Three dimensional Height and distance problem of trigonometry Due to limitation of characters in the title section the original problem is quoted here
A man finds that at a point due south of a tower the angle of elevation of the tower is 60 degrees. He then walks due west 10√6 metres on a horizontal plane and finds that the angle of elevation of the tower at that point is 30 degrees. Then the original distance of the man from the tower,in metres , is?
My attempt(See the picture )
 A: Consider the tower to be PM with base M and height $x$  .And you first walk to A and then  horizontally  $10\sqrt6 $ m to B
now Triangles $\triangle PMA \triangle PMB  $ and $\triangle MBA $ are right angle triangles , with $\angle PAM = 60^\circ$ and $\angle PBM = 30^\circ$
in $\triangle PMA $
$cot(60^\circ) =  \frac{AM}{x} \implies AM = \frac{x}{\sqrt3}$
in $\triangle PMB $
$\cot(30^\circ) = \frac{MB}{x}\implies MB = \sqrt3 .x$
Therfore in $\triangle MBA$
$MB = \sqrt3.x, AB =10\sqrt6$ and $MA = \frac{x}{\sqrt3}$
Apply Pythagoras theorem ,
$(MA)^2+(BA)^2=(MB)^2$ 
$\frac{x^2}{3} +600=3x^2$
$x^2(\frac13-3)=-600$
$x^2 =\frac{(600 )(3)}{8}$
$x= 15 $m
hence the original distance $AM =\frac{15}{\sqrt3}$m
A: Hint: the sides of a $60^o{-}30^o{-}90^o$ triangle are in the ratios $1:\frac{1}{2}:\frac{\sqrt{3}}{2}$ (this follows from the fact that $\sin\,30^o = \frac{1}{2}$). There are two such triangles in your problem each with the tower as one leg and one of the two observation points as the opposite vertex. Compare the ratios of the edges and conclude that the move west tripled the distance from the tower. Hence looking at the plan view of the situation you see a right-angled triangle with one leg and the hypotenuse in the ration $1:3$. Applying Pythagoras, the edges of the plan view triangle are in the ratio $1 : 2\sqrt{2} : 3$ and you are given the length of the longer of the two legs, from which you can calculate the length of the shorter leg, which is what you are asked for.
See https://en.wikipedia.org/wiki/Trigonometric_functions for lots of information and figures involving $\sin$, $\cos$ etc.
A: Let $h$ be the height of the tower, let $x$ be the distance you are trying to find, and let $y$ be the distance to the base of the tower in the second position.
The horizontal triangle is right-angled therefore $$y^2=x^2+(10\sqrt{6})^2$$
You also have, in the first position, $$h=x\tan60\implies x=\frac{h}{\sqrt{3}}$$
In the second position, $$h=y\tan30\implies y=h\sqrt{3}$$
Now substitute these expressions into the first equation and find $h$ then you can find $x$.
