Question

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

• A. 6
• B. 3
• C. $\frac{\sqrt{3}}{2}$
• D. 9

Answer 1

Answer: (A)

Clearly $\Delta =\frac{1}{2}\cdot a\cdot b\sin c$
$\Rightarrow \Delta =\frac{1}{2}\cdot 2\cdot 1\cdot \sin 60^{\circ }$
$\Rightarrow \Delta =\frac{\sqrt{3}}{2}$
$\Rightarrow {\Delta }^2=\frac{3}{4}$
$\Rightarrow 4{\Delta }^2=3$
By sine rule, we get
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
$\Rightarrow \frac{\sin A}{1}=\frac{\sin B}{2}=\frac{\sin 60^{\circ }}{c}$
$\Rightarrow \frac{\sin A}{\sin B}=\frac{1}{2}=\frac{\sin 60^{\circ }}{c}$
Considering last two terms, we get
$\frac{1}{2}=\frac{\frac{\sqrt{3}}{2}}{c}$
$\Rightarrow C=\sqrt{3}\dots$ (ii)
Thus, $4{\Delta }^2+c^2=3+3=6$ [(Using Eqs. (i) and (iii)]