Question

The angles $A,B$ and $C$ of a triangle $ABC$ are in $A.P.$ and $a:b=1:\sqrt{3}.$ If $c=4\text{cm}$, then the area (in sq.cm) of this triangle is:

• A. $4\sqrt{3}$
• B. $2\sqrt{3}$
• C. $\frac{4}{\sqrt{3}}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

The correct option is B $2\sqrt{3}$

Given,
The angles of $△ABC$ are in A.P.
sum of angles of a triangle $=\pi$
$\therefore A+B+C=\pi \&2B=A+C$
$\Rightarrow 3B=\pi \Rightarrow B=\frac{\pi }{3}$
$\Rightarrow A+C=\frac{2\pi }{3}\sigma$
$\Rightarrow \frac{a}{b}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{2R\sin A}{2R\sin B}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{\sin A}{\sin \frac{\pi }{3}}=\frac{1}{\sqrt{3}}$
$\Rightarrow \sin A=\frac{1}{2}$
$\therefore A={30}^{\circ }$
$\Rightarrow a=2,b=2\sqrt{3},c=4$

$\therefore \text{Area}=\frac{1}{2}\times 2\sqrt{3}\times 2=2\sqrt{3}$