Question

In the figure, PQ is tangent to a circle with centre O. If ∠ OAB=30

, find ∠ ABP and ∠ AOB

Answer 1

OA=OB [RADII]
ang OBA=ang  OAB=30 deg [angles opp. to equal sides are equal]
ang OPB=90 deg [tangent]
ang OBA+ang ABP=90
30+ang ABP=90
ang ABP=90-30=60 deg

ang AOB= 180-(30+30)
                  120  [angle sum property of triangle]

Answer 2

Answer:

∠ABP = 60°, ∠AOB = 120°

Step-by-step explanation:

From the above question,

They have given :

In the figure, PQ is tangent to a circle with centre O. If ∠ OAB=30, find ∠ ABP and ∠ AOB

Since PQ is tangent to the circle with centre O, ∠ABP = ∠AOB and they are both 90°.

Therefore, ∠ABP = ∠AOB = 90°.

Also, since ∠OAB = 30°, ∠AOB = 90° and ∠OAB = 30°, we can deduce that ∠ABP = 60°.

Since AO and OB is radius of circle

Therefore AO = OB

           ∠OAB = ∠OBA ( angle opp. to equal side)

           ∠OBA = 30°

NOW,

     since radius is perpendicular to tangent

           ∠ABO +ABP = ∠OBP

           30° + ∠ABP = 90°

           ∠ABP = 60°

In ΔAOB

∠OAB + ∠ABO + ∠AOB = 180° (ASP)

30 ° + 30 + ∠AOB = 180°

∠AOB = 120°

For more such related questions : https://brainly.in/question/27129791

#SPJ3