heights and distances

Asked by  | 29th Sep, 2008, 06:01: PM

Expert Answer:

Let PQ be the tower and the height of the tower =  h

Let the first point of observation be A . From A the angle of elevation is 32deg and let B be the second point of observation 100 mt closer to the tower be B . From  B the angle of elevation is 63 deg

SoFrom triangle PAQ:

 h/x = tan 63 = 1.9626

i.e h = 1.9626 x

Similarly  from triangle PBQ

h / (100+x) = tan 32 = 0.6248

i.e 1.9626 x  =  0/6248 (100+x)

(1.9626- .6248) x   =  62.48

1.3378 x = 62.48

x = 62.48/ 1.3378

X= 46.70

See figure attached

Answered by  | 30th Sep, 2008, 11:23: AM

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