# Height and distance with directions

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

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?

• Taussig, what is his initial distance from the tower? – pi-π Nov 19 '16 at 9:35
• Edited. His initial distance from the tower is $d$. – N. F. Taussig Nov 19 '16 at 10:13
• But the answer doesn't match. The answer in my book is 63.63m. – pi-π Nov 19 '16 at 10:45
• The answer for the height of the tower is 36.74m – pi-π Nov 19 '16 at 12:17
• The answers I obtain for the initial distance of the man from the tower, the height of the tower, and his distance from the tower when he takes the second measurement are, respectively, $d = \sqrt{450}~\text{m} \approx 21.21~\text{m}$, $h = \sqrt{1350}~\text{m} \approx 36.74~\text{m}$, and $x = 3d \approx 63.64~\text{m}$. The book's answers are inconsistent. Since the angle of elevation is initially $60^\circ$, the height of the tower must exceed the man's initial distance from the tower. – N. F. Taussig Nov 19 '16 at 12:32

Since the figure for this question will be formed in 3D and I have to adjust it in 2D I will have to represent the $90^\circ$ like that.
$$\tan 60^\circ =AB/BC$$ $$therefore~BC=AB/{\sqrt3 } \tag{1}$$ $$\tan 30^\circ =AB/DB$$ $$therefore~DB=\sqrt{3}AB \tag{2}$$ $${BC^2}+{DC^2}={DB^2}$$ $${AB^2}/{3}+3600=3{AB^2} (as\hspace{1 mm} DC=60m)$$ $$therefore\hspace{1 mm} AB=\sqrt {1350}=36.742m$$ $$BC=AB/{\sqrt3}~(eqn. 1)$$ $$therefore\hspace{1 mm} BC=21.21m$$ $$AB=height\hspace{1 mm} of\hspace{1 mm} tower=36.742m$$ $$BC = initial \hspace{1 mm}distance\hspace{1 mm} from \hspace{1 mm}tower = 21.21m$$