CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
1225
loader
C
34
loader
D
none
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image