Question

If in a triangle $A B C, A B=5$ units, $\angle B=\cos ^{-1}\left(\frac{3}{5}\right)$ and radius of circumcircle of $\triangle A B C$ is $5$ units, then the area (in sq. units) of $\triangle A B C$ 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, $R=5 \Rightarrow \frac{c}{\sin c}=2 R$
$\Rightarrow \frac{5}{10}=\sin c $
$\Rightarrow C=30^{\circ}$
Now, $\frac{b}{\sin B}=2 R $
$\Rightarrow b=2(5)\left(\frac{4}{5}\right)=8$
Now, by cosine formula
$\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c} $
$\Rightarrow \frac{3}{5}=\frac{a^{2}+25-64}{2(5) a} $
$\Rightarrow a^{2}-6 a-3 g=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, $\Delta=\frac{a b c}{4 R}$
$=\frac{(3+4 \sqrt{3})(8)(5)}{4(5)}=2(3+4 \sqrt{3})$
$\Rightarrow \Delta=(6+8 \sqrt{3})$