The angle of elevation of the top of a tower as observed from the from a point in a horizontal plane through the foot of the tower is 32° when the observer moves towards the tower a distance of 100 M height find the angle of elevation of the top to be 60 degree find the height of the tower and the distance of the first position from the tower. (Tan32 = 0.6248, tan63 = 1.9626)
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let cb=x
tan 32°=AB/BD
0.6248=AB/100+x
62.48+0.6248x=48
tan 63°=AB/CB
1.9°626x -0.628x=62.48
1.3342x=62.48
1.3342x=62.48
x=62.48/1.3342
46.8295608=46.8
There AB=62.48+0.6298*46.8
91.72064 = 91.72
tan 32°=AB/BD
0.6248=AB/100+x
62.48+0.6248x=48
tan 63°=AB/CB
1.9°626x -0.628x=62.48
1.3342x=62.48
1.3342x=62.48
x=62.48/1.3342
46.8295608=46.8
There AB=62.48+0.6298*46.8
91.72064 = 91.72
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