$I(1,0)$ is the centre of in circle of triangle $ABC$, the equation of $BI$ is $x - 1 = 0$ and equat

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$I(1,0)$ is the centre of in circle of triangle $ABC$, the equation of $BI$ is $x - 1 = 0$ and equation of $CI$ is $x - y - 1 = 0$, then angle $BAC$ is :

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Sol. [Key Concept: Incentre $I$ with the two bisector lines $BI,CI$ gives angles $B,C$; then $\angle A = \pi - (B + C)$.]$$\angle BIC = \frac{3\pi}{4} = \frac{\pi}{2} + \frac{A}{2} \Rightarrow A = \frac{\pi}{2} $$Hence, $\angle BAC = \frac{\pi}{2}$

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