Question

kindly solve this.....
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Answer 1

Let x-y plane represent ground.
Set up coordinates:Three observable vertices of the square tower:$A(0,0,h)$, $B(a,0,h)$ and $C(0,a,h)$, where $h$ is the height of tower and $a$ is the side of square.
By symmetry arguments the abscissa and ordinate of the point of observation can be assumed equal.
Let coordinates of points of observation be $P(-x,-x,0)$ ,so the closest point of observation is $A(60^{\circ})$ and the farther points are $B,C(45^{\circ})$.
Slope of a line joining two points in 3D is given by:
$\tan\theta=\dfrac{z_2-z_1}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}\\\text{Between }P \text{ and } A,\\\tan60^{\circ}=\dfrac{h}{\sqrt{(x)^2+(x)^2}}\\\Rightarrow h=\sqrt6x\qquad \boxed{1}\\\text{Between }P \text{ and } (B\text{ or }C),\\\tan45^{\circ}=\dfrac{h}{\sqrt{(x+a)^2+x^2}}\\\Rightarrow (x+a)^2+x^2=h^2\\\Rightarrow (\sqrt6x+\sqrt6a)^2+(\sqrt6x)^2=6h^2\\\text{Put eq.}\boxed{1}\\\Rightarrow (h+\sqrt6a)^2+h^2=6h^2\\\Rightarrow h+\sqrt6a=\sqrt5h\\\Rightarrow h(\sqrt5-1)(\sqrt5+1)=\sqrt6a(\sqrt5+1)\\\Rightarrow \boxed{\dfrac{h}{a}=\dfrac{\sqrt6(\sqrt5+1)}{4}}\\\\Q.E.D$

Answer 2

Hope it helps! ^_^
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