Question

From a point P , tangents PA and PB are drawn to a circle C(O,r). If OP = 2r , show that ∆APB is equilateral.​

Answer 1

Answer:

OA=OB=r

OP=2r

In ΔOAP it is right angled at A

OA2+AP2=OP2

AP2=OP2−OA2=9r2−r2=3r2

AP=3r

Similarly BP=3r

In ΔOAP,tanθ=3rr=31→θ=30∘

α=90∘−30∘=60∘

In ΔOAT

sinα=rAT   23=rAT  

AT=2

hence triangle apb is equilateral

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