Question

In a $△ABC,\angle A=\frac{2\pi }{3};b-c=3\sqrt{3}$ and $\Delta =\frac{9\sqrt{3}}{2}{cm}^2$, then $a=$

• A. $6\sqrt{3}cm$
• B. $9cm$
• C. $18cm$
• D. $6cm$

Answer 1

Answer: (B)

$9cm$

Area $=\frac{bc\sin A}{2}$ $=\frac{c(3\sqrt{3}+c)}{2}.\frac{\sqrt{3}}{2}=\frac{9\sqrt{3}}{2}$

Hence,

$3\sqrt{3}c+c^2=18$
$\Rightarrow c^2+3\sqrt{3}c-18=0$

$c=\frac{-3\sqrt{3}\pm \sqrt{27+72}}{2}=\frac{-3\sqrt{3}\pm 3\sqrt{11}}{2}$

$\Rightarrow c=\frac{3\sqrt{11}-3\sqrt{3}}{2}$

Also given, $b-c=3\sqrt{3}$
$b-\left(\frac{3\sqrt{11}-3\sqrt{3}}{2}\right)=3\sqrt{3}$
$\Rightarrow b=\frac{3\sqrt{11}+3\sqrt{3}}{2}$

Now,

$\cos A=\frac{b^2+c^2-a^2}{2bc}$
$=\frac{2(99+27)-4a^2}{2(99-27)}=\frac{252-4a^2}{144}=-\frac{1}{2}$

$\Rightarrow 252-4a^2=-72$
$\Rightarrow 4a^2=324$
$\Rightarrow 2a=18$
$\Rightarrow a=9cm$