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Please solve this question 
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Asked by vikasg13.hardware | 28 Jun, 2018, 06:14: AM
answered-by-expert Expert Answer
Refer to the attached diagram in which we have drawn the common tangent TPR to the two given circles.
Let PZ=x so that YZ=2x.
XY is tangent to the smaller circle at Q.
angle ZQY= angle ZPQ by the tangent chord or alternate segment theorem.
But angle ZQY = angle PZQ - angle ZYQ = angle QPR - angle RPX, once again by the aforesaid theorem.
Moreover, angle QPR - angle RPX = angle QPX which implies that PQ bisects angle YPX=90°; therefore angle YPQ=45°.
Now, YQ² =YZ*YP=2x*3x=6x² or YQ=√6*x.
Now, we apply sine rule in triangle YPQ whereby, sin(YQP)/sin(45°) =3x/(√6*x) =√3/2.
Clearly, angle YQP is obtuse; hence angle YQP=120° .
Thus, angle PYX =180 -(120+45) =15°
Answered by Arun | 10 Jul, 2018, 11:36: AM

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