2016 Question Paper Solution | Question 1

This 1-mark question is from the chapter Circles. Concept tested: Tangents and circles and alternate segment theorem.This question appeared in 2016 CBSE Board paper..

Question 1: PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA.

Video Explanation

Explanatory Answer | Q1 2016 Board Paper

AB is a diameter to the circle and O is the centre of the circle.
Join OC.

OA and OC are radii to the circle. So, OA = OC
∴ ΔAOC is an isosceles triangle.

In an isosceles triangle, two sides and correspondingly two opposite angles are equal.
Sides OC and OA are equal. Angles opposite the two sides ∠OAC and ∠OCA will be equal.
∠OAC = ∠CAB = 30°
So, ∠OAC = ∠OCA = 30°

∠OCP is the angle between the tangent and the radius.
∴ ∠OCP = 90°

∠ACP = ∠OCP - ∠OCA = 90° – 30° = 60°

Alternative Appraoch

∠ACP is the angle between a tangent PQ and a chord AC.
By Alternate Segment Theorem, the angle between a tangent and a chord will be the same as the angle subtended by the chord in the alternate segment

Angle subtended by the chord in the alternate segment is ∠ABC

Because AB is the diameter to the circle, ∠ACB = 90°
Therefore, ∠ABC = 180 - ∠ACB - ∠CAB
∠ABC = 180 - 90 - 30 = 60°

Therefore, ∠ACP = 60°