  Question

Two circles of radii 4 cms & 1 cm touch each other externally and $$\displaystyle \theta$$ is the angle contained by their direct common tangents Then $$\displaystyle \sin \theta$$ =

A
2425  B
1225  C
34  D
none  Solution

The correct option is B $$\displaystyle \frac{24}{25}$$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 $$BR$$; $$\angle QAP=\angle RBP=90^{ 0 }$$[At point of contact, radius and tangent perpendicular to each other]. So of the above, we have two right triangles, $$QAP$$ and $$RBP$$ As $$\angle QAP=\angle RBP=90^{ 0 }$$ $$\angle APQ=\angle BPR$$ [Common] So the two triangles, $$APQ$$ and $$BPR$$ are similar [AA similarity] Hence, $$\frac { PQ }{ AQ } =\frac { PR }{ BR } \\ \Rightarrow \frac { QR+PR }{ AQ } =\frac { PR }{ BR } \\ \Rightarrow \frac { 5+x }{ 4 } =\frac { x }{ 1 } \\ \Rightarrow x=\frac { 5 }{ 3 } \\ \Rightarrow PR=\frac { 5 }{ 3 }$$  So taking $$\angle BPR=\theta$$, $$\sin { \theta } =\frac { BR }{ PR } =\frac { 1 }{ \frac { 5 }{ 3 } } =\frac { 3 }{ 5 }$$  So the angle between two tangents =$$2\theta$$ Thus,$$\sin { 2\theta } =2\sin { \theta } \cos { \theta } =2\times \frac { 3 }{ 5 } \times \sqrt { 1-\sin ^{ 2 }{ \theta } } =2\times \frac { 3 }{ 5 } \times \frac { 4 }{ 5 } =\frac { 24 }{ 25 }$$Hence, $$\sin { \theta } =\frac { 24 }{ 25 }$$.Mathematics

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