Question

A person walks along a straight road and observes that the greatest angle subtended by two objects is $\alpha$; from the point where this greatest angle is subtended he walks distance c along the road, and finds that the two objects are now in a straight line which makes an angle $\beta$ with the road; prove that the distance between the objects is
$c\sin \alpha \sin \beta \sec \frac{\alpha +\beta }{2}\sec \frac{\alpha -\beta }{2}$ or $\frac{2c\sin \alpha \sin \beta }{\cos \alpha +\cos \beta }$

Answer 1

RS is a road on which there is a point A at which two objects P and Q subtend the greatest angle $\angle$, then by geometry, we know that the circle through P, Q and A will touch the road RS at the point A. If the line joining the objects PQ cuts the road at B, then AB = c and $\angle ABP=\beta$ as given now suppose $\angle QAB=\theta$ so that $\angle QPA=\theta$. We have to find PQ whereas we are given AB = c. We will apply sine rule on ${\Delta }^s$ in which these two aides occur and eliminate the common side. Consider ${\Delta }^{s}PQA$ and BQA with common side QA which is to be eliminated.
From $\Delta =PAB$, $(\theta +\alpha )+\beta +\theta ={180}^o$
$\therefore \theta ={90}^o-\frac{\alpha +\beta }{2}$ ...(A)
From $\Delta PQA$, $\frac{PQ}{\sin \alpha }=\frac{AQ}{\sin \theta }$ ...(1)
From $\Delta BQA$,
$\frac{AB}{\sin ({180}^o-(\beta +\theta \left)\right)}=\frac{AQ}{\sin \beta }$ ...(2)
Eliminating AQ from (1) and (2), we get
$\frac{PQ}{c}\frac{\sin (\beta +\theta )}{\sin \alpha }=\frac{\sin \beta }{\sin \theta }$, by dividing
$PQ=\frac{c\sin \alpha \sin \beta }{\sin \theta \sin (\beta +\theta )}$
Now put for unknown $\theta$ from (A)
$PQ=\frac{c\sin \alpha \sin \beta }{\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)}=\frac{2c\sin \alpha \sin \beta }{\cos \alpha +\cos \beta }$