If the angles of a triangle be as $5 : 10 : 21$, and the side opposite the smaller angle be $3 \;$ cm, find the other sides.
Given: $\;$ $A : B : C = 5 : 10 : 21$
Let $k$ be the constant of proportionality.
Then, $\;$ $A = 5k$, $\;$ $B = 10 k$, $\;$ $C = 21 k$
In $\triangle ABC$, $\;$ $A + B + C = 180^\circ$
i.e. $\;$ $5 k + 10 k + 21 k = 180^\circ$
i.e $\;$ $36 k = 180^\circ$ $\implies$ $k = 5^\circ$
$\implies$ $A = 5 k = 25^\circ$, $\;$ $B = 10 k = 50^\circ$, $\;$ $C = 21 k = 105^\circ$
In $\triangle ABC$, by sine rule
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
i.e. $\;$ $b = \dfrac{a \; \sin B}{\sin A} = \dfrac{3 \times \sin \left(50^\circ\right)}{\sin \left(25^\circ\right)} = \dfrac{3 \times 0.7760}{0.4226} = 5.4378 \;$ cm
and $\;$ $c = \dfrac{a \; \sin C}{\sin A} = \dfrac{3 \times \sin \left(105^\circ\right)}{\sin \left(25^\circ\right)} = \dfrac{3 \times 0.9659}{0.4226} = 6.8568 \;$ cm