Use the sine rule in a non-right-angled triangle

Using the sine rule

In triangle ABC, angle A is 48°, angle B is 67° and side c is 18 cm. The sine rule links each side to the sine of its opposite angle. Match the pairs before substituting, find a missing third angle when needed, and round only the final stated answer

1. Match opposite pairs Pair each lower-case side with the upper-case angle opposite it \(\displaystyle \frac{a}{\sin A}=\frac{c}{\sin C}\)
2. Find a missing angle Use the angle sum before forming the sine-rule equation when an opposite angle is not given \(\displaystyle C=180^{\circ}-48^{\circ}-67^{\circ}=65^{\circ}\)
3. Find a missing side Multiply the known side by the ratio of the wanted sine to the known sine \(\displaystyle a=18\frac{\sin48^{\circ}}{\sin65^{\circ}}\approx14.8\)
4. Find a missing angle Isolate the wanted sine and apply inverse sine in degree mode. Check whether each angle fits the triangle angle sum \(\displaystyle B_1=\sin^{-1}\left(\frac{b\sin A}{a}\right),\quad B_2=180^{\circ}-B_1\)

Worked example

Triangle PQR has angle P of 53° and angle Q of 72°. Side r, opposite angle R, is 22 m. Find side p to 1 decimal place

1. \(\displaystyle R=180^{\circ}-53^{\circ}-72^{\circ}=55^{\circ}\)The opposite angle R is found from the triangle angle sum
2. \(\displaystyle \frac{p}{\sin53^{\circ}}=\frac{22}{\sin55^{\circ}}\)The side p and side r are paired with their opposite angles
3. \(\displaystyle p=22\frac{\sin53^{\circ}}{\sin55^{\circ}}\approx21.44899\ \mathrm{m}\)Rearranging the ratio gives the unrounded side value
4. \(\displaystyle p\approx21.4\ \mathrm{m}\)Round only the final side to 1 decimal place

Common mistakes

Difficulty levels

Questions about this topic

In this chapter