Question

In the given figure, O is the centre of the circle and $\angle BCO={30}^{\circ }$. Find x and y.

Answer 1

ANSWER:
In the given figure, OD is parallel to BC.
∴ ∠ BCO = ∠ COD (Alternate interior angles)
⇒ ∠COD=30°
...(1)
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the
remaining part of the circle.
Here, arc CD subtends ∠ COD at the centre and ∠ CBD at B on the circle.
∴ ∠ COD = 2∠ CBD
⇒ ∠ CBD=30°/2=15°
∴ y=15°
(from (1))
...(2)
Also, arc AD subtends ∠ AOD at the centre and ∠ ABD at B on the circle.
∴ ∠ AOD = 2∠ ABD
⇒ ∠ ABD=90°/2=45°
...(3)
In ∆ ABE,
x + y + ∠ ABD + ∠ AEB = 180 ∘
⇒ x + 15 ∘ + 45 ∘ + 90 ∘ = 180 ∘
⇒ x = 180 ∘ − (90 ∘ + 15 ∘ + 45 ∘ )
⇒ x = 180 ∘ − 150 ∘
⇒ x = 30 ∘
(Sum of the angles of a triangle)
(from (2) and (3))
Hence, x = 30 ∘ and y = 15 ∘ .