Q21. In the given figure, O is the centre of a circle and ∠BCO = 30°. Find x and y.
Answers
ANSWER:
Given:
- ∠BCO = 30°
- ∠AOD = ∠AEC = 90°
To Find:
- Value of x and y
Construction(refer attachment):
- Join A to D
- Join C to D
Solution:
We are given that,
⇒ ∠AOD = 90°
We can see that, ∠AOD is made from the chord AD at centre.
But, ∠ABD is also made from the chord AD but at circumference.
We know that, the angle subtended by a chord at the centre is double than the angle subtended by the same chord at the circumference.
So,
⇒∠ABD = 1/2 × ∠AOD = 1/2 × 90° = 45°
Hence,
⇒ ∠ABD = 45°. --------(1)
We are given that,
⇒ ∠AOD = ∠AEC = 90°
This means AE OD and AE BC
So, OD // BC.
Now, as OD // BC and OC is the transversal.
Therefore,
⇒ ∠BCO = ∠COD = 30° (Alternate interior angles)
Hence,
⇒ ∠COD = 30° --------(2)
Now, we take chord CD.
We can see that, ∠COD is made from the chord CD at centre.
But, ∠CBD is also made from the chord CD but at circumference.
We know that, the angle subtended by a chord at the centre is double than the angle subtended by the same chord at the circumference.
So,
⇒∠CBD = 1/2 × ∠COD = 1/2 × 30° = 15°
Hence,
⇒ ∠CBD = 15°. --------(3)
But,
⇒ ∠CBD = y
Therefore,
⇒ y = 15°
Now, we take ∆ABE.
We can see that,
⇒ ∠ABE = ∠ABD + ∠CBD
So, from (1) & (3),
⇒ ∠ABE = 45° + 15°
Hence,
⇒ ∠ABE = 60° --------(4)
And,
As, ∠AEC = 90°
So,
⇒ ∠AEC = ∠AEB = 90° (linear pair)
Hence,
⇒∠AEB = 90° --------(5)
Now, we know that, by Angle sum Property if triangles,
⇒ ∠AEB + ∠ABE + ∠BAE = 180°
So, from (4) & (5),
⇒ 90° + 60° + ∠BAE = 180°
⇒ 150° + ∠BAE = 180°
So,
⇒ ∠BAE = 180° - 150°
⇒ ∠BAE = 30°
But,
⇒ ∠BAE = x
So,
⇒ x = 30°
Therefore, the value of x and y are 30° and 15° respectively.
