Question

In a $△ABC,A=\frac{2\pi }{3},b-c=3\sqrt{3}cm$ and area $\left(△ABC\right)=\sqrt{9\sqrt{3}}2c{m}^2$. Then 'a' is

• A. $\sqrt{27+12\sqrt{3\sqrt{32}}}$
• B. $9$ cm
• C. $18$ cm
• D. $7$ cm

Answer 1

The correct option is A $\sqrt{27+12\sqrt{3\sqrt{32}}}$

We have;

In $\Delta ABC;$

$A=\frac{2\pi }{3};b-c=3\sqrt{3}cm⟹(b-c{)}^2=27$ (1)

$\Delta =\sqrt{9\sqrt{32}}{cm}^2=\frac{1}{2}bc\sin \frac{2\pi }{3}$

$⟹\frac{1}{2}bc\times \frac{\sqrt{3}}{2}=\sqrt{9\sqrt{32}}$

$⟹bc=4\sqrt{3\sqrt{32}}$

Taking equation 1;

$b^2+c^2-2bc=27$

$⟹b^2+c^2-2\times 4\sqrt{3\sqrt{32}}=27$

$⟹b^2+c^2=27+8\sqrt{3\sqrt{32}}$

Applying Cosine rule;

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$⟹\cos \frac{2\pi }{3}=\frac{27+8\sqrt{3\sqrt{32}}-a^2}{2\times 4\sqrt{3\sqrt{32}}}⟹-4\sqrt{3\sqrt{32}}=27+8\sqrt{3\sqrt{32}}-a^2$

$⟹a^2=27+12\sqrt{3\sqrt{32}}$

$⟹a=\sqrt{27+12\sqrt{3\sqrt{32}}}$