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From the top of a tower h m high, the angles 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

Solution:

Given, from the top of a tower h m high, the angles of depression of two objects which are in line with the foot of the tower are α and β (β > α)

We have to find the distance between the two objects.

Let AB be the tower

The height of the tower = h m

Angle of depression, ∠ABD = 𝛼

Angle of depression, ∠ACD = β.

Let us consider,

BC = x

CD = y

By the properties of a triangle, we know that alternate angles of a triangle are equal.

So, ∠ABD = ∠BAX = 𝛼

∠ACD = ∠CAY = β.

Considering triangle ABD,we get

tan𝛼 = AD/BD

tan𝛼 = h/BC + CD

tan𝛼 = h/x+y

y = h/tan𝛼 - x -------------------(1)

Considering triangle ACD,we get

tanβ = AD/CD

tanβ = h/y

y = h/tanβ -----------------------(2)

Now, by comparing (1) and (2), we get,

h/tan𝛼 - x = h/tanβ

x = h/tan𝛼 - h/tanβ

x = h(1/tan𝛼 - 1/tanβ)

x = h(cot𝛼 - cotβ)

Therefore, the required distance is h(cot𝛼 - cotβ).

✦ Try This: From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8

NCERT Exemplar Class 10 Maths Exercise 8.4 Problem 14

From the top of a tower h m high, the angles 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

Summary:

From the top of a tower, the angles of depression of two objects, which are in line with the foot of the tower are α and β (β > α). The distance between the two objects is h(cot𝛼 - cotβ)

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