Question

From an external point P, tangent PA and PB are drawn to a circle with centre O. angle PAB=50 then find angle AOB. please solve this question on ppr

Answer 1

We know that PA= PB (Tangents drawn from the same external point)
So, Angle PAB = Angle PBA= 50° (Given)

Now, ∠PAO + ∠PBO= 90° (Tangents are perpendicular to radius)
∠BAO= ∠PAO - ∠PAB
= 90- 50
= 40°

In triangle AOB
∠AOB + ∠ABO + ∠BAO = 180°
∠AOB + 40 + 40 = 180
∠AOB= 180- 80
∠AOB= 100°

Answer 2

PA &PB are two tangents
Angle PAB=50°
Angle PAO=90(tangents radius thm)
angle BAO=90-50=40

In∆AOB,
OA=OB(radius)
therefore,angle OAB=OBA(angles opposite to equal sides are equal)
ANGLE OBA=40

In∆AOB,
Angle AOB+angle BAO+angle ABO=180
angle AOB =180-40-40
angle AOB=100

I hope it will help you
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