In Fig. 9.7, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to
a. 25°
b. 30°
c. 40°
d. 50°

Solution:

Given, PA and PB are tangents to the circle with centre O.

Also, ∠APB = 50°

We have to find the measure of angle OAB.

We know that the tangents through an external point to a circle are equal.

So, tangents PA = PB

In triangle PAB,

Since, PA = PB, PAB is an isosceles triangle.

In an isosceles triangle, two sides and two angles are equal.

i.e., PA = PB

Also, ∠PAB = ∠PBA —------------------ (1)

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

Now, ∠APB + ∠PAB + ∠PBA = 180°

From (1),

50° + ∠PAB + ∠PAB = 180°

2∠PAB = 180° - 50°

2∠PAB = 130°

∠PAB = 130°/2

∠PAB = 65°

since ∠PAB = ∠PBA, ∠PBA = 65°

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

So, ∠PAO = ∠PBO = 90°

From the figure,

∠PAO = ∠PAB + ∠OAB

90° = 65° + ∠OAB

∠OAB = 90° - 65°

∠OAB = 25°

Therefore, the measure of the angle OAB is 25°

✦ Try This: In Fig. 9.7, if TP and TQ are tangents to the circle with centre O such that ∠PTQ = 70°, then ∠OPQ is equal to

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

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NCERT Exemplar Class 10 Maths Exercise 9.1 Problem 8

In Fig. 9.7, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to a. 25°, b. 30°, c. 40°, d. 50°

summary:

In Fig. 9.7, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to 25°

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