In Fig.10.14, ∠ACB = 40º. Find ∠OAB.
Solution:
Given, ∠ACB = 40º
We have to find ∠OAB
We know that in a circle the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
∠AOB = 2∠ACB
∠AOB = 2(40º)
∠AOB = 80º ----------------------- (1)
Considering triangle AOB,
OA = OB = radius of the circle
We know that the angles opposite to the equal sides are equal.
∠OBA = ∠OAB ------------------- (2)
By angle sum property of the triangle,
∠AOB + ∠OBA + ∠OAB = 180º
From (1) and (2),
80º + ∠OAB + ∠OAB = 180º
2∠OAB = 180º - 80º
2∠OAB = 100º
∠OAB = 100º/2
Therefore, ∠OAB = 50º
✦ Try This: ABCD is a cyclic trapezium and AB∥CD. If AB is the diameter of the circle and ∠CAB = 30º, then the value of ∠ADC is
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.3 Problem 16
In Fig.10.14, ∠ACB = 40º. Find ∠OAB.
Summary:
A circle is a two-dimensional figure formed by a set of points that are at a constant or at a fixed distance (radius) from a fixed point (center) in the plane. In Fig.10.14, ∠ACB = 40º, then ∠OAB = 50º
☛ Related Questions:
- A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and ∠ADC = 130º. Find ∠BAC
- Two circles with centres O and O′ intersect at two points A and B. A line PQ is drawn parallel to OO . . . .
- In Fig.10.15, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. F . . . .