Question

in the given figure PA and PB are tangents of the circle with center O . if angle AOB=100° then angle APO IS​

Answer 1

Answer:

$.$

Step-by-step explanation:

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