From the top of a hill the angles of depression of two consecutive kilometer stones, due east are found to be 30° and 45° respectively. Find the distance of the two stones from the foot of the hill.

Given,

A is the top of the tower and B the foot. C and D be two consecutive kilometer stones with depression angles 30° and 45° respectively.

Since stones are consecutive kilometer stones hence distance between them = 1 km.

From figure,

∠XAD = ∠ADB = 30° [Alternate angles are equal]

∠XAC = ∠ACB = 45° [Alternate angles are equal]

CD = 1 km

DB = x + 1

From right angled ΔABC, we get

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 45^{\circ} = \dfrac{AB}{CB} \\[1em] \Rightarrow 1 = \dfrac{h}{x} \\[1em] \Rightarrow h = x .$

From right angled ΔADB, we get

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 30^{\circ} = \dfrac{AB}{DB} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{x + 1} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{x}{x + 1} \\[1em] \Rightarrow x + 1 = \sqrt3 x \\[1em] \Rightarrow \sqrt{3}x - x = 1 \\[1em] \Rightarrow x(\sqrt{3} - 1) = 1 \\[1em] \Rightarrow x = \dfrac{1}{\sqrt3 - 1} \\[1em] \Rightarrow x = \dfrac{1}{\sqrt3 - 1} \times \dfrac{\sqrt3 + 1}{\sqrt3 + 1} \\[1em] \Rightarrow x = \dfrac{\sqrt3 + 1}{(\sqrt{3})^2 - (1)^2} \\[1em] \Rightarrow x = \dfrac{2.73}{2} = 1.36 \text{ km}$

DB = x + 1 = 1.36 + 1 = 2.36 km.

Hence, the distance of two stones from hill are 1.36 km and 2.36 km.