Question

From a point P, two tangents PA and PB are drawn to a circle C(0, r). If OP = 2r, then find ∠.​

Answer 1

Answer:

InΔOAP:

OA=r

OP=2r

⇒PA

2

=OP

2

−OA

2

=4r

2

−r

2

=3r

2

⇒PA=

3

r

∴PA=PB=

3

r

(∵Tangents from an external point are equal in length )

Now, tan(∠APO)=

AP

OA

=

r

3

r

=

3

1

∠APO=30°

Similarly

∠BPO=30°

∴∠APB=60°

In ΔABP,

Using cosine rule :

cos(∠APB)=

2AP.BP

AP

2

+BP

2

−AB

2

⇒cos(60°)=

2(

3

r)

2

(

3

r)

2

+(

3

r)

2

−AB

2

⇒

2

1

=

2(3r

2

)

6r

2

−AB

2

⇒3r

2

=6r

2

−AB

2

⇒AB=

3

r

∴AB=AP=BP=

3

r

Hence, ABP is an equilateral triangle

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