Let ABC be an isosceles triangle in which A i | vedclass

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(B) Given $A = (-1, 0)$ and $B$ is on the positive $x$-axis,let $B = (b, 0)$ where $b > 0$.Since $AB = AC$ and $\angle A = 120^{\circ}$,the base angle

Vedclass · Accepted Answer

$36$

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(B) Given $A = (-1, 0)$ and $B$ is on the positive $x$-axis,let $B = (b, 0)$ where $b > 0$.Since $AB = AC$ and $\angle A = 120^{\circ}$,the base angles are $\angle B = \angle C = 30^{\circ}$.Using the sine rule in $	riangle ABC$: $\frac{AB}{\sin 30^{\circ}} = \frac{BC}{\sin 120^{\circ}}$.$\frac{AB}{1/2} = \frac{4\sqrt{3}}{\sqrt{3}/2}$ $\Rightarrow 2AB = 8$ $\Rightarrow AB = 4$.Since $A = (-1, 0)$ and $B = (b, 0)$ with $b > 0$,$AB = |b - (-1)| = b + 1 = 4$,so $b = 3$.Thus,$B = (3, 0)$.The slope of line $BC$ is $	an(180^{\circ} - 30^{\circ}) = -	an 30^{\circ} = -\frac{1}{\sqrt{3}}$.The equation of line $BC$ is $y - 0 = -\frac{1}{\sqrt{3}}(x - 3) \Rightarrow x + \sqrt{3}y = 3$.Solving $x + \sqrt{3}y = 3$ and $y = x + 3$:Substitute $x = y - 3$ into the first equation: $(y - 3) + \sqrt{3}y = 3$ $\Rightarrow y(1 + \sqrt{3}) = 6$ $\Rightarrow y = \frac{6}{\sqrt{3} + 1} = 3(\sqrt{3} - 1)$.Then $x = 3(\sqrt{3} - 1) - 3 = 3\sqrt{3} - 6$.So $\alpha = 3(\sqrt{3} - 2)$ and $\beta = 3(\sqrt{3} - 1)$.$\frac{\beta^4}{\alpha^2} = \frac{[3(\sqrt{3} - 1)]^4}{[3(\sqrt{3} - 2)]^2} = \frac{81(3 - 2\sqrt{3} + 1)^2}{9(3 - 4\sqrt{3} + 4)} = \frac{9(4 - 2\sqrt{3})^2}{7 - 4\sqrt{3}} = \frac{9 \cdot 4(2 - \sqrt{3})^2}{7 - 4\sqrt{3}} = \frac{36(4 - 4\sqrt{3} + 3)}{7 - 4\sqrt{3}} = \frac{36(7 - 4\sqrt{3})}{7 - 4\sqrt{3}} = 36$.

Let $ABC$ be an isosceles triangle in which $A$ is at $(-1, 0)$,$\angle A = \frac{2\pi}{3}$,$AB = AC$ and $B$ is on the positive $x$-axis. If $BC = 4\sqrt{3}$ and the line $BC$ intersects the line $y = x + 3$ at $(\alpha, \beta)$,then $\frac{\beta^4}{\alpha^2}$ is :

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