From a point due south of a tower, a man observes the angle of elevation of the top of the tower to be 60°. He then walks 60m due west on the horizontal plane and observes the angle of elevation to be 30°. Find the height of the tower and his initial distance from the tower..
My attempt
I have the figure but I could not solve it to get the answer. please anyone solve this figure.
solving the figure, the initial distance.from the tower to the man does not match. The initial distance between the tower and the man is 63.63m. Where is the error
Best Answer
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?