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

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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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☛ Related Questions:

• In Fig. 10.6, if ∠OAB = 40º, then ∠ACB is equal to : a. 50º, b. 40º, c. 60º, d. 70°
• 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 = . . . .