an aeroplane at an altitude of200m above a river abserves that the angle of deprassion of opposite point on the bans are 45°and 60° find in meters, the width of the river. (tack√3=1•73).
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285.76metersghh fjbff
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the position of aeroplane and river banks can be imagined like a triangle ABC where B and C are the river banks and aeroplane is at A. AD is altitude on BC
angle ABC = 45°
angle ACB = 60°
AD = 200 m
in ∆ ABD
tan 45 = AD/BD
1 = AD/BD
BD = AD = 200 m
in ∆ ACD
tan 60 = AD/DC
√3 = AD/DC
DC = AD/√3
= 200/√3
= 200*√3/3
= 200*1.73/3
= 346/3
= 115.33 m
width of river
BC = BD + DC
= 200 + 115.33
= 315.33 m
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