Q110 / 160 · TS EAMCET 2017 · Mathematics

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

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

Solution

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

Hence, the correct answer is option: 'A'.

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