Question

With the usual notation, in $△ABC$, if $\angle A+\angle B={120}^{\circ }$, $a=\sqrt{3}+1$ and $b=\sqrt{3}-1$, then the ratio $\angle A:\angle B$, is :

• A. $5:3$
• B. $3:1$
• C. $9:7$
• D. $7:1$

Answer 1

The correct option is D $7:1$

$\angle A+\angle B={120}^{\circ }\cdots \left(1\right)$
As, $\angle A+\angle B+\angle C={180}^{\circ }$
$\therefore \angle C={60}^{\circ }$
As we know,
$\tan \frac{\angle A-\angle B}{2}=\frac{a-b}{a+b}\cdot \cot \left(\frac{\angle C}{2}\right)$
$\Rightarrow \tan \frac{\angle A-\angle B}{2}=\frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+(\sqrt{3}-1)}\cot {30}^{\circ }$
$\Rightarrow \tan \frac{\angle A-\angle B}{2}=1$
$\Rightarrow \frac{\angle A-\angle B}{2}={45}^{\circ }$
$\Rightarrow \angle A-\angle B={90}^{\circ }\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right),$ we get
$\angle A={105}^{\circ }$ and $\angle B={15}^{\circ }$
Hence, $\angle A:\angle B=7:1$