O is rhe centre of the circle PA and PB are tangent if PAB = 70° then find APB.
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Answer:
Join OB.
We know that the radius and tangent are perpendicular at their point of contact.
∴ ∠OBP=∠OAP=90
o
Now, In a quadrilateral AOBP
⇒ ∠AOB+∠OBP+∠APB+∠OAP=360
o
[ Sum of four angles of a quadrilateral is 360
o
. ]
⇒ ∠AOB+90
o
+60
o
+90
o
=360
o
⇒ 240
o
+∠AOB=360
o
⇒ ∠AOB=120
o
.
Since OA and OB are the radius of a circle then, △AOB is an isosceles triangle.
⇒ ∠AOB+∠OAB+∠OBA=180
o
⇒ 120
o
+2∠OAB=180
o
[ Since, ∠OAB=∠OBA ]
⇒ 2∠OAB=60
o
∴ ∠OAB=30
o
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