Finding the height of a building given two angles of elevation

Question

Um observador vê um prédio, no nível do solo, construído em terreno plano, sob um ângulo de 60º. Afastando-se do edifício mais 30m, passa a ver o edifício sob ângulo de 45º. Qual é a altura do prédio?

Translation?: An observer views a building, constructed on flat terrain, from ground level, at an angle of $60^\circ$. Moving away another $30$ meters, he now sees it from an angle of $45^\circ$. What is the height of the building?

answer: $\frac{30\sqrt{3}}{\sqrt{3}-1}$

Answer 1

Sorry, but I do not know Portuguese, so this answer is in English.

Let the initial distance between the observer and the building be $x$m.

Then, from the figure, $$\begin{align} \tan 60^0 &= \dfrac{h}{x} \\ \tan 45^0 &= \dfrac{h}{x+30} \end{align}$$

So, we get, $$h = \sqrt{3}x$$ $$h = x + 30$$

Now, eliminate $x$ and you will get,

$$h = \dfrac{30\sqrt{3}}{\sqrt{3}-1}$$

Answer 2

Just $$\frac{h}{h-30}=\sqrt3,$$ which gives the answer.