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Home » From the top of a hill, the angles of depression of two consecutive kilometer stones, due east are found to be 300 and 450 respectively. Find the distance of two stones from the foot of the hill.
From the top of a hill, the angles of depression of two consecutive kilometer stones, due east are found to be 300 and 450 respectively. Find the distance of two stones from the foot of the hill.
ML Agarwal Heights and distances class 10 Maths
From the top of a hill, the angles of depression of two consecutive kilometer stones, due east are found to be 300 and 450 respectively. Find the distance of two stones from the foot of the hill.

Solution:

Consider A and B as the position of two consecutive kilometre stones

Here AB = 1 km = 1000 m

Take Distance BC = x m

Distance AC = (1000 + x) m

In right angled triangle BCD

CD/BC = tan 450

So we get

CD/BC = 1

CD = BC = x

In right angled triangle ACD

DC/AC = tan 300

x/ (x + 1000) = 1/√3

By cross multiplication

√3x = x + 1000

(√3 – 1)x = 1000

x = 1000/ (√3 – 1)

We can write it as

x = [1000/(√3 – 1) × (√3 + 1)/ (√3 + 1)]

x = [1000 (√3 + 1)/ (3 – 1)]

x = [1000 (√3 + 1)/ 2]

x = 500 (1.73 + 1)

So we get

x = 500 × 2.73

x = 1365 m

Here the distance of first stone from the foot of hill = 1365 m

Distance of the second stone from the foot of hill = 1000 + 1365 = 2365 m

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