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प्रश्न
From the top of a 60m high building the angles of depression of the top and bottom of a lamp post are 30° and 60° respectively. Find the distance on the ground between the building and the lamp post and the difference in their heights.
उत्तर
Let AB be the building. Then, AB = 60 m.
Let the height of the lamp post (CD) be h.
Let the distance between the building and the lamp post be x.
In ΔACB,
`tan60^circ = "AB"/"BC"`
∴ `sqrt(3) = 60/X`
∴ `X = 60/sqrt(3) = 20sqrt(3) = 20 × 1.732 = 34.64` ...(1)
Thus, the distance between the building and the lamp post is 34. 64 m
In ΔADE,
`tan30^circ = "AE"/"DE"`
∴ `1/sqrt(3) = (60 - h)/X`
∴ `X = sqrt(3)(60 - h)` ..(2)
From (1) and (2):
`sqrt(3)(60 - h) = 20sqrt(3)`
60 - h = 20
h = 40
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संबंधित प्रश्न
In the figure given, from the top of a building AB = 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to 30o and 60o respectively. Find:
1) The horizontal distance between AB and CD.
2) The height of the lamp post.
Evaluate without using trigonometric tables.
`2((tan 35^@)/(cot 55^@))^2 + ((cot 55^@)/(tan 35^@)) - 3((sec 40^@)/(cosec 50^@))`
A man observes the angle of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60 m, the angle of elevation changes to 60°. Find the height of the building correct to the nearest metre.
In the given figure, from the top of a building AB = 60 m hight, the angle of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find:
- the horizontal distance between AB and CD.
- the height of the lamp post.
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