From the top of a tower hm high, angle of depression of two objects, which are in line with the foot of the tower are α and β (β > α). Find the distance between the two objects
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Let the distance between two objects is X m.
and CD = y m.
Given that,
angle BAX = α angle ABD, [alternate angle]
angle CAY = β angle ACD [alternate angle]
and the height of tower, AD = he
Now, in ΔACD,
tan β = AD/CD = h/y
y = h/tanβ
__________________1
and in ΔABD,
tan α = AD/BD ≈ AD/BC + CD
tan α = h/x + y ≈ x + y = h / tan α
y = h / tan α - x
_________________ 2
From Eqs. (1) and (2),
h / tan β = h / tan α - x
x = h/tan α - h/tan β
= h {1/tan α - 1/tan β}
= h(cot α - cot β) [• cot θ = 1 / tan θ]
which is the required distance between the two objects.
Hence Proved..!!!!
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