Question

16. In the figure, PA and PB are tangents to a circle with centre O. If angle AOB = 120°, then find angle OPA.
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Answer 1

Answer:

Step-by-step explanation:<A= 90(RADIUS)

                                            <B=90(radius)

                                             <AOB =120 (given)

                                           <P=?

                             <A+<B+<AOB+<P=360

                               90+90+120+<P=360

                              <P=360-300

                                <P=60

                              ie <OPA=60

                               

Answer 2

Answer:

If two tangents are drawn to a circle from a single external point, then the tangents are equal and the angles formed by the radii and the tangents are supplementary.

⇒∠AOB + ∠APB = 180°..............................(1)

Now,

In ΔOAP AND ΔOBP,

  OA = OB (radii)

  OP = OP (common in both Δs)

 AP = BP (tangents from the same external point)

by AAA similarity,

ΔOAP  IS SIMILAR TO ΔOBP

by CPST we can say ∠OPA = ∠OPB ........................ (2)

From (1),

∠APB = 60°

and ∠OPA + ∠OPB = ∠APB

      2∠OPA = 60°

   ⇒ ∠OPA = 30°

Hence, ∠OPA is 30°.