Question

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 ​

Answer 1

Answer:

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..!!!!

Answer 2

Answer:

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..!!!!

Step-by-step explanation: