Question

From the tpo of a hill the angles of depression of two consecutive kilometre stones, due east are found to be ${30}^0$and ${45}^0$ respectively. Find the distance the two stones from the foot of the hill.

Answer 1

Let $AB$ be the tower.

Let $x$ be the distance between the first stone and the tower.

Therefore,

Distance between the second stone and the tower $=(x+1)$

Now,

For first stone:-

$\theta =45°$

As we know that,

$\tan \theta =\frac{\text{Perpendicular}}{\text{Base}}$

Therefore,

In $△ABC$,

$\tan 45°=\frac{AB}{BC}$

$\Rightarrow AB=x.....\left(1\right)$

Similarly, for second person:-

$\theta =30°$

In $△ABD$,

$\tan 30°=\frac{AB}{BD}$

$\Rightarrow AB=\frac{x+1}{\sqrt{3}}.....\left(2\right)$

From equation $\left(1\right)\&\left(2\right)$, we have

$x=\frac{x+1}{\sqrt{3}}$

$\sqrt{3}x=x+1$

$x=\frac{1}{\sqrt{3}-1}$

$x=\frac{\sqrt{3}+1}{2}km$

Therefore,

$BC=x=\frac{1+\sqrt{3}}{2}km$

$BD=x+1=\frac{3+\sqrt{3}}{2}km$

Hence the distance of the two stones from the foot of the hill is $\frac{1+\sqrt{3}}{2}$ and $\frac{3+\sqrt{3}}{2}$ respectively.