From the top of a hill, the angles of depression of two consecutive kilometre 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.
Given :
From the top of Hill Angle of depression of two consecutive kilometre stones are
45°and 30°
To Find :
Distances of two stones from foot of the Hill.
Solution :
Height of hill AB = h
Distance BC = x km
Distance CD = 1 km
In , ABC $$\displaystyle tan\ \theta \ =\ \frac{opposite}{hypotenuse} \ $$
$$\displaystyle tan\ 45=\ \frac{h}{x} \ $$ (tan 45° = 1)
$$\displaystyle 1\ =\ \frac{h}{x}$$
h = x .....................................................(i)
In, ABD
$$\displaystyle tan\ 30\ =\ \frac{h}{x\ +\ 1} \ $$
$$\displaystyle \frac{1}{\sqrt{3}} \ =\ \frac{h}{x\ +\ 1} \ $$
Cross multiply,
$$\displaystyle x\ +\ 1\ =\ \sqrt{3} h$$
Rewrite,
$$\displaystyle \ \sqrt{3} h\ =\ x\ +\ 1\ ...........................................( ii)$$
Substitute (i) in (ii)
$$\displaystyle \ \sqrt{3} \ x\ =\ x\ +\ 1\ $$
$$\displaystyle \ \sqrt{3} \ x\ -\ x\ =\ 1\ $$
Take x as common,
$$\displaystyle x\ \left(\sqrt{3} \ -\ 1\right) \ =\ 1$$
$$\displaystyle x\ =\ \frac{1}{\sqrt{3} \ -\ 1}$$
Take conjucate,
$$\displaystyle x\ =\ \frac{1}{\sqrt{3} \ -\ 1} \ \times \frac{\sqrt{3} \ +\ 1}{\sqrt{3} \ +\ 1}$$
$$\displaystyle x\ =\ \frac{\sqrt{3} \ +\ 1}{\left(\sqrt{3} \ \ \right)^{2} -\ ( 1)^{2}}$$
$$\displaystyle x\ =\ \frac{\sqrt{3} \ +\ 1}{3\ -\ 1}$$
$$\displaystyle x\ =\ \frac{\sqrt{3} \ +\ 1}{2}$$
√3 = 1.732
$$\displaystyle x\ =\ \frac{\ 1.732+\ 1}{2}$$
$$\displaystyle x\ =\ \frac{\ 2.732}{2}$$
x = 1.366
Distance BC =1.366 km
Distance of first stone from foot of hill = 1.366 km
Distance of second stone from foot of hill :
$$\displaystyle BD\ =\ BC\ +\ CD$$
$$\displaystyle BD\ =\ 1.366\ +\ 1$$
Distance = 2.366 km
Distance of second stone from foot of hill = 2.366 km
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