Given the lengths of two sides $a$ and $b$, and the angle between them $C$, you can use this formula for the area of a triangle. \[\text{area } = \frac{1}{2}ab\sin C\text{.}\]
Given $a=10$, $b=12$ and $C=54^{\circ}$, find the area of the triangle $ABC$.
\begin{align} \text{Area } &= \frac{1}{2} \times 10 \times 12 \times \sin 54^{\circ}\\ &= \frac{1}{2} \times 120 \times \sin 54^{\circ}\\\\ &= 60\sin 54^{\circ}\\ &= 48.5 \text{ (to 3 sig.fig.)} \end{align}
Given the area of a triangle is $12$, and the sides $a$ and $b$ are $7$ and $4$ respectively, find the angle between them.
Label the angle between $a$ and $b$ as $C$. Now use the area formula:
\begin{align} 12 &= \frac{1}{2} \times 7\times 4\times \sin C\\ 12 &= \frac{1}{2} \times 28 \times \sin C\\ 12 &= 14 \sin C\\ \sin C &= \frac{6}{7}\\ C &= \sin^{-1}\left(\frac{6}{7}\right)\\ C &= 59^{\circ} \text{ (to 2 sig.fig.)} \end{align}