Asked by Anonymous
The angle of elevation of the top of a tower is 45degree from a point 10m above the water level of a lake.From the same point, the angle of depression of its image in the lake is 60degree.Find the height of the t
Answers
Answered by
Steve
Draw a diagram. Let point M be where the image of the top of the tower appears.
P = point of observation
Q = point on water below P
T = top of tower
B = base of tower at water level
so, the height h of the tower is BT
So, now we have
PM = 10/√3
Since light reflects at equal angle, angle TMB is also 60°
Now we have
h/MB = tan60° = √3
But, MB = QB-QM - h-10-10/√3
so,
h/(h-10-10/√3) = √3
h = 10(2+√3) ≈ 37.32 m
P = point of observation
Q = point on water below P
T = top of tower
B = base of tower at water level
so, the height h of the tower is BT
So, now we have
PM = 10/√3
Since light reflects at equal angle, angle TMB is also 60°
Now we have
h/MB = tan60° = √3
But, MB = QB-QM - h-10-10/√3
so,
h/(h-10-10/√3) = √3
h = 10(2+√3) ≈ 37.32 m
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