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
$\frac{1}{2}=\frac{\frac{\sqrt{3}}{2}}{c}$
$\Rightarrow C=\sqrt{3} \ldots$ (ii)
Thus, $4 \Delta^{2}+c^{2}=3+3=6$ [(Using Eqs. (i) and (iii)]