Question

From the top of a hill, the angles of depression of two consecutive kilometer stones, due east, are found to be ${30}^{\circ }and{45}^{\circ }$ respectively. Find the distances of the two stones from the foot of the hill.

Answer 1

Let the height of the hill AB = h km

C and D are two stones of the hill at a distance of 1 km .

Angles of depression of C and D 45° and 30° respectively.

Assume that Distance from foot of the hill to first stone AC = x km.

Then, Distance of the second stone from the hill = BD = x + 1 km

In $△CAB$

$\tan {45}^0=\frac{AB}{AC}$

$1=\frac{h}{x}$

$h=x$ --------(1)

In a $△DAB$

$\tan {30}^0=\frac{AB}{BD}$

$\to \frac{1}{\sqrt{3}}=\frac{h}{(x+1)}$

$\to \sqrt{3}\times h=(x+1)$

$\to \sqrt{3}\times h=h+1from\left(1\right)$

$\to \sqrt{3}\times h-h=1$

$\to h=\frac{1}{\surd 3-1}$ (rationalize denominator )

$\to h=\frac{\sqrt{3}+1)}{2}=\frac{1.732+1}{2}$

$\therefore h=\frac{2.732}{2}=\mathrm{1.366.}$
$\therefore x=1.366$.

$\therefore$ Distance of the two stones from the foot of the hill are 1.366 km and 2.366 km (1+1.366 km).