Let the two circles, $C(Q, 4)$ and $C(R, 1)$ touch externally. So the distance $Q R$ (distance between centers) $=5$

Let $A B \& C D$ be the two common tangents meet at $P$. So extending the line of centers, $Q R$, it will also meet at $P$.

Join $A Q$ and $B R ; \angle Q A P=\angle R B P=90^{\circ}$

[At point of contact, radius and tangent perpendicular to each other].

So of the above, we have two right triangles, $Q A P$ and $R B P$ As $\angle Q A P=\angle R B P=90^{\circ}$

$\angle A P Q=\angle B P R[$ Common $]$

So the two triangles, $A P Q$ and $B P R$ are similar [AA similarity]

Hence,

$\frac{P Q}{A Q}=\frac{P R}{B R}$

$\Rightarrow \frac{Q R+P R}{A Q}=\frac{P R}{B R}$

$\Rightarrow \frac{5+x}{4}=\frac{x}{1}$

$\Rightarrow x=\frac{5}{3}$

$\Rightarrow P R=\frac{5}{3}$

So taking $\angle B P R=\theta$,

$\sin \theta=\frac{B R}{P R}=\frac{1}{\frac{5}{3}}=\frac{3}{5}$

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