Question

In Fig. 10.63, two tangents AB and AC are drawn to a circle with centre O such that . Prove that OA = 2AB.

Answer 1

Proved OA = 2AB

•In triangle OBA and triangle OCA

   OB=OC    (Radius of same circle)

   <B =<C     ( Each 90°)

   OA=OA    ( Common)

•Triangle OBA is congruent to

  Triangle OCA            ( RHS)

• <OAB=<OAC  ( CPCT)

• Also, <OAB+<OAC=120°

=>       2<OAB= 120°

=>          <OAB= 60°

•Now, cos(<OAB)=B/H= AB/OA

                      1/2= AB/OA

                      OA=2AB

Answer 2

Explanation:

Given that AB and AC are two tangents drawn to a circle with centre O.

We need to prove that $\mathrm{OA}=2 \mathrm{AB}$

From the figure, we can see that $\angle B A C=120^{\circ}$

By the tangent property, the tangents AB and AC are equal.

Thus, we have, $A B=A C$

Since, OB and OC are radius of the circle.

We have,

$O B=O C$

By angle bisector theorem, we have,

$\angle B A O=\angle C A O=60^{\circ}$

Now, let us consider the triangle ABO,

  $\cos A=\frac{A B}{O A}$

$\cos 60^{\circ}=\frac{A B}{O A}$

Simplifying, we get,

$\frac{1}{2} =\frac{A B}{O A}$

Cross multiplying, we get,

$\mathrm{OA}=2 \mathrm{AB}$

Hence proved.

Learn more:

(1) In figure, two tangents AB and AC are drawn to a circle with centre O such that angle BAC = 60°,if AB=5cm , find BC

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(2) Two tangents AB and AC are drawn to a circle with a centre O so that angle BAC= 120° .prove that OA = 2AB.​

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