Question

from an aeroplane vertically above the straight horizontal road, the angles of depression of the two consecutive kilometer stones on opposite sides of the aeroplane are observed to be 60° and 30° show that height (in meters) of the aeroplane above the road is root 3÷4 km.
PLEASE...

Answer 1

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Answer 2

Answer:

Refer the attached figure

The two stones at point C and D are 1 km apart

So, CD = 1 km

Let CB be x

So, BD = 1-x

AB is the height of the airplane

We are given that  the angles of depression of the two consecutive kilometer stones on opposite sides of the aeroplane are observed to be 60° and 30°

So, ∠ACB =60° and ∠ADC = 30°

In ΔABC

$Tan \theta = \frac{Perpendicular}{Base}$

$Tan 60^{\circ}= \frac{AB}{CB}$

$\sqrt{3}= \frac{AB}{x}$

$\sqrt{3}x= AB$  --A

In ΔABD

$Tan \theta = \frac{Perpendicular}{Base}$

$Tan 30^{\circ}= \frac{AB}{BD}$

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

$\frac{1}{\sqrt{3}}(x-1)= AB$   --B

Equate A and B

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

$3x= 1-x$

$x=\frac{1}{4}$

Substitute the value of x in A

$\sqrt{3}(\frac{1}{4}= AB$  

$\frac{\sqrt{3}}{4}= AB$  

Hence the height of the aeroplane is $\frac{\sqrt{3}}{4}$