Finding Unknown Side
When one side and one acute angle in a right-angled triangle are known, a trig ratio finds another side. In NSW Year 9 Advanced maths, students choose the ratio that links the known side, the unknown side and the angle, then multiply (NESA outcome MA5‑TRG‑C‑01).
You will find sides labelled in diagrams or named by vertices, such as \(AB\) in \(\triangle ABC\), and round to the stated accuracy. Practical problems include a wheelchair ramp, a ladder and a boat's mast. You will find perimeters and areas, split an isosceles triangle into two right-angled triangles, find how far a slipping ladder moves, and show that the area is \(\frac{1}{2}c^2\sin\theta\cos\theta\). It is part of Year 9 Advanced Trigonometry.
Theory
To find an unknown side, label the sides from the given angle, choose sine, cosine or tangent, and multiply or divide. This Year 9 Advanced maths page covers the unknown on the bottom of the ratio, degrees and minutes, angles measured from the vertical, right-angled triangles hidden in other shapes, two-triangle problems and an unknown in two places.
Label the sides from the given angle: hypotenuse (H) opposite the right angle, opposite (O) facing the angle and adjacent (A) next to it. Then choose the ratio that contains the side you know and the side you want.
If the unknown is on the top of the fraction, multiply: \(\cos 25^\circ = \dfrac{x}{8}\) gives \(x = 8\cos 25^\circ\). If it is on the bottom, divide: \(\cos 25^\circ = \dfrac{8}{x}\) gives \(x = \dfrac{8}{\cos 25^\circ}\).
Read the angle carefully. An angle measured from a wall or pole (the vertical) puts the vertical side next to the angle, so it is the adjacent side.
Many problems hide the right-angled triangle inside another shape. Find it first: draw a perpendicular, a height or a diagonal, and use symmetry to halve an angle or a side.
| To find | Know \(H\) | Know \(O\) | Know \(A\) |
|---|---|---|---|
| \(O\) | \(O = H\sin\theta\) | — | \(O = A\tan\theta\) |
| \(A\) | \(A = H\cos\theta\) | \(A = \dfrac{O}{\tan\theta}\) | — |
| \(H\) | — | \(H = \dfrac{O}{\sin\theta}\) | \(H = \dfrac{A}{\cos\theta}\) |
How to find an unknown side
- Sketch the triangle and mark the right angle, the given angle, the known side and the unknown.
- Label H, O and A from the given angle.
- Choose sin, cos or tan to link the known side and the unknown.
- Solve: multiply if the unknown is on top, divide if it is on the bottom.
- Round once, at the end, and give units. The hypotenuse must be the longest side.
Multi-step problems
- Two sides, then a perimeter or area: find each side from the given values, then add or multiply.
- Isosceles triangle: the perpendicular from the apex halves the apex angle and the base. Double the half-base at the end.
- Rectangle: a diagonal makes two right-angled triangles; the sides are \(d\cos\theta\) and \(d\sin\theta\).
- Two triangles sharing a side: find the shared side in the first triangle, keep its full value and use it in the second.
- A ladder that slips: work out the height and the foot distance in each position, then subtract.
- Unknown in two places: write the ratio, multiply out, collect the \(x\)-terms and factorise (Example 4).
Sketch and label. \(YZ\) is opposite the angle, \(XZ\) is the hypotenuse and \(XY\) is adjacent.
| \(\sin 38^\circ 15'\) | \(=\) | \(\dfrac{7.6}{XZ}\) |
| \(XZ\) | \(=\) | \(\dfrac{7.6}{\sin 38^\circ 15'}\) |
| \(\) | \(=\) | \(12.2760\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{12.28}\text{ cm}\) |
| \(\tan 38^\circ 15'\) | \(=\) | \(\dfrac{7.6}{XY}\) |
| \(XY\) | \(=\) | \(\dfrac{7.6}{\tan 38^\circ 15'}\) |
| \(\) | \(=\) | \(9.6405\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{9.64}\text{ cm}\) |
Use the given side and a ratio for each answer, so a rounding slip in one does not carry into the other.
The angle is at the top, so the pole is adjacent to it and \(d\) is opposite.
| \(h\) | \(=\) | \(14.5\cos 23^\circ\) |
| \(\) | \(=\) | \(13.3473\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{13.35}\text{ m}\) |
| \(d\) | \(=\) | \(14.5\sin 23^\circ\) |
| \(\) | \(=\) | \(5.6656\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{5.67}\text{ m}\) |
Check with Pythagoras: \(13.35^2 + 5.67^2 \approx 14.5^2\).
The diagonals of a rhombus cross at right angles, halve each other and halve the angles. So each quarter is a right-angled triangle with hypotenuse \(15\) and angle \(35^\circ\).
| \(\text{half short diagonal}\) | \(=\) | \(15\sin 35^\circ\) |
| \(\text{short diagonal}\) | \(=\) | \(30\sin 35^\circ \approx \mathbf{17.21}\text{ cm}\) |
| \(\text{half long diagonal}\) | \(=\) | \(15\cos 35^\circ\) |
| \(\text{long diagonal}\) | \(=\) | \(30\cos 35^\circ \approx \mathbf{24.57}\text{ cm}\) |
| \(\text{Area}\) | \(=\) | \(\dfrac{1}{2} \times 30\sin 35^\circ \times 30\cos 35^\circ\) |
| \(\) | \(=\) | \(211.4308\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{211}\text{ cm}^2\) |
Use the unrounded diagonals in the area. Remember to double each half-diagonal.
Write the tangent ratio, then solve the equation: collect the \(x\)-terms on one side and factorise.
| \(\tan 50^\circ\) | \(=\) | \(\dfrac{x + 4}{x}\) |
| \(x\tan 50^\circ\) | \(=\) | \(x + 4\) |
| \(x\tan 50^\circ - x\) | \(=\) | \(4\) |
| \(x(\tan 50^\circ - 1)\) | \(=\) | \(4\) |
| \(x\) | \(=\) | \(\dfrac{4}{\tan 50^\circ - 1}\) |
| \(\) | \(=\) | \(20.8601\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{20.9}\) |
Check: \(\dfrac{24.86}{20.86}\) \(\approx 1.19\) \(\approx \tan 50^\circ\). Type the denominator in brackets on the calculator.
Common pitfalls
Frequently asked questions
How do you find a missing side with trigonometry?
Label the hypotenuse, opposite and adjacent sides from the given angle. Choose the ratio that contains the known side and the unknown side, write it as an equation, then multiply or divide to find the unknown.
When do I divide instead of multiply in trigonometry?
Divide when the unknown side is on the bottom of the fraction. For example cos 25 degrees = 8 over x gives x = 8 divided by cos 25 degrees.
How do you find the hypotenuse using trigonometry?
Use sine if you know the opposite side, or cosine if you know the adjacent side. The hypotenuse is always on the bottom, so divide: H = O divided by sin of the angle, or H = A divided by cos of the angle.
What if the angle is given with the wall instead of the ground?
Then the wall is the side next to the angle, so it is the adjacent side. The distance along the ground is opposite the angle.
How do you find the sides of a rhombus with trigonometry?
Draw both diagonals. They cross at right angles, halve each other and halve the angles, making four right-angled triangles with the rhombus side as the hypotenuse.