Question

In a $△ABC,\angle A=\frac{\pi }{3}$ and $b:c=2:3$.If $\tan \theta =\frac{\sqrt{3}}{5},0<\theta <\frac{\pi }{2},$ then

• A. $B={60}^0+\theta$
• B. $C={60}^0+\theta$
• C. $B={60}^0-\theta$
• D. $C={60}^0-\theta$

Answer 1

The correct option is B $C={60}^0+\theta$

Given $\angle A=\frac{\pi }{3}$
Now, $\tan \frac{C-B}{2}=\frac{c-b}{c+b}\cot \frac{A}{2}$
$\Rightarrow \tan \frac{C-B}{2}=\frac{c-b}{c+b}\cot \frac{\pi }{6}$
$\Rightarrow \tan \frac{C-B}{2}=\frac{3-2}{3+2}\sqrt{3}$
$\Rightarrow \tan \frac{C-B}{2}=\frac{1}{5}\sqrt{3}$
$\Rightarrow \tan \frac{C-B}{2}=\frac{\sqrt{3}}{5}=\tan \theta$
$\therefore C-B=2\theta$
and $C+B={180}^0-A={180}^0-{60}^0={120}^0$
On solving we get $C+B+C-B={120}^0+2\theta$
$\Rightarrow 2C={120}^0+2\theta$
or $C={60}^0+\theta$