Hii...Guysss plzzzz answer it ....Plzz answer Q20 plzzzz...... Created by Sanjana111111. Math

Hii...Guysss plzzzz answer it ....Plzz answer Q20 plzzzz...

sushant2505 HELLO DEAR,

JOIN. A. TO. B

WE HAVE

op = diameter

or

OP+OQ=DIAMETER

Or,

Radius + PQ = diameter (that is OQ=radius)

Or,

PQ = diameter – radius

Or,

PQ = radius

Or,

OQ = PQ = radius

Thus OP is the hypotenuse of the right angle
triangle AOP.

So, in ∆AOP sin of angle P = OP /AO = 1/2

So, P = 30°

Hence, APB = 60°

Now, As in ∆AOP, AP=AB

So, PAB = PBA =60°

Hence,

∆ABP is equilateral triangle.

I HOPE ITS HELP YOU DEAR,
THANKS

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Verified answer

PA and PB are the tangents to the circle.

∴ OA ⊥ PA

⇒ ∠OAP = 90°

In ΔOPA,

sin ∠OPA = OA OP = r/2r [Given OP is the diameter of the circle]

⇒ sin ∠OPA = 1/2 = sin 30 ⁰

⇒ ∠OPA = 30°

Similarly, it can be proved that ∠OPB = 30°.

Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°

In ΔPAB,

PA = PB [lengths of tangents drawn from an external point to a circle are equal]

⇒∠PAB = ∠PBA ............(1) [Equal sides have equal angles opposite to them]

∠PAB + ∠PBA + ∠APB = 180° [Angle sum property]

⇒∠PAB + ∠PAB = 180° – 60° = 120° [Using (1)]

⇒2∠PAB = 120°

⇒∠PAB = 60° .............(2)

From (1) and (2)

∠PAB = ∠PBA = ∠APB = 60°

∴ ΔPAB is an equilateral triangle.

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