If $\cos A = \dfrac{17}{22}$ and $\cos C = \dfrac{1}{14}$, find the ratio of $a : b : c$.
Given: $\;$ $\cos A = \dfrac{17}{22}$, $\;$ $\cos C = \dfrac{1}{14}$
Now, $\;$ $\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\dfrac{17}{22}\right)^2} = \dfrac{\sqrt{195}}{22}$ $\;\;\; \cdots \; (1a)$
and $\;$ $\sin C = \sqrt{1 - \cos^2 C} = \sqrt{1 - \left(\dfrac{1}{14}\right)^2} = \dfrac{\sqrt{195}}{14}$ $\;\;\; \cdots \; (1b)$
In $\triangle ABC$, $\;$ $A + B + C = \pi$
i.e. $\;$ $B = \pi - \left(A + C\right)$
$\begin{aligned}
\therefore \; \sin B & = \sin \left[\pi - \left(A + C\right)\right] \\\\
& = \sin \left(A + C\right) \\\\
& = \sin A \cos C + \cos A \sin C \\\\
& = \dfrac{\sqrt{195}}{22} \times \dfrac{1}{14} + \dfrac{17}{22} \times \dfrac{\sqrt{195}}{14} \\\\
& = \dfrac{9 \sqrt{195}}{154} \;\;\; \cdots \; (1c)
\end{aligned}$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
$\begin{aligned}
\implies a : b : c & = \sin A : \sin B : \sin C \\\\
& = \dfrac{\sqrt{195}}{22} : \dfrac{9 \sqrt{195}}{154} : \dfrac{\sqrt{195}}{14} \;\; \left[\text{by equations (1a), (1b), (1c)}\right] \\\\
& = \dfrac{1}{22} : \dfrac{9}{154} : \dfrac{1}{14} \\\\
& = \dfrac{7}{154} : \dfrac{9}{154} : \dfrac{11}{154}
\end{aligned}$
$\therefore \;$ $a : b : c = 7 : 9 : 11$