Sine Rule - SS2 Mathematics Lesson Note

Consider the acute and obtuse angled triangles below:

The sine rule states that for any acute-angled or obtuse-angled triangles, \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\) or \(\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\)

Example: Given a \(\mathrm{\Delta}ABC\) as shown above, where \(a = 9.3cm,\ \angle A = 81{^\circ}\) and \(\angle C = 42{^\circ}\). Find the length of sides \(b\) and \(c\).

Solution

First, calculate the \(\angle B\)

\[\angle B = 180{^\circ} - (\angle A + \angle C)\]

\[\angle B = 180{^\circ} - (81{^\circ} + 42{^\circ})\]

\[\angle B = 180{^\circ} - 123{^\circ}\]

\[\angle B = 57{^\circ}\]

From the Sine Rule, \(\frac{a}{\sin A} = \frac{b}{\sin B}\)

\[\frac{9.3}{\sin 81} = \frac{b}{\sin 57}\]

\[\frac{9.3}{0.9877} = \frac{b}{0.8387}\]

\[b = \frac{9.3 \times 0.8387}{0.9877} \approx 7.9cm\]

From the Sine Rule, \(\frac{a}{\sin A} = \frac{c}{\sin C}\)

\[\frac{9.3}{\sin 81} = \frac{c}{\sin 42}\]

\[\frac{9.3}{0.9877} = \frac{c}{0.6691}\]

\[c = \frac{9.3 \times 0.6691}{0.9877} \approx 6.37cm\]