Question

A man on a hill observes that three towers on a horizontal plane subtend equal angles on his eye and that the angles of depression of their bases are $\theta ,\varphi$ and $\psi$. Prove that if a, b and c are their heights, then
$\frac{sin(\theta -\varphi )}{csin\psi }+\frac{sin(\varphi -\psi )}{asin\theta }+\frac{sin(\psi -\theta )}{bsin\varphi }=0$

Answer 1

Let $\alpha$ be the equal angle subtended by the towers at the man at P. The height of P be taken as x. Both x and $\alpha$ are unknown and they have to be eliminated. $\alpha$ comes in $\Delta APB$ and x comes in $\Delta APO$ and both have side AP common which is to be eliminated. By sine in $\Delta ABP.$
$\frac{a}{sin\alpha }=\frac{AP}{cos(\theta -\alpha )}$ ...(1)
Now $\frac{x}{sin\theta }=\frac{AP}{sin{90}^o}$ $\therefore AP=\frac{x}{sin\theta },\Delta OAP$
Putting in (1) we get
$\frac{a}{sin\alpha }=\frac{x}{sin\theta cos(\theta -\alpha )}$
$\therefore \frac{xsin\alpha }{asin\theta }=cos(\theta -\alpha )$
$\frac{xsin\alpha }{bsin\varphi }=cos(\varphi -\alpha )$
$\frac{xsin\alpha }{csin\psi }=cos(\psi -\alpha )$
$T_1=\frac{sin(\theta -\varphi )}{csin\psi }=\frac{sin\left[\right(\theta -\alpha )-(\varphi -\alpha \left)\right]}{csin\psi }$
$=\frac{sin(\theta -\alpha )cos(\varphi -\alpha )-cos(\theta -\alpha )sin(\varphi -\alpha )}{csin\psi }$
$=\frac{1}{csin\psi }\left[sin\right(\theta -\alpha )\frac{xsin\alpha }{bsin\varphi }-sin(\varphi -\alpha \left)\frac{xsin\alpha }{asin\theta }\right]$
$=\frac{xsin\alpha }{abcsin\theta sin\varphi sin\psi }\left[sin\theta sin\right(\theta -\alpha )-sin\varphi sin(\varphi -\alpha \left)\right]$
$T_1=k\left[sin\theta sin\right(\theta -\alpha )-sin\varphi sin(\varphi -\alpha \left)\right]$
Similarly write $T_2$ and $T_3$ by symmetry and then
add. $T_1+T_2+T_3=0$ as $\sum (a-b)=0$