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Question

Two circles of radii 4 cms & 1 cm touch each other externally and θ is the angle contained by their direct common tangents Then sinθ =

A
2425
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B
1225
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C
34
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D
none
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Solution

The correct option is B 2425
Let the two circles, C(Q,4) and C(R,1) touch externally.
So the distance QR (distance between centers) = 5
Let AB & CD be the two common tangents meet at P.
So extending the line of centers, QR, it will also meet at P.
Join AQ and BR; QAP=RBP=900
[At point of contact, radius and tangent perpendicular to each other].
So of the above, we have two right triangles, QAP and RBP
As QAP=RBP=900
APQ=BPR [Common]
So the two triangles, APQ and BPR are similar [AA similarity]
Hence,

PQAQ=PRBRQR+PRAQ=PRBR5+x4=x1x=53PR=53

So taking BPR=θ,

sinθ=BRPR=153=35
So the angle between two tangents =2θ

Thus,

sin2θ=2sinθcosθ=2×35×1sin2θ=2×35×45=2425

Hence, sinθ=2425.

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