^{-1}\left(\dfrac{O}{A}\right)\). Give it in degrees and minutes if asked.
| You know | You want | Use |
|---|---|---|
| angle, distance along the ground | height | \(h = d\tan\theta\) |
| angle, height | distance along the ground | \(d = \dfrac{h}{\tan\theta}\) |
| angle, height | length of a string or line of sight | \(L = \dfrac{h}{\sin\theta}\) |
| height, distance | angle | \(\theta\) \(= \tan^{-1}\left(\dfrac{h}{d}\right)\) |
How to solve an elevation or depression problem
- Draw the horizontal ground, the vertical object and the line of sight. Mark the observer.
- Mark the angle from the horizontal. For a depression, move it to ground level using alternate angles.
- Find the right-angled triangle. Adjust for eye height, a window or a platform: work with the difference in heights.
- Choose the ratio and solve. Use the inverse ratio for an unknown angle.
- Answer the question asked: add back an eye height, subtract two heights, or turn a distance into a speed.
Common set-ups
- Two objects on opposite sides of a tower: find each ground distance and add.
- A boat or car approaching a cliff: the two distances are on the same side; subtract. Divide by the time for a speed.
- Two angles from one point (bottom and top of a flagpole on a roof): find both heights and subtract.
- A balloon rising or falling: find the height at each angle; the change divided by the rate gives the time.
- Two buildings: the angle of depression between their tops uses the difference in their heights.
- Walking towards a tower: write the height in two ways and solve for the unknown distance (Example 3).
- Show-that formulas: keep \(h\), \(\alpha\) and \(\beta\) as letters, write each distance as \(\dfrac{h}{\tan\alpha}\), then combine.
(i) How far is \(P\) from the foot of the cliff, to the nearest metre?
(ii) Buoy \(Q\) is \(150\) m further out, in line with \(P\). Find the angle of depression of \(Q\), to the nearest minute.
(i) The angle of depression equals the angle of elevation at \(P\) (alternate angles). The height is opposite and \(d\) is adjacent.
| \(\tan 17^\circ\) | \(=\) | \(\dfrac{64}{d}\) |
| \(d\) | \(=\) | \(\dfrac{64}{\tan 17^\circ} = 209.3345\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{209}\text{ m}\) |
(ii) Use the unrounded distance. \(Q\) is \(d + 150\) m from the foot.
| \(\tan\theta\) | \(=\) | \(\dfrac{64}{209.3345\ldots + 150}\) |
| \(\theta\) | \(=\) | \(\tan^{-1}\left(\dfrac{64}{359.3345\ldots}\right)\) |
| \(\) | \(=\) | \(10.0988\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{10^\circ 6'}\) |
To get minutes, use the degrees-minutes key, or multiply the decimal part by \(60\). The further buoy has the smaller angle, as expected.
The horizontal from \(W\) splits the tower into two right-angled triangles. The lower one has the known \(18\) m, so it gives the street width \(w\).
| \(\tan 15^\circ\) | \(=\) | \(\dfrac{18}{w}\) |
| \(w\) | \(=\) | \(\dfrac{18}{\tan 15^\circ} = 67.1769\ldots\) |
| \(\text{above } W\) | \(=\) | \(w\tan 27^\circ = 34.2283\ldots\) |
| \(H\) | \(=\) | \(34.2283\ldots + 18\) |
| \(\) | \(\approx\) | \(\mathbf{52.2}\text{ m}\) |
The \(18\) m below the window is added back at the end.
Neither triangle has two known sides. Write \(h\) in two ways and set them equal.
| \(h\) | \(=\) | \(d\tan 35^\circ\) |
| \(h\) | \(=\) | \((d + 80)\tan 21^\circ\) |
| \(d\tan 35^\circ - d\tan 21^\circ\) | \(=\) | \(80\tan 21^\circ\) |
| \(d\) | \(=\) | \(\dfrac{80\tan 21^\circ}{\tan 35^\circ - \tan 21^\circ} = 97.0752\ldots\) |
| \(h\) | \(=\) | \(d\tan 35^\circ = 67.9728\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{68.0}\text{ m}\) |
Check with the other triangle: \((97.08 + 80)\tan 21^\circ \approx 68.0\).
Find the horizontal distance to the plane each time (height opposite, distance adjacent), then subtract.
| \(x_1\) | \(=\) | \(\dfrac{1500}{\tan 48^\circ} = 1350.6060\ldots\) |
| \(x_2\) | \(=\) | \(\dfrac{1500}{\tan 26^\circ} = 3075.4557\ldots\) |
| \(x_2 - x_1\) | \(=\) | \(1724.8496\ldots\text{ m}\) |
| \(\text{speed}\) | \(=\) | \(\dfrac{1724.8\ldots}{20} = 86.2424\ldots\text{ m/s}\) |
| \(\) | \(=\) | \(86.2424\ldots \times 3.6\text{ km/h}\) |
| \(\) | \(\approx\) | \(\mathbf{310}\text{ km/h}\) |
\(1\) m/s \(= 3.6\) km/h, because \(3600\) s \(= 1\) h and \(1000\) m \(= 1\) km.
Common pitfalls
Frequently asked questions
What is an angle of elevation?
It is the angle measured up from the horizontal to the line of sight when you look up at an object, such as the top of a tree or a building.
What is an angle of depression?
It is the angle measured down from the horizontal to the line of sight when you look down at an object, such as a boat seen from a cliff top.
Why is the angle of depression equal to the angle of elevation?
The horizontal lines at the two ends are parallel, and the line of sight crosses both. The angle of depression at the top and the angle of elevation at the bottom are alternate angles, so they are equal.
How do you include a person's eye height?
Solve the right-angled triangle from eye level to get the height above the eyes, then add the eye height to get the full height of the object.
How do you find a height when the distance is unknown?
Use two angles of elevation from two points a known distance apart. Write the height in terms of the unknown distance in each triangle, set the two expressions equal and solve.