Question

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

• A. $\frac{24}{25}$
• B. $\frac{12}{25}$
• C. $\frac{3}{4}$
• D. $none$

Answer 1

Answer: (A)

$\frac{24}{25}$

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, $\sin \theta =\frac{24}{25}$.