Question

the two tangents drawn from an external point p to a circle with center o perpendicular
​

Answer 1

Step-by-step explanation:

Since tangent and radius are perpendicular at the point of contact, we have:

∠OAP=∠OBP=90

∘

.

So, ∠AOB+∠BPA=180

∘

…(1)

In △OAB, OA=OB, so, ∠OAB=∠OBA=10

∘

.

So, ∠AOB=180

∘

−10

∘

−10

∘

=160

∘

…(2)

From (1) and (2), we get ∠BPA=180

∘

−160

∘

=20

o

.