Question

25 Points....

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The horizontal distance between two trees of different heights is 30 m. The angle of depression of the top of the first tree when seen from the top of the second tree is 30°. If the height of the second tree is 40 m, find the height of the first tree.

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Answer 1

From Figure:

Let the height of the first tree = AB = h.

Let the height of the second tree = CE = 40 m.

The distance between the trees = BC = 30 m.

In ΔADE,

⇒ Tan 30 = DE/AD

⇒ (1/√3) = (40 - h)/30

⇒ 30 = √3(40 - h)

⇒ 30 = 40√3 - √3h

⇒ 30 = 69.28 - 1.732h

⇒ 39.28 = 1.732h

⇒ ~h = 22.68 m.

Therefore, the height of the tree = 22.68 m.

Hope it helps!

Answer 2

Given,horizontal distance between 2 trees

BC = 30m

since ,BC is parallel to AD

so then AD = BC = 30m

let the height of first tree be' h m'

given,height of second tree CE = 40 m

perpendicular distance from the top of first tree will be DE = ( 40 - h )m

since , we have given horizontal distance between two trees and we have to find the height of first tree. so then we solve either it for tanA or cotA .

let us solve it for cotA

since ,angle A = 30°

cotA = AD / DE

cot 30° = 30 / (40 - h )

√3 = 30 / ( 40 - h )

40 - h = 30/ √3

40 - h = (30 × √3) / (√3 × √3)

40 - h = 30√3/ 3

40 - h = 10√3

h = (40 - 10√3) m or

h = 40 - 17.32= 22.68m

Answer:
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height of the first tree = 22.68m

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