Question

In a $△ABC,\frac{a}{b}=2+\sqrt{3}$ and $\angle C={60}^{\circ }.$ Then the ordered pair $(\angle A,\angle B)$ is equal to:

• A. $({45}^{\circ },{75}^{\circ })$
• B. $({75}^{\circ },{45}^{\circ })$
• C. $({105}^{\circ },{15}^{\circ })$
• D. $({15}^{\circ },{105}^{\circ })$

Answer 1

The correct option is C $({105}^{\circ },{15}^{\circ })$

$\frac{a}{b}=\frac{2+\sqrt{3}}{1}$
By componendo and dividendo,
$\frac{a+b}{a-b}=\frac{3+\sqrt{3}}{1+\sqrt{3}}=\sqrt{3}$
Now, By Napier's law,
$\tan \frac{A-B}{2}=\frac{a-b}{a+b}\cot \frac{C}{2}$
$\Rightarrow \tan \frac{A-B}{2}=1$
$\Rightarrow \frac{A-B}{2}={45}^{\circ }$
$\Rightarrow A-B={90}^{\circ }$
Also, $A+B={120}^{\circ }$
$\therefore A={105}^{\circ },B={15}^{\circ }$