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

Answer 1

Answer: (C)

As, $\cos B=\frac{3}{5}\Rightarrow B=53^{\circ }$
As, $R=5\Rightarrow \frac{c}{\sin c}=2R$
$\Rightarrow \frac{5}{10}=\sin c$
$\Rightarrow C=30^{\circ }$
Now, $\frac{b}{\sin B}=2R$
$\Rightarrow b=2(5)\left(\frac{4}{5}\right)=8$
Now, by cosine formula
$\cos B=\frac{a^2+c^2-b^2}{2ac}$
$\Rightarrow \frac{3}{5}=\frac{a^2+25-64}{2(5)a}$
$\Rightarrow a^2-6a-3g=0$
$\therefore a=\frac{6\pm \sqrt{192}}{2}=\frac{6\pm 8\sqrt{3}}{2}$
$\Rightarrow 3+4\sqrt{3}$ (Reject $a=3-4\sqrt{3})$
Now, $\Delta =\frac{abc}{4R}$
$=\frac{(3+4\sqrt{3})(8)(5)}{4(5)}=2(3+4\sqrt{3})$
$\Rightarrow \Delta =(6+8\sqrt{3})$