Question:
The angles $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ of a triangle $\mathrm{ABC}$ are in A.P. and $\mathrm{a}: \mathrm{b}=1: \sqrt{3}$. If $\mathrm{c}=4 \mathrm{~cm}$, then the area (in sq. cm) of this triangle is :
Correct Option: , 3
Solution:
$\angle \mathrm{B}=\frac{\pi}{3}$, by sine Rule
$\sin \mathrm{A}=\frac{1}{2}$
$\Rightarrow \mathrm{A}=30^{\circ}, \mathrm{a}=2, \mathrm{~b}=2 \sqrt{3}, \mathrm{c}=4$
$\Delta=\frac{1}{2} \times 2 \sqrt{3} \times 2=2 \sqrt{3} \mathrm{sq} . \mathrm{cm}$
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