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Let $A B C$ be an isosceles triangle in which $A$ is at $(-1,0), \angle A=\frac{2 \pi}{3}, A B=A C$ and $B$ is on the positive $\mathrm{x}$-axis. If $\mathrm{BC}=4 \sqrt{3}$ and the line $\mathrm{BC}$ intersects the line $y=x+3$ at $(\alpha, \beta)$, then $\frac{\beta^4}{\alpha^2}$ is :
$85$
$36$
$45$
$75$
Solution
$\frac{\mathrm{c}}{\sin 30^{\circ}}=\frac{4 \sqrt{3}}{\sin 120^{\circ}}$ [By sine rule]
$2 c=8 \Rightarrow c=4$
$ \mathrm{AB}=|(\mathrm{b}+1)|=4 $
$ \mathrm{~b}=3, \mathrm{~m}_{\mathrm{AB}}=0 $
$ \mathrm{~m}_{\mathrm{BC}}=\frac{-1}{\sqrt{3}} $
$ B C:-y=\frac{-1}{\sqrt{3}}(x-3) $
$ \sqrt{3} \mathrm{y}+\mathrm{x}=3 $
$ \text { Point of intersection : } y=x+3, \sqrt{3} y+x=3 $
$ (\sqrt{3}+1) \mathrm{y}=6 $
$ \mathrm{y}=\frac{6}{\sqrt{3}+1} $
$ x=\frac{6}{\sqrt{3}+1}-3 $
$ =\frac{6-3 \sqrt{3}-3}{\sqrt{3}+1} $
$ =3 \frac{(1-\sqrt{3})}{(1+\sqrt{3})}=\frac{-6}{(1+\sqrt{3})^2} $
$ \frac{\beta^4}{\alpha^2}=36 $
