Paragraph: A triangle ABC is inscribed in the circle $x^2+y^2=100 . \mathrm{B}$ and C have coordinates $(6,8)$ and $(-8,6)$ respectively.
Question: What is $\angle \mathrm{BAC}$ equal to?

• A

  $\frac{\pi}{2}$
• B

  $\frac{\pi}{3}$ or $\frac{2 \pi}{3}$
• Correct answer
  C

  $\frac{\pi}{4}$ or $\frac{3 \pi}{4}$
• D

  $\frac{\pi}{6}$ or $\frac{5 \pi}{6}$

Solution

Let O is centre of circle $\mathrm{x}^2+\mathrm{y}^2=100$, radius $=\mathrm{OB}=$ $\mathrm{OC}=10$ units. $\begin{aligned} & \mathrm{BC}=\sqrt{(6+8)^2+(8-6)^2}=\sqrt{200}=10 \sqrt{2} \\ & \mathrm{OB}^2+\mathrm{OC}^2=\sqrt{100+100}=\sqrt{200}=10 \sqrt{2} \\ & \text { In } \triangle \mathrm{BOC}, \mathrm{OB}^2+\mathrm{OC}^2=\mathrm{BC}^2=(10 \sqrt{2})^2=200 \end{aligned}$
So $\angle \mathrm{BOC}=90^{\circ}$ (By converse of Pythagoras theorem) $\therefore \quad \angle \mathrm{BAC}=45^{\circ} \text { or } 180^{\circ}-45^{\circ}=\frac{3 \pi}{4}$ (Angle subtended by a chord at centre is double than the angle at any point on circle) Option (c) is correct.

Hence, the correct answer is option: 'C'.