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In Fig. 9.8, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to
a. 20°
b. 40°
c. 35°
d. 45°

In Fig. 9.8, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to

Solution:

Given, PQR is the tangent at Q to a circle whose centre is O.

AB is a chord parallel to PR.

Also, ∠BQR = 70°

We have to find the measure of angle AQB.

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

So, ∠OQR = ∠OQP = 90°

From the figure,

∠OQR = ∠OQB + ∠BQR

90° = ∠OQB + 70°

∠OQB = 90° - 70°

∠OQB = 20°

Let the line from O meet AB at C.

Considering triangles AQM and BQM,

We know that the perpendicular from the centre of the circle to the chord bisects the chord.

So, AC = BC

∠QMA = ∠QMB = 90°

QM = QM = common side

By SAS criterion, the triangles AQM and BQM are similar.

So, ∠MQA = ∠MQB = 20°

From the figure,

∠AQB = ∠MQA + ∠MQB

∠AQB = 20° + 20°

∠AQB = 40°

Therefore, the measure of the angle AQB is 40°

✦ Try This: In the given figure, O is the center of the circle. PQ is the tangent to the circle at A. If ∠PAB = 58°, then ∠AQB equals

In the given figure, O is the center of the circle. PQ is the tangent to the circle at A. If ∠PAB = 58°, then ∠AQB equals

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

NCERT Exemplar Class 10 Maths Exercise 9.1 Problem 10

In Fig. 9.8, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to a. 20°, b. 40°, c. 35°, d. 45°

Summary:

In Fig. 9.8, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to 40°

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