Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $A={30}^0$ and the area of $\Delta$ is $\frac{\sqrt{3}{a}^2}{4}$, then which of the following is correct?

• A. $B=4C$
• B. $C=4B$
• C. $B=2C$
• D. $C=2B$

Answer 1

The correct option is A $B=4C$

Sol: Formula used:

1. Sine Rule: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

2. Area of a $\Delta ABC=\frac{1}{2}bc\sin A$

3. $\cos (B-C)-\cos (B+C)=2\sin B.\sin C$

We have $A=30°$

Area of $\Delta ABC=\frac{\sqrt{3}}{4}{a}^2=\frac{1}{2}bc\sin A$

$⟹\frac{\sqrt{3}}{4}{a}^2=\frac{1}{2}bc\sin {30}^o$

$⟹\frac{\sqrt{3}}{4}{a}^2=\frac{1}{4}bc⟹bc=\sqrt{3}{a}^2$ ......(1)

Using Sine Rule:

$\frac{a}{\sin 30°}=\frac{b}{\sin B}=\frac{c}{\sin C}$

$⟹b=2a\sin B\text{and}c=2a\sin C$

Putting the value of $b$ and $c$ in equation (1)

$⟹2a\sin B.2a\sin C=\sqrt{3}{a}^2$

$⟹2\sin B.\sin C=\frac{\sqrt{3}}{2}$

$⟹\cos (B-C)-\cos (B+C)=\frac{\sqrt{3}}{2}$

$[\because A=30°⟹A+B+C=180°⟹B+C=150°]$

$⟹\cos (B-C)-\cos 150°=\frac{\sqrt{3}}{2}$

$⟹\cos (B-C)+\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$

$⟹\cos (B-C)=0$

$⟹B-C=\frac{\pi }{2}=90°$ ...... (2) [$\frac{3\pi }{2}$ cannot be its solution as B+C= 150°:which means their difference cannot be270°]

$B+C=150°$ ...... (3)

Adding equation (2) and (3) we get

$2B=240°⟹B=120°$

$C=150°-120°=30°$

Hence, $B=4C$