In Fig. 10.21, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB.

Solution:

Given, O is the centre of the circle

BD = OD

CD ⊥ AB

We have to find ∠CAB

In triangle OBD,

Given, BD = OD

OD = OB = radius of the circle

OB = OD = BD

Therefore, ODB is an equilateral triangle

We know that in an equilateral triangle each angle is equal to 60 degrees.

∠BOD = ∠OBD = ∠ODB = 60°

In triangles MBC and MBD,

Common side = MB

∠CMB = ∠BMD = 90°

We know that in a circle any perpendicular drawn on a chord also bisects the chord.

CM = MD

SAS criterion states that if two sides of one triangle are proportional to two sides of another triangle and their included angles are congruent, then the triangles are similar.

By SAS criterion, the triangles MBC and MBD are similar.

The Corresponding Parts of Congruent Triangles are Congruent (CPCTC) theorem states that when two triangles are similar, then their corresponding sides and angles are also congruent or equal in measurements.

By CPCTC,

∠MBC = ∠MBD

∠MBC = ∠OBD = 60°

Since AB is a diameter of a circle

∠ACB = 90°

In triangle ACB,

By angle sum property of a triangle,

∠CAB = ∠CBA + ∠ACB = 180°

∠CAB + 60° + 90° = 80°

∠CAB + 60° = 180° - 90°

∠CAB = 90° - 60°

Therefore, ∠CAB = 30°

✦ Try This: In the given diagram, O is the center of the circle and CD is a tangent. ∠CAB and ∠ACD are supplementary to each other. ∠OAC = 30°. Find the value of ∠OCB?

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

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NCERT Exemplar Class 9 Maths Exercise 10.4 Problem 14

In Fig. 10.21, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB.

Summary:

In Fig. 10.21, O is the centre of the circle, BD = OD and CD ⊥ AB. The value of ∠CAB = 30°

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☛ Related Questions:

• Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD
• AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of . . . .
• In Fig. 10.20,O is the centre of the circle, ∠BCO = 30°. Find x and y