Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.$A=\frac{\pi }{3}$ and $b:c=2:3.$. If $\tan a=\frac{\sqrt{3}}{5},0<a<\frac{\pi }{2},$ then?

• A. $B={60}^{\circ }+\alpha$
• B. $C={60}^{\circ }+\alpha$
• C. $B={60}^{\circ }-\alpha$
• D. $C={60}^{\circ }-\alpha$

Answer 1

Answer: (B), (C)

$B={60}^{\circ }-\alpha$
$C={60}^{\circ }+\alpha$

Given, $A=\frac{\pi }{3}$
Hence
$\cot \frac{A}{2}$ $=\sqrt{3}$
Hence
$\tan \left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}.\sqrt{3}$
$\tan \left(\frac{B-C}{2}\right)=\frac{2-3}{2+3}.\sqrt{3}$
$=\frac{-\sqrt{3}}{5}$
$\tan \left(\frac{C-B}{2}\right)=\frac{\sqrt{3}}{5}$
$=\tan \left(\alpha \right)$
Hence,
$\frac{C-B}{2}=\alpha$
$C-B=2\alpha$.
Hence,
$C=\theta +\alpha$
$B=\theta -\alpha$.
If $\theta ={60}^0$, we get Option B and Option C.