2D Applications
2D applications use right-angled triangles to answer practical questions, often with an angle of elevation, measured up from the horizontal, or an angle of depression, measured down from it. In NSW Year 10 Advanced maths the problems take several steps, can combine two triangles, and include slopes given as a ratio such as 1 in \(n\) (NESA outcome MA5‑TRG‑C‑02).
You will find heights and distances from angles of elevation and depression, including how far an observer moves when the angle changes. Problems include a snapped post, a netball shot, headlights on a road, a crane's jib, a driveway and a wheelchair ramp checked against rules, a cable car, a dam wall, a plane's speed and the sun's angle on a window eave. Harder questions use two angles to find the height of a cliff. It is part of Year 10 Advanced Trigonometry.
Every question with a fully worked solution.
- 2D Applications - Video - Angles of Elevation and Depression Watch
Theory
Real problems hide a right-angled triangle inside a picture of a cliff, a ramp, a crane or a shadow. Draw the triangle, mark the angle of elevation or depression, then use sine, cosine or tangent. This Year 10 Advanced maths page covers slopes written as \(1\) in \(n\), heights measured from eye level or a pivot, two observation points, speeds and problems where you add or subtract two lengths.
The angle of elevation is the angle you look up from the horizontal. The angle of depression is the angle you look down from the horizontal. Both are always measured from a horizontal line, never from a vertical wall.
The angle of depression from \(A\) to \(B\) equals the angle of elevation from \(B\) to \(A\). The two horizontals are parallel, so these are alternate angles.
A slope of \(1\) in \(n\) rises \(1\) unit for every \(n\) units measured horizontally, so \(\tan\theta = \dfrac{1}{n}\). A distance along a slope, ramp, cable or rope is a hypotenuse.
The sun's angle of elevation sets the length of a shadow: a vertical object of height \(h\) casts a shadow of length \(\dfrac{h}{\tan\theta}\) on level ground.
With angle \(\theta\), height \(h\) opposite and horizontal distance \(d\) adjacent:
How to solve a 2D trigonometry problem
- Sketch the situation. Draw the horizontal, the vertical and the line of sight, ramp or cable.
- Mark the angle in the right place. Elevation and depression are measured from the horizontal; use alternate angles to move a depression angle to the ground.
- Adjust heights to the level of the angle: subtract an eye height, or find the height above a pivot and add the pivot height later.
- Solve the right-angled triangle with sine, cosine or tangent.
- Answer the question asked: a total, a difference, a speed or a yes/no comparison with a rule.
Common set-ups
- Two positions (an observer moves, a rope is let out, a shadow grows): find the length for each position, then subtract. Never subtract the angles first.
- Angle between two lines of sight (e.g. a billboard): find each angle from eye level, then subtract.
- Two unknowns, two triangles: write one equation from each triangle, such as \(h = d\tan\alpha\) and \(h - 20 = d\tan\beta\), then subtract to remove \(h\) and solve for \(d\).
- Trapezium cross-sections (dams, channels): split with vertical heights into a rectangle and right-angled triangles. A V-shape splits along its centre line.
- Rules and limits: find the angle of a "\(1\) in \(n\)" rule or the run at the steepest allowed angle, then compare. A steeper ramp needs less horizontal space.
- Turning objects (clock hands, cranes): a minute hand turns \(6^\circ\) per minute. Find the angle with the horizontal, then use \(r\sin\theta\) for the height above the centre.
- Shadows and eaves: the sun's ray past the top of an object or the tip of an eave makes the angle of elevation with the horizontal.
Angle of depression at the balcony \(=\) angle of elevation at the car (alternate angles). At the car, \(18.5\) is opposite and \(d\) is adjacent.
| \(\tan 23^\circ 40'\) | \(=\) | \(\dfrac{18.5}{d}\) |
| \(d\) | \(=\) | \(\dfrac{18.5}{\tan 23^\circ 40'}\) |
| \(\) | \(=\) | \(42.2108\ldots\) |
| \(\therefore\ d\) | \(\approx\) | \(\mathbf{42.2}\text{ m}\) |
Put the angle where it belongs: the angle of depression is not the angle at the top of the building between the wall and the line of sight.
Find the boat's distance from the cliff at each angle, subtract, then divide by the time.
| \(FP\) | \(=\) | \(\dfrac{60}{\tan 14^\circ} = 240.64\ldots\) |
| \(FQ\) | \(=\) | \(\dfrac{60}{\tan 25^\circ} = 128.67\ldots\) |
| \(x\) | \(=\) | \(240.64\ldots - 128.67\ldots\) |
| \(\) | \(=\) | \(111.97\ldots\text{ m}\) |
Divide by the \(40\) seconds, then multiply m/s by \(3.6\) to get km/h.
| \(\text{speed}\) | \(=\) | \(\dfrac{111.97\ldots}{40}\text{ m/s}\) |
| \(\) | \(=\) | \(2.7994\ldots \times 3.6\text{ km/h}\) |
| \(\) | \(\approx\) | \(\mathbf{10.1}\text{ km/h}\) |
Both angles are measured from the ground at \(O\). Find the height of each point, then subtract.
| \(\text{building}\) | \(=\) | \(40\tan 31^\circ = 24.034\ldots\) |
| \(\text{top of pole}\) | \(=\) | \(40\tan 37^\circ = 30.142\ldots\) |
| \(p\) | \(=\) | \(30.142\ldots - 24.034\ldots\) |
| \(\) | \(=\) | \(6.107\ldots\) |
| \(\therefore\ p\) | \(\approx\) | \(\mathbf{6.1}\text{ m}\) |
The \(6^\circ\) between the two lines of sight is not in a right-angled triangle, so it cannot give \(p\) directly.
(i) Find the angle the road makes with the horizontal, correct to the nearest minute.
(ii) A cyclist rides \(600\) m along the road. How much height does she gain, correct to the nearest metre?
(i) \(1\) in \(8\) is a rise of \(1\) for a horizontal run of \(8\): opposite over adjacent.
| \(\tan\theta\) | \(=\) | \(\dfrac{1}{8}\) |
| \(\theta\) | \(=\) | \(7.1250\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{7^\circ 8'}\) |
(ii) The \(600\) m is measured along the road, so it is the hypotenuse. The height gained is opposite \(\theta\).
| \(h\) | \(=\) | \(600\sin\theta\) |
| \(\) | \(=\) | \(600\sin 7.1250\ldots^\circ\) |
| \(\) | \(=\) | \(74.420\ldots\) |
| \(\therefore\ h\) | \(\approx\) | \(\mathbf{74}\text{ m}\) |
Use the stored angle, not \(7^\circ 8'\) typed back in. Note \(600 \div 8 = 75\) is the rise for \(600\) m measured horizontally, which is not what the question gives.
Common pitfalls
Frequently asked questions
What is the difference between angle of elevation and angle of depression?
The angle of elevation is measured upwards from the horizontal to an object above you. The angle of depression is measured downwards from the horizontal to an object below you. From two points looking at each other, the two angles are equal.
Why is the angle of depression equal to the angle of elevation?
The horizontal lines at the two points are parallel and the line of sight crosses both of them, so the angles are alternate angles, which are equal.
How do you find the height of a building using trigonometry?
Stand a known distance \(d\) from the building and measure the angle of elevation \(\theta\) of the top. The height above your eye is \(d\tan\theta\). Add your eye height to get the height above the ground.
What does a gradient of 1 in 20 mean?
It means the slope rises \(1\) unit for every \(20\) units measured horizontally. The angle with the horizontal is \(\tan^{-1}\left(\dfrac{1}{20}\right) \approx 2.9^\circ\).
How do you find speed in a trigonometry question?
Use trigonometry to find the distance travelled, usually by finding two distances and subtracting. Divide by the time taken. To change metres per second to kilometres per hour, multiply by 3.6.