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
Anonymous:
Is the answer is 160
100 it should be
You minus 50 from 90
Then it's correct
Yeah, but we need to find angle AOB
Angle BAO= 40
I have doubt in it, as i was doing with another method
you answer is correct
Answers
Answered by
4
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°
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°
Answered by
4
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
pls make it brainlist
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
pls make it brainlist
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