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# In Fig. 10.7, if ∠DAB = 60º, ∠ABD = 50º, then ∠ACB is equal to:

a. 60º

b. 50º

c. 70º

d. 80º

**Solution:**

It is given that

∠DAB = 60º

∠ABD = 50º

As ∠ADB = ∠ACB … (1) [angles in same segment of a circle are equal]

In triangle ABD

Using the angle sum property

∠ABD + ∠ADB + ∠DAB = 180º

Substituting the values

50 + ∠ADB + 60 = 180º

By further calculation

∠ADB + 110 = 180

So we get

∠ADB = ∠ACB = 70º

Therefore, ∠ACB is equal to 70º.

**✦ Try This:** If ∠DAB = 50º, ∠ABD = 40º, then ∠ACB is equal to:

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 10

**NCERT Exemplar Class 9 Maths Exercise 10.1 Problem 7**

## In Fig. 10.7, if ∠DAB = 60º, ∠ABD = 50º, then ∠ACB is equal to: a. 60º, b. 50º, c. 70º, d. 80º

**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. 10.7, if ∠DAB = 60º, ∠ABD = 50º, then ∠ACB is equal to 70º

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