# 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. He sees another man due South at an angle of...

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. He sees another man due South at an angle of depression 17 degrees.

justaguide | College Teacher | (Level 2) Distinguished Educator

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The observer is at a height of 15 m. He sees two men, one of them is towards the west at an angle of depression 31 degrees and the other is towards the south at an angle of depression 17 degrees.

The base of the tower, the man towards the west and the man towards the south form a right angled triangle.

The distance of the man towards the west from the base of the triangle can be calculated from the angle of depression which is 31 degrees and the height of the tower which is 15m as 15/tan 31 = 24.96 m. Similarly the man towards the south is 15/tan 17 = 49.06 m from the base.

Now use the Pythagorean Theorem to estimate the distance between the two men. It is given by sqrt[24.96^2 + 49.06^2] = sqrt 3030.36 = 55.04 m

The distance between the two men is 55.04 m

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.