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Question

In the given figure, O is the centre of the circle and ∠BCO = 30°. Find x and y.

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Solution

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° (from (1))

y=15° ...(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 (Sum of the angles of a triangle)
⇒ x + 15 + 45 + 90 = 180 (from (2) and (3))
⇒ x = 180 − (90+ 15 + 45)
⇒ x = 180 − 150
⇒ x = 30

Hence, x = 30 and y = 15.

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