Question

Due south of a tower which is leaning towards north, there are two stations at distance $x,y,x,y$ respectively from its foot. If $\alpha ,\beta$ respectively be the angles of elevation of the top of the tower at these stations, show that the inclination of the tower to the horizon is
${\cot }^{-1}\frac{y\cot \alpha -x\cot \beta }{y-x}$

Answer 1

$AQ=x,BQ=y$ and $PQ$ is a tower leaning towards North at an angle $\theta$. Apply $m-n$ theorem on $\Delta PBQ$.
$(y-x+x)\cot \alpha =x\cot \beta (y-x)\cot (\pi -\theta )$
or $y\cot \alpha -x\cot \beta =(y-x)\cot \theta$
$\therefore \theta =$ etc.