Discussion Forum

In $\triangle ABC$ , if a=1,b=2 $\angle C=60^{0}$ , then $4 \triangle^{2}+c^{2}$=

Answer:

Explanation:

Clearly $\triangle =\frac{1}{2}a.b \sin c$

$\Rightarrow$    $\triangle =\frac{1}{2}.2.1. \sin 60^{0}$

$\Rightarrow$   $\triangle =\frac{\sqrt{3}}{2}$

$\Rightarrow$  $\triangle ^{2}=\frac{3}{4}$

$\Rightarrow$     $4\triangle^{2}=3$   .........(i)

By Sine rule, we get

   $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

$\Rightarrow$    $\frac{\sin A}{1}=\frac{\sin B}{2}=\frac{\sin 60^{0}}{c}$

$\Rightarrow$  $\frac{\sin A}{\sin B}=\frac{1}{2}=\frac{\sin 60^{0}}{c}$

 Considering last two  terms, we get

    $\frac{1}{2}$= $\frac{\frac{\sqrt{3}}{2}}{c}$

$\Rightarrow$        $c=\sqrt{3}$    ..........(ii)

 Thus , $4 \triangle^{2}+c^{2}$=3+3=6   [(using Eqs.(i) and (ii)]