From a point P, two tangents PA and PB are drawn to a circle C(0, r). If √3OP = Diameter, find angle APB.
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From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = diameter of the circle, show that △APB is equilateral.
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OA=OB=r
OP=2r
In ΔOAP it is right angled at A
OA
2
+AP
2
=OP
2
AP
2
=OP
2
−OA
2
=9r
2
−r
2
=3r
2
AP=
3
r
Similarly BP=
3
r
In ΔOAP,tanθ=
3
r
r
=
3
1
→θ=30
∘
α=90
∘
−30
∘
=60
∘
In ΔOAT
sinα=
r
AT
2
3
=
r
AT
AT=
2
3
r
AT=BT=
2
3
r
AB=
3
r
In ΔAPB AP=AB=BP=
3
r
Hence it is equilateral.
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