Question

If in a triangle $ABC,$ $AB=5$ units, $\angle B={\cos }^{-1}\left(\frac{3}{5}\right)$ and radius of circumcircle of $△ABC$ is $5$ units, then the area (in sq. units) of $△ABC$ is

• A. $10+6\sqrt{2}$
• B. $6+8\sqrt{3}$
• C. $8+2\sqrt{2}$
• D. $4+2\sqrt{3}$

Answer 1

The correct option is B $6+8\sqrt{3}$

In $△ABC\cos B=\frac{3}{5},R=5$
$\therefore \sin B=\frac{4}{5}$

$\because \frac{5}{\sin C}=\frac{a}{\sin A}=\frac{b}{\left(\frac{4}{5}\right)}=10$
$\Rightarrow C={30}^{\circ }$ and $b=8$
$\because \sin A=\sin (B+C)\Rightarrow \sin A=\sin B\cos C+\cos B\sin C\Rightarrow \sin A=\frac{4}{5}\cdot \frac{\sqrt{3}}{2}+\frac{3}{5}\cdot \frac{1}{2}=\frac{3+4\sqrt{3}}{10}$
$\therefore △=\frac{1}{2}bc\sin A=\frac{1}{2}\cdot 8\cdot 5\cdot \left(\frac{3+4\sqrt{3}}{10}\right)=6+8\sqrt{3}$