In Fig. 10.6, if ∠OAB = 40º, then ∠ACB is equal to :
a. 50º
b. 40º
c. 60º
d. 70°
Solution:
In triangle OAB
OA = OB (radius of a circle)
∠OAB = ∠OBA
∠OBA = 40º (angles opposite to equal sides are equal)
Using the angle sum property
∠AOB + ∠OBA + ∠BAO = 180º
Substituting the values
∠AOB + 40º + 40º = 180º
By further calculation
∠AOB + 80º = 180º
∠AOB = 180º - 80º
∠AOB = 100º
As 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
Substituting the values
100º = 2 ∠ACB
Dividing both sides by 2
∠ACB = 50º
Therefore, ∠ACB is equal to 50º.
✦ Try This: If ∠OAB = 50º, then ∠ACB is equal to :
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.1 Problem 6
In Fig. 10.6, if ∠OAB = 40º, then ∠ACB is equal to : a. 50º, b. 40º, c. 60º, d. 70°
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. 10.6, if ∠OAB = 40º, then ∠ACB is equal to 50º
☛ Related Questions:
- In Fig. 10.7, if ∠DAB = 60º, ∠ABD = 50º, then ∠ACB is equal to: a. 60º, b. 50º, c. 70º, d. 80º
- ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = . . . .
- In Fig. 10.8, BC is a diameter of the circle and ∠BAO = 60º. Then ∠ADC is equal to : a. 30º, b. 45º, . . . .