Application of Trigonometry My question is-
From an aeroplane vertically over a straight road,the angles of depression of two consecutive kilometer-stones on the same side are 45 degrees and 60 degrees.Find the height of the aeroplane from the road.
Any solution to solve this question would be greatly appreciated.
 A: Let $A$ be the position of the airplane, and let $B$ be the point on the road directly below the plane. Let $C$ be the location of the nearer kilometre marker, and let $D$ be the position of the further one. Make a suitable labelled diagram. 
We are told that $\angle BAC$ is $30^\circ$ ($60^\circ$ below the horizontal) and that $\angle BAD$ is $45^\circ$. 
We have
$$\frac{BC}{h}=\tan(30^\circ)=\frac{1}{\sqrt{3}}.$$
Similarly,
$$\frac{BD}{h}=\tan(45^\circ) =1.$$
Thus $BC=\frac{h}{\sqrt{3}}$ and $BD=h$. But $BD-BC=1$. This gives the equation
$$h\left(1-\frac{1}{\sqrt{3}}\right)=1,$$
and now we can solve for $h$. 
A: I assume this is homework, so I'll give you a hint.
Make a drawing and identify two triangles:


*

*the first is a right triangle, with a horizontal leg on the street (say its length is $l$);

*the second is an obtuse triangle, adjacent to the first, with a side on the street (consecutive to $l$), which is the distance between the two kilometer stones.


Both have the position of the airplane as their upper vertex, and their height will be the height of the airplane on the street, call it $h$.
Note that both $l$ and $L \equiv l+1$ can be expressed as a function of $h$ and of one angle of the triangles you drew. This leads to a simple equation for $h$.
