Question

In the fig. 'C' is the centre of the circle
PA and PB are tangents to the circle

If Angle
APB =50°
then find angle ACB.
(2 Points)
50
B​

Answer 1

Answer:

We know that the radius and tangent are perpendicular at their point of contact.

∴  OA⊥PA and OB⊥PB

∴  ∠OAP=∠OBP=90o

In quadrilateral OAPB,

⇒  ∠OAP+∠APB+∠OBP+∠AOB=360o

⇒  90o+50o+90o+∠AOB=360o

⇒  230o+∠AOB=360o

⇒  ∠AOB=360o−230o

⇒  ∠AOB=130o

In △AOB,

⇒  OA=OB                   [ Radii of same circle ]

⇒  ∠OAB=∠OBA          [ Base angles of equal sides are also equal ]

Now,

⇒  ∠OAB+∠OBA+∠AOB=180o

⇒  ∠OAB+∠OAB+130o=180o

⇒  2∠OAB=180o−130o

⇒  2∠OAB=50o