Question

In AABC, a = 8 and b = 8. If the area of AABC is 16, find m2C.

Answer 1

Area of triangle is given as,

$A=\frac{1}{2}ab\sin \theta$

So,

$\begin{gathered} 16=\frac{1}{2}ab\sin C \\ 16=\frac{1}{2}\times8\times8\times\sin C \\ \frac{16}{32}=\sin C \\ C=\sin ^{-1}\frac{1}{2} \\ C=30^{\circ} \end{gathered}$