The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA=110º , find ∠CBA [see Fig. 9.21]

Solution:

Given, O is the centre of a circle.

AB is the diameter of the circle

AB is extended to P and PC is a tangent to the circle at point C.

Given, ∠PCA = 110º

We have to find ∠CBA

We know that angle in a semicircle is always equal to 90°

So, ∠BCA = 90°

We know that the radius of a circle is perpendicular to the tangent at the point of contact.

i.e., OC ⟂ PC

So, ∠OCP = 90°

From the figure,

∠PCA = ∠BCA + ∠PCB

110° = 90° + ∠PCB

∠PCB = 110° - 90°

∠PCB = 20°

By alternate segment theorem,

We know that the angle between the tangent and the chord of a circle is equal to the angle made by the chord in the alternate segment.

∠PCB = ∠CAB

So, ∠CAB = 20°

Considering triangle ABC,

We know that the sum of all three interior angles of a triangle is equal to 180°

∠BCA + ∠CBA + ∠CAB = 180°

90° + ∠CBA + 20° = 180°

110° + ∠CBA = 180°

∠CBA = 180° - 110°

Therefore, ∠CBA = 70°

✦ Try This: In the figure, PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA.

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

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NCERT Exemplar Class 10 Maths Exercise 9.4 Problem 12

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA=110º , find ∠CBA [see Fig. 9.21]

Summary:

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA=110º, then ∠CBA = 70°

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