Question

A semicircular arch AB of length 2L and a vertical tower PQ are situated in the same vertical plane. The feet A and B of the arch and the base Q of the tower are at the same horizontal level, with B between A and Q. A man at A finds the tower hidden from his view due to the arch. He starts crawling up the arch and just sees the topmost point P of the tower after covering a distance $\frac{L}{2}$ along the arch. He crawls further to the topmost point of the arch and notes the angle of elevation of P to be $\theta$. Compute the height of the tower in terms of L and $\theta$.

Answer 1

$2L=\pi r$ (semi circumference)
$\therefore \frac{L}{2}=\frac{\pi r}{4}$
Since $\frac{l}{r}=\theta$ $\therefore \frac{L/2}{r}=\frac{\pi r}{4}$
i.e., when the man is at M then arc AM subtend and angle of $\frac{\pi }{4}$ at O. From M we can just see the top of tower so that MP is a tangent to arc at M
$\therefore$ OM is normal.
$\therefore \angle PMO={90}^o$ or $\angle PMN={45}^o=\angle CMO$
We have to find the height PQ of tower in term of $\theta$ and L, were
$2L=\pi r$ or $r=\frac{2L}{\pi }$ ...(1)
Let $PD=x$ so that $PQ=x+r$ ...(2)
No $CD=x\cot \theta$ in $\Delta FCD$.
$MN=PN\cot {45}^o=PN$
$MK+KN=PD+DN$
$\frac{r}{\sqrt{2}}+CD=x+(DQ-NQ)$
$\frac{r}{\sqrt{2}}+x\cot \theta =x+(r-\frac{r}{\sqrt{2}})$
$\therefore r(\sqrt{2}-1)=x(1-\cot \theta )$
$\therefore x=\frac{r(\sqrt{2}-1)}{1-\cot \theta }$
$\therefore PQ=x+r=\frac{r(\sqrt{2}-1)}{1-\cot \theta }+r$
or $PQ=\frac{\sqrt{2}-\cot \theta }{1-\cot \theta }=\frac{2L}{\pi }\left(\frac{\sqrt{2}-\cot \theta }{1-\cot \theta }\right)$ by (1)