To get the distance of a point $A$ from a point $B$, a line $BC$ and the angles $ABC$ and $BCA$ are measured and found to be $287 \; cm$ and $55^\circ 32'10''$ and $51^\circ 8' 20''$ respectively. Find the distance $AB$.
Given: $\;$ $\angle ABC = B = 55^\circ 32' 10''$, $\;$ $\angle BCA = C = 51^\circ 8' 20''$, $\;$ $BC = 287 \; cm$
In $\triangle ABC$, $\;$ $A + B + C = 180^\circ$
$\begin{aligned}
\therefore \; A & = 180^\circ - \left(B + C\right) \\\\
& = 180^\circ - \left(55^\circ 32' 10'' + 51^\circ 8' 20''\right) \\\\
& = 73^\circ 19' 30''
\end{aligned}$
In $\triangle ABC$, by sine rule, $\;$ $\dfrac{AB}{\sin C} = \dfrac{BC}{\sin A}$
$\therefore \;$ $AB = \dfrac{BC \; \sin C}{\sin A}$
$\begin{aligned}
i.e. \; AB & = \dfrac{287 \times \sin \left(51^\circ 8' 20''\right)}{\sin \left(73^\circ 19' 30''\right)} \\\\
& = \dfrac{287 \times 0.7787}{0.9579} \\\\
& = 233.31 \; cm
\end{aligned}$