In a circle with centre O, chords PR and QS meet at the point T, when produced, and PQ is a diameter. If \(\angle\)ROS = 42º, then the measure of \(\angle\)PTQ is

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Question

This question was previously asked in

SSC CGL 2021 Tier-I (Held On : 11 April 2022 Shift 2)

Answer 1

Given:

∠ROS = 42º

Concept used:

The sum of the angles of a triangle = 180°

Exterior angle = Sum of opposite interior angles

Angle made by an arc at the centre = 2 × Angle made by the same arc at any point on the circumference of the circle

Calculation:

Join RQ and RS

According to the concept,

∠RQS = ∠ROS/2

⇒ ∠RQS = 42°/2 = 21° .....(1)

Here, PQ is a diameter.

So, ∠PRQ = 90° [∵ Angle in the semicircle = 90°]

In ΔRQT, ∠PRQ is an exterior angle

So, ∠PRQ = ∠RTQ + ∠TQR

⇒ 90° = ∠RTQ + 21° [∵ ∠TQR = ∠RQS = 21°]

⇒ ∠RTQ = 90° - 21° = 69°

⇒ ∠PTQ = 69°

∴ The measure of ∠PTQ is 69°

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