appications of trignometry

Asked by nishant27_c | 10th Feb, 2010, 11:07: AM

Expert Answer:



Consider a top view, as shown in figure,

If the height of tower is h, and horizontal distance from station A to the tower bottom is AC, the distance from station A to the top of the tower is AD. Since we are using top view points C and D will coincide.

Also the horizontal distance from station B to the bottom of tower is BC and to the top is BD.

Now sin a = h/AD; sin b = h/BD


sin2 a = h2/AD2;       sin2 b = h2/BD2 ..................(1)

AD2 = AC2 + h2; and BD2 = BC2 + h2

 But BC2 = AB2 + AC2

BD2 = AB2 + AC2 + h2

Put for AC2 + h2 = AD2 = h2/sin2a from (1)

BD2 = AB2 + h2/sin2a


sin2 b = h2/BD2 , becomes,

sin2 b = h2/ (AB2 + h2/sin2a)

sin2 b(AB2 + h2/sin2a) = h2

AB2 sin2 b  + h2sin2b/sin2a = h2

AB2 sin2 b  =  h2(1 - sin2b/sin2a) 

AB2 sin2 b sin2a =  h2( sin2a - sin2b)

 h2 = AB2 sin2 b sin2a/(sin2a - sin2b)

h = AB sin b sina/(sin2a - sin2b)




Answered by  | 12th Feb, 2010, 06:38: PM

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