JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Heights and Distances

Let the poles stand a distance $x$ apart and let the crossing point be at height $h$. By similar triangles, the line from the top of the $20\text{m}$ pole to the foot of the other gives $\frac{h}{20}=\frac{y}{x}$, where $y$ is the horizontal distance from the $20\text{m}$ pole, and the line from the top of the $80\text{m}$ pole to the foot of the first gives $\frac{h}{80}=\frac{x-y}{x}$. Adding these, $\frac{h}{20}+\frac{h}{80}=\frac{y}{x}+\frac{x-y}{x}=1$, so $h\cdot \frac{4+1}{80}=1$ and $h=16\text{m}$. A neat feature is that this height does not depend on how far apart the poles are.