Question

In $\Delta ABC,A=\frac{2\pi }{3},b-c=3\sqrt{3}cm$ and
$ar\left(\Delta ABC\right)=\frac{9\sqrt{3}}{2}c{m}^2$. Solve for side 'a'.

Answer 1

Area of triangle ABC
$=\frac{bcsinA}{2}$
$=\frac{bc}{2}.\frac{\sqrt{3}}{2}$
$=\frac{9\sqrt{3}}{2}$
Hence
$bc=18$.
And
$b-c=3\sqrt{3}$
Squaring on both the sides we get
$b^2+c^2-2bc=27$
$b^2+c^2-36=27$
$b^2+c^2=63$
Now
$cosA=\frac{b^2+c^2-a^2}{2bc}$
Or
$a^2=b^2+c^2-2bc.cosA$.
$=63-36.\frac{-1}{2}$
$=63+18$
$=81$
Hence
$a=9units$.