Consider a triangle $\mathrm{ABC}$ having the vertices $\mathrm{A}(1,2), \mathrm{B}(\alpha, \beta)$ and $\mathrm{C}(\gamma, \delta)$ and angles $\angle A B C=\frac{\pi}{6}$ and $\angle B A C=\frac{2 \pi}{3}$. If the points $\mathrm{B}$ and $\mathrm{C}$ lie on the line $y=x+4$, then…

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Consider a triangle $\mathrm{ABC}$ having the vertices $\mathrm{A}(1,2), \mathrm{B}(\alpha, \beta)$ and $\mathrm{C}(\gamma, \delta)$ and angles $\angle A B C=\frac{\pi}{6}$ and $\angle B A C=\frac{2 \pi}{3}$. If the points $\mathrm{B}$ and $\mathrm{C}$ lie on the line $y=x+4$, then $\alpha^2+\gamma^2$ is equal to ________

Aicharya · Accepted Answer

Equation of line passes through point $\mathrm{A}(1,2)$ which makes angle $\frac{\pi}{6}$ from $y=x+4$ is $\begin{aligned} & y-2=\frac{1 \pm \tan \frac{\pi}{6}}{1 \mp \tan \frac{\pi}{6}}(x-1) \\ & y-2=\frac{\sqrt{3} \pm 1}{\sqrt{3} \mp 1}(x-1) \\ & \begin{array}{c|c} \oplus & \oplus \\ \mathrm{y}-2=(2+\sqrt{3})(\mathrm{x}-1) & \mathrm{y}-2=(2-\sqrt{3})(\mathrm{x}-1) \end{array} \\ & \text { solve with } \mathrm{y}=\mathrm{x}+4 \quad \text { solve with } \mathrm{y}=\mathrm{x}+4 \\ & x+2=(2+\sqrt{3}) x-2-\sqrt{3} \quad x+2=(2-\sqrt{3}) x-2+\sqrt{3} \\ & x=\frac{4+\sqrt{3}}{1+\sqrt{3}} \quad x=\frac{4-\sqrt{3}}{1-\sqrt{3}} \\ & \end{aligned}$ $\begin{aligned} & \alpha^2+\gamma^2=\left(\frac{4+\sqrt{3}}{1+\sqrt{3}}\right)^2+\left(\frac{4-\sqrt{3}}{1-\sqrt{3}}\right)^2 \\ & \alpha^2+\gamma^2=14\end{aligned}$

Consider a triangle $\mathrm{ABC}$ having the vertices $\mathrm{A}(1,2), \mathrm{B}(\alpha, \beta)$ and…

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