1. ## square tower question

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 degrees,60 degrees and 45 degrees .Show that the height of the tower is to the breadth of one of its sides as $\sqrt6(\sqrt5 + 1)$ to $4$

2. Hello jashansinghal
Originally Posted by jashansinghal
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 degrees,60 degrees and 45 degrees .Show that the height of the tower is to the breadth of one of its sides as $\sqrt6(\sqrt5 + 1)$ to $4$
Look at the diagram I have attached.

Because two of the angles are equal ( $45^o$) the point in the plane, $P$, lies on the diagonal produced of the square base of the tower.

Therefore, in the diagram, $\angle CDP = 135^o$.

Also, $\angle DPA = 60^o,\; \angle CPB = 45^o.$

Therefore, if the height and the breadth of the tower are $h$ and $b$ respectively:
$DP = \frac{h}{\tan60^o}=\frac{h}{\sqrt3}$
and
$CP = \frac{h}{\tan45^o}=h$
and
$DC = b$
Therefore, using the Cosine Rule on $\triangle DPC$:
$h^2=\frac{h^2}{3}+b^2-2\cdot\frac{h}{\sqrt3}\cdot b\cdot\cos135^o$
$=\frac{h^2}{3}+b^2+2\cdot\frac{h}{\sqrt3}\cdot b\cdot\frac{1}{\sqrt2}$
Multiply both sides by $3\sqrt3$:
$\Rightarrow 3\sqrt3h^2=\sqrt3h^2+3\sqrt3b^2+3\sqrt2bh$

$\Rightarrow 2\sqrt3h^2-3\sqrt2bh-3\sqrt3b^2=0$

Treating this as a quadratic in $h$, and using the formula:
$\Rightarrow h = \frac{3\sqrt2\pm\sqrt{18+72}}{4\sqrt3}\cdot b$
$=\frac{3\sqrt2+\sqrt{90}}{4\sqrt3}\cdot b$, since the other root is negative

$=\frac{3\sqrt2+3\sqrt2\sqrt5}{4\sqrt3}\cdot b$

$=\frac{3\sqrt2(1+\sqrt5)}{4\sqrt3}\cdot b$

$=\frac{\sqrt6(1+\sqrt5)}{4}\cdot b$

$\Rightarrow h:b = \sqrt6(\sqrt5+1):4$