A moving boat is observed from the top of a 150m high cliff moving away from the cliff angle of depression change from 60to45 in 2min.find the speed of boat
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Answer:
Height of cliff i.e. AB = 150 m
The angle of elevation ab , ∠ACB = 6 °
Distance of boat from the bottom of the cliff = BC
Changed angle of elevation , ∠ADB = 45°
Now, Distance of boat from the bottom of the cliff = BD
Distance traveled by boat in 2 minutes = CD = BD-BC
In ΔABD
tanθ = Perpendicular / Base
tan45° =AB/BC
[∵tan45°=1]
1= 150/BC
BC =150/√3
So distance traveled by boat = 2 min
CD=BD-BC
150 + CD = 150 × root 3
150 + CD = 150 × 1.732 ( given )
150 + CD = 259.8
CD = 259.8 - 150
CD = 109.8 m
Speed = Distance / Time
Speed = 109.8 / 2
Speed of boat = 54.9 meter/minute
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@GauravSaxena01
Height of cliff i.e. AB = 150 m
The angle of elevation ab , ∠ACB = 6 °
Distance of boat from the bottom of the cliff = BC
Changed angle of elevation , ∠ADB = 45°
Now, Distance of boat from the bottom of the cliff = BD
Distance traveled by boat in 2 minutes = CD = BD-BC
In ΔABD
tanθ = Perpendicular / Base
tan45° =AB/BC
[∵tan45°=1]
1= 150/BC
BC =150/√3
So distance traveled by boat = 2 min
CD=BD-BC
150 + CD = 150 × root 3
150 + CD = 150 × 1.732 ( given )
150 + CD = 259.8
CD = 259.8 - 150
CD = 109.8 m
Speed = Distance / Time
Speed = 109.8 / 2
Speed of boat = 54.9 meter/minute
===================
@GauravSaxena01
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sankruno1:
the answer is wrong
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