Math, asked by sreevathsa99, 3 months ago

two tangents pa and pb are drawn from external point. point of p of a circle at centre o . the no of circles passing through p, a, o and b are

Answers

Answered by ameyntawade

Answer:

Step-by-step explanation:

the answer is one I don't want to draw diagram because my diagram sense is worst

Answered by Anonymous

$\huge \red{AηsωeR}$ ✍

Here PA, PB are two tangents from external point P

Since tangent lengths from the same external points are equal

PA = PB

We are given that OP=2r

Also, we know that tangent is perpendicular to the radius through the point of Contact

So ∠PAO & ∠PBO are 90° each.

Since ΔPAO & ΔPBO are congruent by S.S.S Congruence

So ∠1 = ∠2

Let ∠1 = ∠2 = ϴ

Now we will use Trigonometry to find unknown angle ϴ

In right triangle ΔPAO,

Sin ϴ = perpendicular / Hypotenuse = OA/OP = r / 2r = 1/2

Sinϴ = 1/2

⇒ ϴ = 30°

So ∠1 = ∠2 = 30°

⇒ ∠APB = ∠1 + ∠2=60°

Now In ΔAPB,

PA = PB

⇒ ∠PAB = ∠PBA

As angles opposite to equal sides are equal

∠APB +∠PAB + ∠PBA =180°

60° + ∠PAB + ∠PBA =180°

∠PAB + ∠PBA =180° -60° = 120°

2∠PAB =120°

∠PAB =60°= ∠PBA

Since all the angles of ΔAPB are 60°

Thus ΔAPB is an equilateral triangle.

Second Method: Without using Trigonometry

We know that tangent is perpendicular to the radius through the point of Contact

So ∠PAO & ∠PBO are 90° each.

Here we will join AQ. As we are given that OP=2r

And OQ = r because it is the radius

So Q is the midpoint of hypotenuse OP

Now here we will be using a theorem that in right triangle, the line joining right vertex to the middle point of the hypotenuse is half the hypotenuse.

So AQ=r

Since in ΔAOQ, OA=OQ=AQ=r

So ΔAOQ is an equilateral triangle

Thus

∠AOQ=60°

Hence in ΔPAO, by angle sum property ∠1 =30°