Question
Find the area of the triangle $\triangle\text{ABC}$ in which a = 1, b = 2 and $\angle\text{C}=60^{\circ}.$

Answer

Area of the $\triangle\text{ABC}=\frac{1}{2}\sin\text{C}$
$=\frac{1}{2}\times1\times2\times\sin60$
$=\frac{\sqrt{3}}2{}\text{sq. units}$

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