From the top of a hill, the angle of depression of two consecutive kilometer stones, due east are found to be 30º and 45º respectively. Find the distance of the two stones from the foot of the hill.
Answer: A
Fig.
Let AB be the hill whose foot is B and D and C are two kilometer stones.
Therefore DC = 1 km = 1000 m,
Let AB = h and BC = x.
In right angled Δ ABC,
tan 45º = AB/BC
=> 1 = h/x
=> x = h ------------- (1)
In right angled Δ ABC,
tan 30º = AB/BD
=> 1/√3 = h/(x + 1000)
=> x + 1000 = h√3 ----------------(2)
using (1) and (2) we get
x + 1000 = x√3
=> x(√3 – 1) = 1000
Or, x = [1000/(√3 – 1)] × (√3 + 1)/(√3 + 1)
Or, x = 1000 (√3 + 1)/2
= 500(√3 + 1)
= 500 × 2.732
= 1366 m or 1.366 km.
Therefore first km stone is 1.366 km and second km stone is 2.366 km from the foot of the hill.