Question

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

If $A-B={120}^{\circ }$ and $R=8r,$ where $R$ and $r$ have their usual meaning, then $cosC$ equals

• A. $\frac{3}{4}$
• B. $\frac{2}{3}$
• C. $\frac{5}{6}$
• D. $\frac{7}{8}$

Answer 1

The correct option is D $\frac{7}{8}$

Given that, $A-B={120}^{\circ }$
and $R=8r$
$\Rightarrow R=8\left(4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\right)$
$\Rightarrow 1=16(\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right))\sin \frac{C}{2}$
$\Rightarrow 1=16(\frac{1}{2}-\cos (\pi -\frac{C}{2}))\sin \frac{C}{2}$
$\Rightarrow 1=16(\frac{1}{2}-\sin \frac{C}{2})\sin \frac{C}{2}$
$\Rightarrow 16{\sin }^2\frac{C}{2}-8\sin \frac{C}{2}+1=0$
$\Rightarrow \sin \frac{C}{2}=\frac{1}{4}$
$\Rightarrow \sqrt{\frac{1-\cos C}{2}}=\frac{1}{4}$
Therefore, $\cos C=\frac{7}{8}$
Ans: D