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

☛ Related Questions:

• In Fig.10.4, if ∠ABC = 20º, then ∠AOC is equal to: a. 20º, b. 40º, c. 60º, d. 10º
• In Fig.10.5, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to: a. 30º, b. 60º, . . . .
• In Fig. 10.6, if ∠OAB = 40º, then ∠ACB is equal to : a. 50º, b. 40º, c. 60º, d. 70°