As observed from top of a light house 100 meter above sea level the angel of depression of a ship sailing directly towards it change from 30° to 45° determine the distance traveled by the ship during the period of observation
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73.2 is the answer ..use tan of the given angles...check this out
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In Δ abd,
tan 45°=ab\bd
1 = 100\bd
bd = 100 --------- (i)
In Δ abc,
tan 30°=ab\bc
1\√3 = 100\bc
bc = 100√3 --------(ii)
distance between two shipps = (bc-bd)
∵ 100√3-100 =100(√3 - 1)m
∴ the required distance = 100(√3 -1)m
⇒°HOPE IT WILL HELP YOU!!
tan 45°=ab\bd
1 = 100\bd
bd = 100 --------- (i)
In Δ abc,
tan 30°=ab\bc
1\√3 = 100\bc
bc = 100√3 --------(ii)
distance between two shipps = (bc-bd)
∵ 100√3-100 =100(√3 - 1)m
∴ the required distance = 100(√3 -1)m
⇒°HOPE IT WILL HELP YOU!!
Aadi222:
CORRECTION: distance between the previous position and the present position of ship
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