From a point P, two tangents PA and PB are drawn to a circle with centre O

Question 16 Circles Exercise 10.2

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Answer:

Solution:

Given: From a point P outside the circle with centre O, PA and PB are the tangents to the circle such that OP is diameter.

And, AB is joined.

Required to prove: APB is an equilateral triangle

Construction: Join OP, AQ, OA

Proof:

We know that, OP = 2r

⇒ OQ + QP = 2r

⇒ OQ = QP = r

Now in right ∆OAP,

OP is its hypotenuse and Q is its mid-point

Then, OA = AQ = OQ

(mid-point of hypotenuse of a right triangle is equidistance from its vertices)

Thus, ∆OAQ is equilateral triangle. So, ∠AOQ = 60°

Now in right ∆OAP,

∠APO = 90° – 60° = 30°

⇒ ∠APB = 2 ∠APO = 2 x 30° = 60°

But PA = PB (Tangents from P to the circle)

⇒ ∠PAB = ∠PBA = 60°

Hence ∆APB is an equilateral triangle.

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Last updated at Jan. 18, 2022 by Teachoo

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Last updated at Jan. 18, 2022 by Teachoo

This video is only available for Teachoo black users

Support Teachoo in making more (and better content) - Monthly, 6 monthly, yearly packs available!