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
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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