In Fig. 10.16, ∠OAB = 30º and ∠OCB = 57º. Find ∠BOC and ∠AOC.
Solution:
Given, ∠OAB = 30º
∠OCB = 57º
We have to find ∠BOC and ∠AOC
Considering triangle AOB,
OA = Ob = radius of the circle
We know that the angles opposite to the equal sides are equal.
∠OBA = ∠BAO
Given, ∠OAB = 30º
So, ∠OBA = 30º
In triangle AOB,
By angle sum property of a triangle,
∠AOB + ∠OBA + ∠OAB = 180º
∠AOB + 30º + 30º = 180º
∠AOB + 60º = 180º
∠AOB = 180º - 60º
∠AOB = 120º -------------------------- (1)
Considering triangle OCB,
OC = OB = radius of the circle
We know that the angles opposite to the equal sides are equal.
∠OBC = ∠OCB
Given, ∠OCB = 57º
So, ∠OBC = 57º
By angle sum property of a triangle,
∠COB + ∠OBC + ∠OCB = 180º
∠COB + 57º + 57º = 180º
∠COB + 114º = 180º
∠COB = 180º - 114º
∠COB = 66º ------------------------ (2)
From the figure,
∠AOB = ∠AOC + ∠COB
From (1) and (2),
120º = ∠AOC + 66º
∠AOC = 120º - 66º
Therefore, ∠AOC = 54º
✦ Try This: ABCD is a cyclic quadrilateral in which ∠DAC = 27°; ∠DBA = 50° and ∠ADB = 33°. Calculate ∠DBC
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.3 Problem 20
In Fig. 10.16, ∠OAB = 30º and ∠OCB = 57º. Find ∠BOC and ∠AOC.
Summary:
The fixed point is called the origin or center of the circle and the fixed distance of the points from the origin is called the radius. In Fig. 10.16, ∠OAB = 30º and ∠OCB = 57º, then ∠BOC = 66º and ∠AOC = 54º
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