Let A B C be a triangle such that A B=A C . Suppose the tangent to...

Consider the Isosceles triangle with vertex $A, B, C$ such that $A B=A C$
By symmetry $\rightarrow$ circumcentre will lie on $y$ - axis $\therefore \mathrm{D} \equiv(0, \mathrm{k}) \rightarrow \text { circumcenter }$ Radius of circumcircle $=$\mathrm{DA}=\mathrm{DB}$=\mathrm{DC}$ $$\begin{aligned} & \therefore \sqrt{(a-k)^2}=\sqrt{k^2+1} \\ & \Rightarrow a^2+k^2-2 a k=k^2+1 \\ & \Rightarrow a^2-2 a k=1 \end{aligned}$$ Now Circumcircle: $x^2+(y-k)^2=k^2+1$ $\therefore$ Slope of tangent at B , $$\begin{aligned} & 2 x+2(y-k) y^{\prime}=0 \\ & \left.\Rightarrow y^{\prime}\right|_B=\left.\frac{-x}{y-k}\right|_B=\frac{1}{-k} \end{aligned}$$ Also, $\mathrm{m}_{\mathrm{AC}}=-\mathrm{a}$ By given condition, $(-\mathrm{a}) \times\left(-\frac{1}{\mathrm{k}}\right)=-1$ $\Rightarrow a=-k$ From (1) \& (2), $\mathrm{a}^2+2 \mathrm{a}^2=1$ $\Rightarrow \mathrm{a}= \pm \frac{1}{\sqrt{3}}$ If $\mathrm{a}=\frac{1}{\sqrt{3}}$ $\Rightarrow \mathrm{k}=-\frac{1}{\sqrt{3}}$ $$\begin{aligned} & \therefore \tan \theta=m_{A B}=\frac{1}{\sqrt{3}} \quad \Rightarrow \theta=30^{\circ} \\ & \Rightarrow \angle A B C=30^{\circ} \end{aligned}$$