Q:

Height of Tower with Vertical Angles

An observer wishes to determine the height of the tower. He takes sight at the top of the tower from A and B, which are 50ft apart at the same elevation on a direct line with the tower. The VERTICAL ANGLE at point A is 30 degrees and at point B is 40. What is the height of the tower.
The answer is 92.54 ft.
I am not having trouble with the trigonometry. I am having trouble interpreting the problem. When it says "takes sight at the top of the tower" does it mean he goes to top? Furthermore what does it mean by "Vertical Angle"; from what i know its just the opposite angle when two lines cross. I do not get where the two lines ever crossed. I am aware of angles of elevation and depression but i dont think this is it... any hint?

A:

Draw altitudes $AM$ & $BN$ from the points $A$ & $B$ respectively on the horizontal plane & let the horizontal distance of the point $A$ from the observer $O$ be $d$. Now if $h$ is the height of tower then we have 
In right $\Delta AMO$ $$\tan 30^o=\frac{AM}{OM}=\frac{h}{d}$$ $$\implies d=h\cot 30^o=h\sqrt{3}$$
In right $\Delta BNO$ $$\tan 40^o=\frac{BN}{ON}=\frac{h}{d-50}$$ $$\implies d-50=h\cot 40^o \implies h\sqrt{3}-50=h\cot 40^o$$ $$\implies \color{blue}{h=\frac{50}{\sqrt{3}-\cot 40^o}\approx 92.54165784 \space ft}$$

