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$P$ 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
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Verified Answer
The correct answer is:
$2\left(a^2+b^2\right)$
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 $\triangle D E C$,
$\begin{aligned}
D C^2 & =D E^2+E C^2 \\
& =(a+b)^2+(b-a)^2 \\
& =2\left(a^2+b^2\right)
\end{aligned}$
$\tan 45^{\circ}=\frac{a}{A P} \Rightarrow A P=a$
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 $\triangle D E C$,
$\begin{aligned}
D C^2 & =D E^2+E C^2 \\
& =(a+b)^2+(b-a)^2 \\
& =2\left(a^2+b^2\right)
\end{aligned}$
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