Question

$\large\mathbf{HEY\: QUESTION❤}$




From an external point P, tangents PA and PB are drawn to a circle with centre 0. If ∠PAB = 50°, then find ∠AOB?


★ REQUIRED QUALITY ANSWER​

Answer 1

Step-by-step explanation:

Answer

Given, PA=PB, and ∠PAB=50

o

To prove: Measure of ∠AOB

Proof: PA=PB (∵ given)

∠PBA=∠PAB (∵ angle opposite sides are equal)

∴∠PBA=50

o

.

In ΔPAB

∠PBA+∠PAB+∠APB=180

o

(∵ A.S.P)

∠APB=180

o

−100

o

∠APB=80

o

∠AOB+∠APB=180

o

(∴ supplementary angles)

∠AOB=180

o

−80

o

∠AOB=100

o

∴ Hence proved.

stay safe✌

Hit the❤

Answer 2

Answer:

Given, PA=PB, and ∠PAB=50 degree

To prove: Measure of ∠AOB

Proof: PA=PB (∵ given)

∠PBA=∠PAB (∵ angle opposite sides are equal)

∴∠PBA=50 degree

In ΔPAB

∠PBA+∠PAB+∠APB=180 degree

(∵ A.S.P)

∠APB=180 degree −100 degree

∠APB=80 degree

∠AOB+∠APB=180 degree

(∴ supplementary angles)

∠AOB=180 degree −80 degree

∠AOB=100 degree

∴ Hence proved.

solution