Question

$P$ 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 ${45}^0$ each. Then, the square of the distance between the tops of the poles is

• A. $\frac{a^2+b^2}{2}$
• B. $a^2+b^2$
• C. $2(a^2+b^2)$
• D. $4(a^2+b^2)$

Answer 1

Answer: (C)

$2(a^2+b^2)$

Let $AD$ be the pole of height $a$ and $BC$ be the pole of height $B$

In $\Delta$ $APD$, we have
$\tan {45}^o=\frac{AD}{AP}$

$\Rightarrow 1=\frac{a}{AP}$

$\Rightarrow AP=a$

And in $\Delta$ $BPC$, we have,

$\tan {45}^o$ $=\frac{BC}{PB}$

$\Rightarrow 1=\frac{b}{BP}$

$\Rightarrow BP=b$

$\therefore$ $DE=AB$

$=AP+BP$

$=a+b$

and $CE=BC-BE=b-a$

In $\Delta$ $DEC$, applying pythagoras theorem,

$D{C}^2=D{E}^2+E{C}^2$
$=(a+b{)}^2+(b-a{)}^2$

$=a^2+2ab+b^2+a^2-2ab+b^2$
$=2(a^2+b^2)$