In Fig. 10.20,O is the centre of the circle, ∠BCO = 30°. Find x and y.
Solution:
Given, O is the centre of the circle
∠BCO = 30°
We have to find x and y.
Join OB and AC
In triangle BOC,
OC = OB = radius of the circle
We know that the angles opposite to equal sides are equal
∠OBC = ∠OCB = 30°
By angle sum property of a triangle,
∠BOC + ∠OBC + ∠OCB = 180°
∠BOC = 180° - (∠OBC + ∠OCB)
= 180° - (30° + 30°)
= 180° - 60°
= 120°
So, ∠BOC = 120°
We know that an angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
∠BOC = 2∠BAC
∠BAC = 1/2 ∠BOC
∠BAC = 120°/2
∠BAC = 60°
AE is the angle bisector of ∠A
So, ∠BAE = ∠CAE = 30°
∠BAE = x
So, x = 30°
In triangle ABE,
By angle sum property of a triangle,
∠BAE + ∠EBA + ∠AEB = 180°
30° + ∠EBA + 90° = 180°
∠EBA = 180° - 120°
∠EBA = 60°
We know that the chord subtends an angle to the circle equal to half the angle subtended by it to the centre.
∠ABD = 1/2 ∠AOD
Now, ∠ABD + y = 60°
1/2 ∠AOD + y = 60°
90°/2 + y = 60°
45° + y = 60°
y = 60° - 45°
y = 15°
✦ Try This: In a circle of radius 45cm, an arc subtends an angle of 82° at the centre. Find the length of the arc and area of the sector.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.4 Problem 13
In Fig. 10.20,O is the centre of the circle, ∠BCO = 30°. Find x and y.
Summary:
In Fig. 10.20, O is the centre of the circle, ∠BCO = 30°. The values of x and y are 30° and 15°
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