Question

A train is moving at a constant speed at an angle $\theta$ East of north. Observations of the train are made from a fixed point. It is due north at some instant. Ten minutes earlier it bearing was ${\alpha }_1$ west of north, whereas ten minutes afterwards its bearing is ${\alpha }_2$ east of north. Find $\tan \theta$.

Answer 1

Let the train be moving along $PQ$ and $A$ be the point of observation. The train is at $O$ at some instant so it is due north of $A$ then. Ten minutes earlier, it was at $P$ whose bearing from $A$ is ${\alpha }_1$ west of north so that the $-PAQ={\alpha }_1$. Ten minutes afterwards.
it is at $Q$ whose bearing is ${\alpha }_2$ east of north so that $\angle OAQ=={\alpha }_2$.
Since $PO$ and $OQ$ are covered in equal times, we have $OP=OQ=a$, say.
Now applying trigonometrical theorem in $△PAQ$,
We get
$(a+a)\cot \theta =a\cot {\alpha }_2-a\cot {\alpha }_1$
Or $\cot \theta =\frac{1}{2}\left(\cot {\alpha }_2-\cot {\alpha }_1\right)$
$\therefore \tan \theta =\frac{2}{\cot {\alpha }_2-\cot {\alpha }_1}=\frac{2\sin {\alpha }_1\sin {\alpha }_2}{\sin ({\alpha }_1-{\alpha }_2)}$