Question:
In $\Delta A B C$, if $a=5, b=6$ and $C=60^{\circ}$, show that its area is $\frac{15 \sqrt{3}}{2}$ sq. units.
Solution:
Given : $a=5, b=6, c=60^{\circ}$
Area of a triangle $=\frac{1}{2} a b \sin C$
$=\frac{1}{2} \times 5 \times 6 \times \sin 60^{\circ}=15 \times \frac{\sqrt{3}}{2}$ sq. units
hence proved.
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