In triangle ABC, AB = 14.6 cm, BC = 7.6 cm and angle BAC = 31°. Find

1. Apply the sine rule to find $\sin \angle ACB$.

Let $\angle ACB = C$.

Using the sine rule,
$\frac{\sin C}{AB}=\frac{\sin A}{BC}$
$\frac{\sin C}{14.6}=\frac{\sin 31^\circ}{7.6}$
$\sin C=\frac{14.6\sin 31^\circ}{7.6}$
$\sin C=0.9894\ldots$

2. Find the obtuse value of $C$.

The calculator gives the acute angle:
$C=\sin^{-1}(0.9894\ldots)=81.7^\circ \text{ approximately}$

Since $\sin \theta = \sin(180^\circ-\theta)$, the obtuse angle is
$180^\circ-81.7^\circ=98.3^\circ$

1. Find the included angle between $AB$ and $BC$.

From part (a), $\angle ACB = 98.3^\circ$.

$\angle ABC = 180^\circ - 31^\circ - 98.3^\circ = 50.7^\circ$

2. Use the area formula with two sides and the included angle.

$\text{Area of }\triangle ABC = \frac12(AB)(BC)\sin \angle ABC$

$= \frac12(14.6)(7.6)\sin 50.7^\circ$

$= 42.9\text{ cm}^2 \text{, to 3 s.f.}$