Question

Let an acute-angled triangle $ABC$ be inscribed in a circle whose centre is the origin. Let $B=(3,4)$ and $C=(-4,3)$. Then $\angle BAC$ is

• A. $\frac{\pi }{5}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $\frac{\pi }{4}$

Given that a triangle $ABC$ is inscribed in a circle whose center is origin,

As $\angle BAC=\theta ⟹\angle BOC=2\theta$ from the properties of inscribed triangles,

We know that the slopes of $OB$ and $OC$ are $\frac{4}{3}$ and $-\frac{3}{4}$,

the product of their slopes is $-1$,

$\therefore$ the $\angle BOC={90}^o⟹\angle BAC={45}^o=\frac{\pi }{4}$