an areoplane is flying at a height of 210m. flying at this height at some instant the angles of depression of two points in a line in opposite direction on both the banks of the river are 45 and 60 degrees. find the width of the river. (use root3=1.73)
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let the horizontal distance between aeroplane to nearest point is x m
tan 45= 210/X
1=210/X
X=210 m
let the horizontal distance between aeroplane to the farthest point is Y m
tan 60 = 210/Y
√3=210/Y
Y=210/√3
Y=121.25 m
So the width of river= X+Y m
210+121.25
331.25m
tan 45= 210/X
1=210/X
X=210 m
let the horizontal distance between aeroplane to the farthest point is Y m
tan 60 = 210/Y
√3=210/Y
Y=210/√3
Y=121.25 m
So the width of river= X+Y m
210+121.25
331.25m
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