Finding Unknown Sides
Theory
To find an unknown side of a right-angled triangle, name the sides from the given angle, choose sine, cosine or tangent, then multiply or divide. This Year 10 maths page revises Year 9: turned triangles, the unknown on the top or the bottom of the ratio, angles in degrees and minutes, and practical problems such as slides, shadows, ladders and heights on buildings. Curriculum: VC2M9M03 / AC9M9M03 (Year 9, revised).
In a right-angled triangle the sides are named from one of the acute angles, \(\theta\). The hypotenuse (\(H\)) is opposite the right angle and is the longest side. The opposite side (\(O\)) faces \(\theta\). The adjacent side (\(A\)) runs from \(\theta\) to the right angle.
The three trigonometric ratios are \(\sin\theta = \dfrac{O}{H}\), \(\cos\theta = \dfrac{A}{H}\) and \(\tan\theta = \dfrac{O}{A}\), remembered as SOH CAH TOA. Pick the ratio that holds the side you know and the side you want.
Sides are often named by their end points. In \(\triangle ABC\), \(\angle ACB\) is the angle at \(C\), and the side opposite it is \(AB\).
Angles can be given in degrees and minutes: \(1^\circ = 60'\), so \(25^\circ 30'\) is \(25.5^\circ\).
For an acute angle \(\theta\) in a right-angled triangle:
Each ratio rearranged for the unknown 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
- Label the hypotenuse first (opposite the right angle), then the opposite and adjacent sides from the given angle. Turn your head with the triangle if it is tilted.
- 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 with the calculator in degree mode, round once at the end and give the units.
- Check the hypotenuse is the longest side. If a shorter side comes out longer than the hypotenuse, you multiplied when you should have divided.
Practical problems
- Sketch the situation and mark the right angle: a wall or post meets level ground at \(90^\circ\).
- Angle of elevation: the angle up from the horizontal, at the observer. A shadow is the adjacent side to the sun's angle of elevation, and the object is opposite.
- A slope or line (slide, ramp, zip line, string) is the hypotenuse. Its vertical drop or rise is opposite the angle with the horizontal.
- A rectangle's diagonal is the hypotenuse of two right-angled triangles: length \(= d\cos\theta\) and width \(= d\sin\theta\). Work out the area from the unrounded length and width.
- Two positions (a ladder that slips) or two lines of sight: solve each right-angled triangle separately, then subtract (Example 4).
- A perpendicular splits a triangle into two right-angled triangles. Find the shared side, or add the two parts of the base (Example 3).
(i) \(PQ\)
(ii) \(QR\)
Name the sides from the \(33^\circ\) angle at \(R\). \(PR\) is opposite the right angle, so it is the hypotenuse. \(PQ\) faces \(R\), so it is opposite. \(QR\) touches \(R\), so it is adjacent.
(i) Opposite and hypotenuse: use sine.
| \(\sin 33^\circ\) | \(=\) | \(\dfrac{PQ}{15.6}\) |
| \(PQ\) | \(=\) | \(15.6 \sin 33^\circ\) |
| \(\) | \(=\) | \(8.4963\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{8.5}\text{ cm}\) |
(ii) Adjacent and hypotenuse: use cosine.
| \(\cos 33^\circ\) | \(=\) | \(\dfrac{QR}{15.6}\) |
| \(QR\) | \(=\) | \(15.6 \cos 33^\circ\) |
| \(\) | \(=\) | \(13.0832\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{13.1}\text{ cm}\) |
Check: both answers are shorter than the hypotenuse.
(i) the length \(r\) of the rope
(ii) the distance \(d\) from the peg to the foot of the pole.
From the angle at the peg, the pole (\(2.4\) m) is opposite. The rope is the hypotenuse and \(d\) is adjacent. Both unknowns end up on the bottom of the ratio.
(i) Opposite and hypotenuse: use sine.
| \(\sin 52^\circ 40'\) | \(=\) | \(\dfrac{2.4}{r}\) |
| \(r \sin 52^\circ 40'\) | \(=\) | \(2.4\) |
| \(r\) | \(=\) | \(\dfrac{2.4}{\sin 52^\circ 40'}\) |
| \(\) | \(=\) | \(3.0184\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{3.02}\text{ m}\) |
(ii) Opposite and adjacent: use tangent.
| \(\tan 52^\circ 40'\) | \(=\) | \(\dfrac{2.4}{d}\) |
| \(d\) | \(=\) | \(\dfrac{2.4}{\tan 52^\circ 40'}\) |
| \(\) | \(=\) | \(1.8305\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{1.83}\text{ m}\) |
Enter \(52^\circ 40'\) with the degrees-minutes key, or as \(52 + 40 \div 60\). It is not \(52.4^\circ\). Check: the rope (\(3.02\) m) is longer than the pole, as a hypotenuse must be.
\(XW\) splits the triangle into two right-angled triangles. Find the shared side \(XW\) in \(\triangle XYW\), then use it in \(\triangle XZW\).
In \(\triangle XYW\): \(XW\) is opposite \(50^\circ\) and \(XY\) is the hypotenuse.
| \(XW\) | \(=\) | \(14 \sin 50^\circ\) |
| \(\) | \(=\) | \(10.7246\ldots\) |
In \(\triangle XZW\): \(XW\) is opposite \(35^\circ\) and \(XZ\) is the hypotenuse, so \(XZ\) is on the bottom.
| \(\sin 35^\circ\) | \(=\) | \(\dfrac{XW}{XZ}\) |
| \(XZ\) | \(=\) | \(\dfrac{10.7246\ldots}{\sin 35^\circ}\) |
| \(\) | \(=\) | \(18.6978\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{18.7}\text{ cm}\) |
Keep the full value of \(XW\) in the calculator. Rounding it to \(10.7\) first gives \(18.65\ldots\), which rounds the same here but can tip other answers.
Each line of sight makes its own right-angled triangle with the ground and the wall. In both, the height is opposite and the \(18\) m is adjacent. Find both heights, then subtract.
| \(\text{bottom}\) | \(=\) | \(18 \tan 28^\circ = 9.5707\ldots\) |
| \(\text{top}\) | \(=\) | \(18 \tan 37^\circ = 13.5639\ldots\) |
| \(\text{sign}\) | \(=\) | \(13.5639\ldots - 9.5707\ldots\) |
| \(\) | \(=\) | \(3.9932\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{4.0}\text{ m}\) |
Watch out: \(18\tan 9^\circ\) is wrong. The \(9^\circ\) between the two lines is not in a right-angled triangle with the sign.
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. Rearrange, calculate in degree mode, and round at the end.
When do you multiply and when do you divide in trigonometry?
Multiply when the unknown is on the top of the ratio: \(x = 12\tan 40^\circ\). Divide when the unknown is on the bottom: \(x = \dfrac{12}{\cos 40^\circ}\).
How do you know which side is the hypotenuse?
The hypotenuse is the side opposite the right angle. It is always the longest side. It does not depend on which acute angle you use, but the opposite and adjacent sides do.
How do you put degrees and minutes into a calculator?
Use the degrees-minutes-seconds key, often marked \(^\circ\,'\,''\). Or divide the minutes by \(60\) and add them to the degrees: \(38^\circ 45' = 38.75^\circ\).
What does SOH CAH TOA mean?
It is a memory aid for the three ratios. SOH: \(\sin\theta = \dfrac{O}{H}\). CAH: \(\cos\theta = \dfrac{A}{H}\). TOA: \(\tan\theta = \dfrac{O}{A}\).
Why is my trig answer wrong?
Check that the calculator is in degree mode, that minutes were not typed as a decimal, that you chose the ratio for the two sides involved, and that you divided when the unknown was on the bottom.