Is there an easier way of doing this problem:
A square tower stands upon a horizontal plane. From a point in this place from which three of its upper corners are visible their angular elevations are respectively 45°, 60° and and 45°. Show that the height of the tower is to the breadth of one of its sides as $\sqrt{6}(\sqrt{5}+1)$ is to 4.
I'm not sure if my approach is correct since it seems that to get the result using the way I've done it isn't as simple as previous questions have been (a few lines of working). In the digram $a$ is the length of the side of the tower and $h$ is the height of the tower. To start I tried to find $x$ in the diagram: $$ h^2=\frac{a^2}{2}+(x+\frac{\sqrt{2}a}{2})^2 $$ which already isn't very straight forward to find $x$ but I was using Mathematica just to see if this method would eventually work which gave $x$ to be:
$$ x=\frac{1}{2} \sqrt{4 h^2-2 a^2}-\frac{a}{\sqrt{2}} $$ which if we then use the top right triangle that $\tan{60^\circ}=h/x$: $$ \sqrt{3}=\frac{h}{\frac{1}{2} \sqrt{4 h^2-2 a^2}-\frac{a}{\sqrt{2}}}$$ which simplifies to give (well it reduces to this according to Mathematica, I can't seem to easily do it, but since I'm convinced there's an easier way I didn't spend too long on it.): $$ a=\frac{h(\sqrt{5}-1)}{\sqrt{6}}$$ and after a little rearrangement this gives what seems to be the required result: $$ \frac{h}{a}=\frac{\sqrt{6}(\sqrt{5}+1)}{4} $$
Is there any easier way to do this since this since this can't be the simplest way (I seem to be quite good at overcomplicating things a little too often!)?
Good useful pictures!
We might as well let the side of the tower be $1$. Let $h$ be the height of the tower. Because the angle of elevation from the observer to the other two corners is $45^{\circ}$, the distance from the observer to the point on the ground under these corners is $h$, as you labelled it.
The distance you called $x$ is, as you found, $\frac{w}{\sqrt{3}}$. Then the distance from the observer to the centre of the bottom of the tower is
$\frac{w}{\sqrt{3}}+\frac{\sqrt{2}}{2}$.
Now by the Pythagorean Theorem, we have $$w^2=\left(\frac{w}{\sqrt{3}}+\frac{\sqrt{2}}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2.$$ Expand and simplify a little. We get $$w^2=\frac{w^2}{3}+\frac{\sqrt{2}}{\sqrt{3}}w+1.$$ Multiply through by $3$, and manipulate a little. We get $$2w^2-\sqrt{6} w-3=0.$$ Now use the Quadratic Formula. One of the roots is negative, so irrelevant. The other root is equal to $$\frac{\sqrt{6}+\sqrt{30}}{4},$$ which can be, if we wish, written as $\frac{\sqrt{6}(1+\sqrt{5})}{4}$.
This looks to me very close to what you did, apart from the non-use of Mathematica.