A man standing on the deck of a ship, which is 10m above the water level, observes the angle of elevation of the top of a hill as 60 degree and the angle of depression of the base of the hill as 30 degree. Find the distance of the hill fom the ship and the height of the hill.
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Let a man is standing on the Deck of a ship at point a such that AB = 10 m & let CE be the hill
Thus, AB = CD = 10 m
The top and bottom of a hill is E and C.
Given, the angle of depression of the base C of the hill observed from A is 30° and angle of elevation of the top of the hill observed from A is 60 °
Then ∠EAD= 60° &
∠CAE= ∠BCA= 30°. (Alternate ANGLES)
Let AD = BC = x m & DE= h m
In ∆ ADE
tan 60° = Perpendicular / base = DE/AD
√3= h/x [tan 60° = √3]
h = √3x……..(1)
In ∆ ABC
tan 30° = AB /BC
[ tan30° = 1/√3]
1/√3 = 10/x
x= 10√3 m.. …………..(2)
Substitute the value of x from equation (2) in equation (1), we have
h = √3x
h= √3× 10√3= 10 × 3= 30 m
h = 30 m
The height of the hill is CE= CD+ DE= 10 +30= 40 m
Hence, the height of the hill is 40 m & the Distance of the hill from the ship is 10√3 m.
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Hope this will help you...
Thus, AB = CD = 10 m
The top and bottom of a hill is E and C.
Given, the angle of depression of the base C of the hill observed from A is 30° and angle of elevation of the top of the hill observed from A is 60 °
Then ∠EAD= 60° &
∠CAE= ∠BCA= 30°. (Alternate ANGLES)
Let AD = BC = x m & DE= h m
In ∆ ADE
tan 60° = Perpendicular / base = DE/AD
√3= h/x [tan 60° = √3]
h = √3x……..(1)
In ∆ ABC
tan 30° = AB /BC
[ tan30° = 1/√3]
1/√3 = 10/x
x= 10√3 m.. …………..(2)
Substitute the value of x from equation (2) in equation (1), we have
h = √3x
h= √3× 10√3= 10 × 3= 30 m
h = 30 m
The height of the hill is CE= CD+ DE= 10 +30= 40 m
Hence, the height of the hill is 40 m & the Distance of the hill from the ship is 10√3 m.
=================================================================
Hope this will help you...
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