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 $\displaystyle \sqrt6(\sqrt5 + 1)$ to $\displaystyle 4 $

Look at the diagram I have attached.

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

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

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

Therefore, if the height and the breadth of the tower are $\displaystyle h$ and $\displaystyle b$ respectively:$\displaystyle DP = \frac{h}{\tan60^o}=\frac{h}{\sqrt3}$

and$\displaystyle CP = \frac{h}{\tan45^o}=h$

and$\displaystyle DC = b$

Therefore, using the Cosine Rule on $\displaystyle \triangle DPC$:$\displaystyle h^2=\frac{h^2}{3}+b^2-2\cdot\frac{h}{\sqrt3}\cdot b\cdot\cos135^o$$\displaystyle =\frac{h^2}{3}+b^2+2\cdot\frac{h}{\sqrt3}\cdot b\cdot\frac{1}{\sqrt2}$

Multiply both sides by $\displaystyle 3\sqrt3$:$\displaystyle \Rightarrow 3\sqrt3h^2=\sqrt3h^2+3\sqrt3b^2+3\sqrt2bh$

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

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

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

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

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

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

Grandad