Question

Adhirath standing on a horizontal plane spots and aeroplane flying at a distance of 1 km from him and at an elevation of 60 degree. His friend standing on 10 M high Bridge from the same horizontal plane finds the angle of elevation of the same plane to be 30 degree.​

Answer 1

Answer: 1712m

Step-by-step explanation:

Completing the question : Both adhirath and his friend are on opposite sides of the aeroplane. find the distance of the aeroplane from the Man Standing On The Bridge.

1) Let the position of adhirath be at F and his is at Bridge BE  of height 10m.

A is position of aeroplane .

Check out figure for more information.

2)We have,  

AF = 1 km=1000m,

CD=BE=10m

In ΔAFD,

$sin(60\degree)=\frac{AD}{AF}\\ \\=>\frac{\sqrt{3}}{2}=\frac{AD}{1000}\\ \\=>AD=500\sqrt{3}m$

AC = (500√3-10)m

In ΔABC,

$sin(30\degree)=\frac{AC}{AB}\\ \\=>\frac{1}{2}=\frac{500\sqrt{3}-10}{AB}\\ \\=>AB=1000\sqrt{3}-20$

Hence,  distance from his friend to aeroplane 1000*1.732-20=1712m.