Question

in the figure PA and PB are tangents to the circle. If ∠APO=30°, find ∠AOB

Answer 1

Answer:

∠ AOB = 120°

Step-by-step explanation:

in the given figure,

OA is normal to PA and OB is normal to PB.

so, angle ∠A = ∠B = 90°

OP is angle bisector of  ∠ APB.

so , ∠ BPO = 30°

in Δ PAO ,

∠ POA = 60°  ( since sum of angles of a triangle = 180°)

in Δ PBO ,

∠ POB = 60°  ( since sum of angles of a triangle = 180°)

NOW,

∠ AOB = ∠ POA + ∠POB

∠ AOB = 60° + 60°

∠ AOB = 120°

thus you got the answer !

Hope it helps !

Answer 2

Step-by-step explanation:

<A 30°. <B 30°

PAO=POB

<AOB=30+30=60

<AOB=60+60

<AOB=120