Question

from an external point P ,tangents PA and PB are drawn to a circle with centre O.if angle PAB=50°,then find angleAOB

Answer 1

in ∆pab given that

angle PAB is 50°

and angle PAO=90°

PAO= angle PAB+OAB

90=50+OAB

then angle OAB=40°

NOW In trianlge OAB

OA=OB
THUS angle OAB and oba will be equal

then in triangle OAB

OAB+oba+ aob =180

40+40+aob=180

aob=180+80

aob=100°

Answer 2

∠PAB = 50°

∠PBA = 50° (as PA = PB)

⇒ 50 + 50 + ∠P = 180°  (angle sum property)

⇒ ∠P = 80°

Also, ∠O + ∠P = 180°

⇒ ∠O = 180° – 80° = 100