trigonometry
The angle of elevation of the top of a vertical tower from a point $A$, due east of it is $45^{\circ}$. The angle of elevation of the same tower from a point $B$, due south of $A$ is $30^{\circ}$. If the distance between $A$ and $B$ is $54\sqrt2$m the find the height of the tower.
My attempt:
Let the height of tower be $h$ meters. So, the distance of the foot of the tower from $A$ is $h$ and from $B$ is $h\sqrt3$. So, $h^2+(54\sqrt2)^2=(h\sqrt3)^2\implies h=54m$.
Answer is given as $108$m.
Or...
Use the Law of Sines to find the longest side in the triangle with the 100 m side (you know all the angles).
This longest side is also the hypotenuse of a another, right angle triangle where you know the angle opposite the height you want...
Consider the figure below.
The man is initially at point $S$, standing a distance $d$ from the base $B$ of the tower, which has height $h$. He measures the angle of elevation, $\angle BST$, to be $60^\circ$. He then moves $60~\text{m}$ west from $S$ to point $W$, where he measures the angle of elevation, $\angle BWT$, to be $30^\circ$.
Then
\begin{align*}
\tan 60^\circ & = \frac{h}{d}\\
\sqrt{3} & = \frac{h}{d}\\
d\sqrt{3} & = h
\end{align*}
and
\begin{align*}
\tan 30^\circ & = \frac{h}{x}\\
\frac{1}{\sqrt{3}} & = \frac{h}{x}\\
x & = h\sqrt{3}\\
& = (d\sqrt{3})\sqrt{3}\\
& = 3d
\end{align*}
Moreover, since $S$ is directly south of the tower and $W$ is directly west of $S$, $\angle BSW$ is a right angle, so $\triangle BSW$ is a right triangle with hypotenuse $\overline{BW}$. Hence, by the Pythagorean Theorem,
$$d^2 + (60~\text{m})^2 = (3d)^2$$
Can you take it from here?
Best Answer
As already mentioned in the comment, your solution $h=54 m$ is correct.
Below is a 3d-diagram depicting the situation.
Btw., the distance from $B$ to the top of the tower is $108 m$.