# 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, . . . .

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