Question

(i) In Figure (1), O is the centre of the circle. If $\angle \text{OAB}={40}^{\circ }$ and $\angle \text{OCB}={30}^{\circ }$, find $\angle \text{AOC}$.

(ii) In Figure (2), A, B and C are three points on the circle with centre O such that $\angle \text{AOB}={90}^{\circ }$ and $\angle \text{AOC}={110}^{\circ }$. Find $\angle \text{BAC}$.

Answer 1

ANSWER:
(i) Join BO.
In Δ BOC, we have:
OC = OB (Radii of a circle)
⇒ ∠ OBC = ∠ OCB
∠ OBC = 30 °
...(i)
In Δ BOA, we have:
OB = OA (Radii of a circle)
⇒ ∠ OBA = ∠ OAB [∵ ∠ OAB = 40 °]
⇒ ∠ OBA = 40 °
...(ii)
Now, we have:
∠ ABC = ∠ OBC + ∠ OBA
= 30 ° + 40 ° [From (i) and (ii)]
∴ ∠ ABC = 70 °
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on
the circumference.
i.e., ∠ AOC = 2 ∠ ABC
= (2 × 70 ° ) = 140 °
(ii)
Here, ∠ BOC = {360 ° - (90 ° + 110 ° )}
= (360 ° - 200 °) = 160 °
We know that ∠ BOC = 2 ∠ BAC
⇒∠BAC=∠BOC/2=160°/2=80°
Hence, ∠ BAC = 80 °