Find the distance between the men?An observer at the top of a tower of height 15m sees a man due West of him at an angle of depression 31 degrees....

Let H be the position of the observer and O the point
vertically down the observer on the ground. Let W and S bethe position of the men on the
ground as described.

Now HOW, HOS and and HOS are the 3
right angles in 3 mutually perpendicular planes.

In the
right angled triangle HOW, OH = 15.  OW/OH = tan(90- 31). So OW = OH tan(90-31) = 15
tan59

In the right angled triange HOS, OS/OH tan(90- 17).
So OS = OH tan 73 = 15 tan73.

In the right angled triangle,
WOS,

WS^2 = OW^2+OS^2

WS^2
=(15tan59)^2+(15tan73)^2

WS = 15sqrt(tan^2 59+tan^2 73) =
55.05m

Therefore the distance between the
men  is 55.05m.