In the given figure, O is the centre of a circle.
Question:

In the given figure, O is the centre of a circle. If AOB = 100° and ∠AOC = 90°, then ∠BAC = ?
(a) 85°

(b) 80°
(c) 95°
(d) 75°

 

Solution:

(a) 85°
We have:
∠BOC + ∠BOA + ∠AOC = 360°
⇒ ∠BOC + 100° + 90° = 360°
⇒ ∠BOC = (360° – 190°) = 170°

$\therefore \angle \mathrm{BAC}=\left(\frac{1}{2} \times \angle \mathrm{BOC}\right)=\left(\frac{1}{2} \times 170^{\circ}\right)=85^{\circ}$

 

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