Finding Unknown Sides
To find an unknown side in a right-angled triangle, pick the ratio that links the known angle, the known side and the unknown side, then rearrange. If the unknown is on the bottom, as in \(\cos38^\circ15'=\dfrac{6.5}{x}\), divide rather than multiply. In NSW Year 10 Advanced maths the triangles sit inside larger shapes and problems take several steps (NESA outcome MA5‑TRG‑C‑01).
You will find sides in triangles with angles in degrees and minutes, then work through two connected triangles, isosceles triangles, a rhombus, a trapezium and a regular octagon. Problems include the width of a river, an A-frame tent, a skillion roof, a Ferris wheel seat, a swinging pendulum, a flagpole brace and a chord. Harder questions prove a formula for a length and use a perimeter or area to find the sides. It is part of Year 10 Advanced Trigonometry.
Every question with a fully worked solution.
- Finding Unknown Sides - Video - Using Trig Ratios to Find Missing Side Lengths Watch
Theory
To find an unknown side in a right-angled triangle, name the sides from the given angle, choose sine, cosine or tangent, then multiply or divide. This Year 10 Advanced maths page covers angles in degrees and minutes, the unknown on the top or the bottom of the ratio, and multi-step problems where you first build the right-angled triangle inside another shape.
In a right-angled triangle the sides are named from a chosen acute angle \(\theta\). The hypotenuse is opposite the right angle. The opposite side faces \(\theta\). The adjacent side runs from \(\theta\) to the right angle.
The ratios \(\sin\theta = \dfrac{O}{H}\), \(\cos\theta = \dfrac{A}{H}\) and \(\tan\theta = \dfrac{O}{A}\) (SOH CAH TOA) are the same for every right-angled triangle with that angle, because all such triangles are similar. To find a side, 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: \(x = 12\sin 35^\circ\). If it is on the bottom, it swaps places with the ratio: \(x = \dfrac{12}{\sin 35^\circ}\).
Angles are often given in degrees and minutes, where \(1^\circ = 60'\). Keep the full calculator value through every step and round once, at the end.
Many problems hide the right-angled triangle inside another shape. Your first job is to find it: draw a height, a diagonal or a perpendicular, and use symmetry to halve an angle or a side.
Relative to the angle \(\theta\):
Rearranged for each 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
- Find the right-angled triangle. Draw a height, a diagonal or a perpendicular if it is hidden inside another shape.
- Label the hypotenuse, opposite and adjacent sides from the angle you are using.
- Choose the ratio that contains the known side and the unknown side.
- Rearrange: multiply if the unknown is on the top, divide if it is on the bottom.
- Calculate in degree mode, keep the full value and round once at the end, with units.
Finding the hidden right angle
- Isosceles triangle: the height from the apex halves the apex angle and the base.
- Rhombus: the diagonals cross at right angles, halve each other and halve the angles of the rhombus.
- Regular polygon with \(n\) sides: each side makes an angle of \(\dfrac{360^\circ}{n}\) at the centre. The line from the centre to the midpoint of a side halves that angle and meets the side at right angles.
- Circle: the perpendicular from the centre to a chord halves the chord, and every radius is the same length. Pythagoras' theorem often finds the radius first.
- Trapezium or composite shape: drop a perpendicular to split it into a rectangle and a right-angled triangle.
- Two triangles sharing a side: find the shared side in the first triangle and carry its unrounded value into the second.
- Heights from a centre or pivot: work out the vertical part with \(r\cos\theta\), then add it to or subtract it from the known height.
- No triangle with two known sides: write the same length in two ways, or two lengths in terms of one unknown, and form an equation (Example 4).
(i) the length \(L\) of the ramp
(ii) the horizontal distance \(d\) it covers.
The rise is opposite the angle. \(L\) is the hypotenuse, so use sine. \(d\) is adjacent, so use tangent. Both unknowns land on the bottom of the fraction.
(i)
| \(\sin 9^\circ 36'\) | \(=\) | \(\dfrac{1.2}{L}\) |
| \(L\) | \(=\) | \(\dfrac{1.2}{\sin 9^\circ 36'}\) |
| \(\) | \(=\) | \(7.1955\ldots\) |
| \(\therefore\ L\) | \(\approx\) | \(\mathbf{7.20}\text{ m}\) |
(ii)
| \(\tan 9^\circ 36'\) | \(=\) | \(\dfrac{1.2}{d}\) |
| \(d\) | \(=\) | \(\dfrac{1.2}{\tan 9^\circ 36'}\) |
| \(\) | \(=\) | \(7.0948\ldots\) |
| \(\therefore\ d\) | \(\approx\) | \(\mathbf{7.09}\text{ m}\) |
Enter \(9^\circ 36'\) with the degrees-minutes key: it is \(9.6^\circ\), not \(9.36^\circ\). Check: \(L\) is longer than \(d\), as the hypotenuse must be.
In both positions the ladder is the hypotenuse and the height up the wall is opposite the angle, so height \(= 3.6\sin\theta\).
| \(h_1\) | \(=\) | \(3.6\sin 68^\circ\) |
| \(\) | \(=\) | \(3.3378\ldots\) |
| \(h_2\) | \(=\) | \(3.6\sin 55^\circ\) |
| \(\) | \(=\) | \(2.9489\ldots\) |
| \(h_1 - h_2\) | \(=\) | \(0.3889\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{0.39}\text{ m}\) |
Subtract the full calculator values and round once, at the end. The top slides down about \(0.39\) m.
(i) the length \(s\) of a sloping wall
(ii) the depth \(h\) of the channel.
Drop a perpendicular from a bottom corner. By symmetry the extra width is shared equally between the two sides:
| \(\dfrac{1.8 - 1.0}{2}\) | \(=\) | \(0.4\text{ m}\) |
In the small right-angled triangle, \(0.4\) m is adjacent to \(64^\circ\), \(s\) is the hypotenuse and \(h\) is opposite.
(i)
| \(\cos 64^\circ\) | \(=\) | \(\dfrac{0.4}{s}\) |
| \(s\) | \(=\) | \(\dfrac{0.4}{\cos 64^\circ}\) |
| \(\) | \(=\) | \(0.9124\ldots\) |
| \(\therefore\ s\) | \(\approx\) | \(\mathbf{0.91}\text{ m}\) |
(ii)
| \(\tan 64^\circ\) | \(=\) | \(\dfrac{h}{0.4}\) |
| \(h\) | \(=\) | \(0.4\tan 64^\circ\) |
| \(\) | \(=\) | \(0.8201\ldots\) |
| \(\therefore\ h\) | \(\approx\) | \(\mathbf{0.82}\text{ m}\) |
Neither triangle has two known sides. In each one \(h\) is opposite the angle and the ground distance is adjacent, so write both distances in terms of \(h\) with tangent.
| \(BF\) | \(=\) | \(\dfrac{h}{\tan 47^\circ}\) |
| \(AF\) | \(=\) | \(\dfrac{h}{\tan 32^\circ}\) |
| \(AF - BF\) | \(=\) | \(15\) |
Substitute and take out the common factor \(h\):
| \(h\left(\dfrac{1}{\tan 32^\circ} - \dfrac{1}{\tan 47^\circ}\right)\) | \(=\) | \(15\) |
| \(h\) | \(=\) | \(\dfrac{15}{\dfrac{1}{\tan 32^\circ} - \dfrac{1}{\tan 47^\circ}}\) |
| \(\) | \(=\) | \(22.4611\ldots\) |
| \(\therefore\ h\) | \(\approx\) | \(\mathbf{22.5}\text{ m}\) |
Check: \(BF \approx 20.95\) m and \(AF \approx 35.95\) m, which differ by \(15\) m.
Common pitfalls
Frequently asked questions
How do you find a missing side using trigonometry?
Label the hypotenuse, opposite and adjacent sides from the given angle. Choose sine, cosine or tangent so the ratio contains the side you know and the side you want. Then multiply or divide to get the unknown, and round at the end.
When do you multiply and when do you divide in trigonometry?
If the unknown side is on the top of the ratio, multiply: \(x = 12\sin 35^\circ\). If it is on the bottom, divide: \(x = \dfrac{12}{\sin 35^\circ}\).
How do you enter degrees and minutes on a calculator?
Use the degrees-minutes-seconds key, often marked \(^\circ\,'\,''\). Or divide the minutes by \(60\) and add them to the degrees: \(9^\circ 36'\) is \(9.6^\circ\), not \(9.36^\circ\).
If the angle gets steeper, does the side get longer or shorter?
It depends which side is fixed. For a fixed height, such as a pole braced by a wire, a steeper angle gives a shorter wire and a closer anchor point, because \(H = \dfrac{O}{\sin\theta}\) and \(\sin\theta\) grows as \(\theta\) grows.
Why is my trigonometry answer wrong?
The most common reasons are a calculator in radian mode, minutes typed as a decimal, the wrong ratio for the two sides involved, or multiplying when the unknown was on the bottom of the fraction.
How do you find a side if the triangle is not right-angled?
Look for a way to make a right angle: draw a height, a perpendicular or a diagonal, or use symmetry. Then solve the right-angled triangle you have made. The sine rule and cosine rule handle the remaining cases.