Mixed Right-Angled Triangle Problems
Pull all your right-angled triangle skills together for NSW Year 12 Mathematics Standard 1 (NESA outcome MST-12-S1-04). Each problem needs one of three tools: Pythagoras' theorem, a trigonometric ratio to find a side, or a trigonometric ratio with the inverse key to find an angle.
You will sketch and label the triangle, decide which tool suits the information you have, and solve real problems with angles of elevation and depression, as well as true or compass bearings — from ladders and shadows to lookouts and hiking trips.
Theory
Every right-angled triangle problem is solved with one of three tools: Pythagoras’ theorem, SOH CAH TOA for a side, or SOH CAH TOA with the inverse for an angle. This Year 12 Standard 1 (NSW) guide shows how to choose the right one.
Start every problem by drawing (or copying) the triangle, marking the right angle, and writing down what is given and what you want. The hypotenuse is opposite the right angle and is always the longest side.
If you know two sides and want the third, and no angle is involved, use Pythagoras: \(c^2=a^2+b^2\). If you know one angle and one side and want a side, use \(\sin\), \(\cos\) or \(\tan\). If you know two sides and want an angle, use the ratio and then the inverse key.
Elevation, depression and bearings questions use the same triangles. Angles of elevation and depression are measured from the horizontal. The angle of depression from the top equals the angle of elevation from the bottom, because the two horizontals are parallel (alternate angles), so the angle can be put inside the triangle at ground level.
True bearings are measured clockwise from north and written with three figures (e.g. \(050^\circ\text{T}\)); a compass bearing such as \(\text{N}40^\circ\text{E}\) is \(040^\circ\text{T}\), and \(\text{S}30^\circ\text{W}\) is \(180^\circ+30^\circ=210^\circ\text{T}\). The back bearing (the return direction) differs by \(180^\circ\): add \(180^\circ\) if the bearing is less than \(180^\circ\), otherwise subtract \(180^\circ\).
Pythagoras (no angle involved):
SOH CAH TOA (an angle is involved):
Inverse (finding an angle):
How to solve any right-angled triangle problem
- Draw the triangle, mark the right angle, and label what is given and what you want.
- Choose the tool: Pythagoras (no angle), SOH CAH TOA (a side), or SOH CAH TOA with the inverse (an angle).
- Solve, keeping full calculator values until the end.
- Round and check: the hypotenuse is the longest side and an acute angle is between \(0^\circ\) and \(90^\circ\).
Sketch the triangle. No angle is involved and the sloping surface is the hypotenuse, so use Pythagoras and subtract.
| \(h^2\) | \(=\) | \(4.5^2-4.2^2\) |
| \(h^2\) | \(=\) | \(20.25-17.64=2.61\) |
| \(h\) | \(=\) | \(\sqrt{2.61}=1.61\ldots\) |
| \(h\) | \(\approx\) | \(1.6\text{ m}\) |
The ramp rises about \(1.6\text{ m}\).
Sketch the triangle. \(6.5\text{ m}\) is opposite \(28^\circ\) and the cable is the hypotenuse, so use sine. \(x\) is underneath, so divide.
| \(\sin 28^\circ\) | \(=\) | \(\dfrac{6.5}{x}\) |
| \(x\) | \(=\) | \(\dfrac{6.5}{\sin 28^\circ}\) |
| \(x\) | \(=\) | \(13.84\ldots\) |
| \(x\) | \(\approx\) | \(13.8\text{ m}\) |
The cable is about \(13.8\text{ m}\) long.
Angle of depression. From the top of a \(24\text{ m}\) building the angle of depression of a bus stop is \(38^\circ\). The angle of elevation from the stop is also \(38^\circ\) (alternate angles), and \(d\) is underneath, so divide.
| \(\tan 38^\circ\) | \(=\) | \(\dfrac{24}{d}\) |
| \(d\) | \(=\) | \(\dfrac{24}{\tan 38^\circ}=30.71\ldots\) |
| \(d\) | \(\approx\) | \(31\text{ m}\) |
The stop is about \(31\text{ m}\) from the building.
Sketch the triangle. Two sides are known and an angle is wanted: the post is opposite and the shadow is adjacent, so use \(\tan^{-1}\).
| \(\tan\theta\) | \(=\) | \(\dfrac{3.6}{5.2}\) |
| \(\theta\) | \(=\) | \(\tan^{-1}\!\left(\dfrac{3.6}{5.2}\right)=34.69\ldots^\circ\) |
| \(0.69\ldots^\circ\) | \(=\) | \(0.69\ldots\times 60'\approx 42'\) |
(i) \(35^\circ\) (ii) \(34^\circ 42'\)
Sketch the trip. Use Pythagoras for the distance; for the bearing, find the angle west of north with \(\tan^{-1}\).
| \(d\) | \(=\) | \(\sqrt{4^2+9^2}=\sqrt{97}=9.84\ldots\) |
| \(\tan\theta\) | \(=\) | \(\dfrac{4}{9}\) |
| \(\theta\) | \(=\) | \(23.96\ldots^\circ\approx 24^\circ\) |
| \(\text{bearing}\) | \(=\) | \(360^\circ-24^\circ=336^\circ\) |
| \(\text{back bearing}\) | \(=\) | \(336^\circ-180^\circ=156^\circ\text{T}\) |
(i) \(9.8\text{ km}\) (ii) \(336^\circ\text{T}\) (iii) \(156^\circ\text{T}\)
Common pitfalls
Frequently asked questions
How do I know whether to use Pythagoras or trigonometry?
If no angle is given or wanted, and you know two sides, use Pythagoras. If an angle is given or wanted, use sine, cosine or tangent.
How do you find a shorter side using Pythagoras?
Square the hypotenuse, subtract the square of the known shorter side, then take the square root. For example, the square root of 17 squared minus 8 squared is 15.
When do I multiply and when do I divide in trigonometry?
Write the ratio as an equation. If the unknown side is on top of the fraction, multiply both sides by the bottom number. If the unknown is underneath, divide the known side by the sine, cosine or tangent of the angle.
How do I find the angle of elevation of the sun from a shadow?
The object is opposite the angle and the shadow is adjacent. Use tan inverse of the height divided by the shadow length.
How do I use an angle of depression?
The angle of depression from the top equals the angle of elevation from the bottom (alternate angles). For example, from the top of a 24 m building with an angle of depression of 38 degrees, tan 38 degrees = 24 / d, so d = 24 / tan 38 degrees, which is about 31 m.
Should I round my answer in the middle of a calculation?
No. Keep the full calculator value until the final step, then round once to the accuracy the question asks for. Rounding early can make the final answer wrong.