Question

In $∆ABC,\mathrm{if}a=5,b=6\mathrm{and}C=60°$, show that its area is $\frac{15\sqrt{3}}{2}\mathrm{sq}$. units.

Answer 1

$$\begin{gathered} \mathrm{Given}:a=5,b=6,c=60° \\ \mathrm{Area}\mathrm{of}\mathrm{a}\mathrm{triangle}=\frac{1}{2}ab\sin C \\ =\frac{1}{2}\times 5\times 6\times \sin 60°=15\times \frac{\sqrt{3}}{2}\mathrm{sq}.\mathrm{units} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$