= x=(3000 - 1000, 3) = ( 3000 - 1000 x1.73)
= (3000 -1730) = 1270.
Hence, the required distance between the two aeroplanes
1270 metres.
A man standing on the deck of a ship, which is 10 m above the wat
level, observes the angle of elevation of the top of a hill as 60°, and t!
angle of depression of the base of the hill as 30. Find the distance
the hill from the ship and the height of the hill. [CBSE 2006, 1
Let AB be the deck and CD be the hill.
D
Let the man be at B.
Then, AB = 10 m.
10) m
Answers
Answer:
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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