$P$ is a point on the segment joining the feet of two vertical poles of heights $a$ and $b$. The angles of elevation of the tops of the poles from $P$ are $45^{\circ}$ each. Then, the square of the distance between the tops of the poles is

• A

  $\frac{a^2+b^2}{2}$
• B

  $a^2+b^2$
• Correct answer
  C

  $2\left(a^2+b^2\right)$
• D

  $4\left(a^2+b^2\right)$

Solution

In $\triangle A P D$, $\tan 45^{\circ}=\frac{a}{A P} \Rightarrow A P=a$
and in $\triangle B P C$, $\tan 45^{\circ}=\frac{b}{P B} \Rightarrow P B=b$ $\therefore D E=a+b$ and $C E=b-a$ In $\triangle D E C$, $\begin{aligned} D C^2 & =D E^2+E C^2 \\ & =(a+b)^2+(b-a)^2 \\ & =2\left(a^2+b^2\right) \end{aligned}$

Hence, the correct answer is option: 'C'.