# A square tower stands upon a horizontal plane. From a point in this plane, from which three of its upper corners are visible, their angular elevations are respectively 45 degree, 60 degree and 45 degree. Show that the height of the tower is to the breadth of one of its sides as 61/2(51/2+1) to 4.

Let height of tower  =  h
And
As base and top is a square so , Length  =  breadth =  l
So,
AB  =  BC  =  CD =  DA  = A'B'  =  B'C'  =  C'D'  =  D'A'  =  l
Now we show our diagram , As :

Here ABCD is a square base and A'B'C'D' is a square top and P is a point of observation of three corners.
And given
Angular elevations angles are
$\angle$ BPB' =  45$°$
And
$\angle$ APA' =  60$°$
And
$\angle$ DPD' =  45$°$
And Here
Because two of the angles are equal ( 45$°$ ) the point in the plane, P, lies on the diagonal produced of the square base of the tower. As diagonal CA produced to meet point " P "
Then
$\angle$ BAC =  45$°$                         ( As we know diagonal of square bisect it's vertices angle )
So,
$\angle$ PAB =  135$°$                       ( As here $\angle$ BAC and $\angle$ PAB are supplementary as they formed by a straight line CP )

Now In $∆$ APA'

We get

tan 60$°$

$\sqrt{3}$$\frac{h}{PA}$                                                              ( As we know tan 60$°$  = $\sqrt{3}$ )
So,
PA = $\frac{h}{\sqrt{3}}$

Now In $∆$ BPB'

We get

tan 45$°$

1 = $\frac{h}{PB}$                                                              ( As we know tan 45$°$  = 1 )
So,
PB = h

Now In $∆$ PAB

We apply cosine rule and get

PB2  =  PA2 + AB2 - 2 $×$ PA $×$Ab $×$ Cos 135$°$

Now we substitute all values and get

h2 = ( $\frac{h}{\sqrt{3}}$ )2l2 - 2 $×$ $\frac{h}{\sqrt{3}}$ $×$ l $×$ $-\frac{1}{\sqrt{2}}$ )                               ( AS we know Cos 135$°$$-\frac{1}{\sqrt{2}}$  )

h2$\frac{{h}^{2}}{3}$  +  l2$\frac{2hl}{\sqrt{6}}$

$\frac{{h}^{2}}{3}$  +  l2$\frac{2hl}{\sqrt{6}}$  - h2  = 0

- $\frac{2{h}^{2}}{3}$  +  l2$\frac{2hl}{\sqrt{6}}$  = 0

$\frac{2{h}^{2}}{3}$  -  l2$\frac{2hl}{\sqrt{6}}$  = 0
Taking LCM , we get

2h- $\sqrt{6}$hl - 3l2  = 0
So,
We know quadratic formula As  :
x  =
Here
a  = 2  , b  = - $\sqrt{6}$and  c  =  - 3l2
So,
h  =
Now we neglect negative value as h can't be negative , So

h  =
So,
$\frac{h}{l}$  =

$\frac{h}{l}$  =
So,
Ratio of height of tower :  Breadth of one of it's side  =   :  4             ( Hence proved )

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