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# In Fig.10.5, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to:

a. 30º

b. 60º

c. 90º

d. 45º

**Solution:**

We know that

Diameter subtends a right angle to the circle

∠BCA = 90º … (1)

It is given that

AC = BC

As the angles opposite to equal sides are equal

∠ABC = ∠CAB … (2)

In triangle ABC using the angle sum property

∠CAB + ∠ABC + ∠BCA = 180º

From equations (1) and (2)

∠CAB + ∠CAB + 90º = 180º

2∠CAB = 180 - 90

2∠CAB = 90

Dividing both sides by 2

∠CAB = 45º

Therefore, ∠CAB is equal to 45º.

**✦ Try This:** If ∠DEF = 60º, then ∠DOF is equal to:

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

**NCERT Exemplar Class 9 Maths Exercise 10.1 Problem 5**

## In Fig.10.5, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to: a. 30º, b. 60º, c. 90º, d. 45º

**Summary:**

In Fig.10.5, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to 45º

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