an observer flying in an aeroplane at an altitude of 900m observes 2 ships in front of him, which are in the same direction at an angles of depression of 60 and 30 respectively. find the distance between the two ships
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see diagram. P is the plane. C and D are ships.
A = angle of depression made by ship C at P = 60 deg
B = angle of depression made by ship D at plane P = 30 deg
CD = ?
angle A = angle PEC as they are alternate angles between parallel lines and PC is the transversal line.
angle B = angle PDE , alternate angles between parallel lines with PD being the transversal line.
in triangle PCE, Tan A = 900/EC => EC = 900/ tan 60°
in triangle PDE , Tan B = 900/ED => ED = 900 / tan 30°
CD = distance between ships = ED - EC = 1039.23 meters
A = angle of depression made by ship C at P = 60 deg
B = angle of depression made by ship D at plane P = 30 deg
CD = ?
angle A = angle PEC as they are alternate angles between parallel lines and PC is the transversal line.
angle B = angle PDE , alternate angles between parallel lines with PD being the transversal line.
in triangle PCE, Tan A = 900/EC => EC = 900/ tan 60°
in triangle PDE , Tan B = 900/ED => ED = 900 / tan 30°
CD = distance between ships = ED - EC = 1039.23 meters
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