Question

an eagle is flying at a constant height of 1000root3 meters from the ground level . a boy standing on the ground observes the angle of elevation of the eagles as 45. after 15 seconds it's angle becomes 30. find it's speed in km/hr

Answer 1

Answer:
`240√(3)km/h`

Explanation:

Let AB denote the height of eagle and and after 15 seconds it will become DE.

Also denote C is the point where boy is standing.

Now since both the triangles ABC and DEC are right angle triangle.

Now, in triangle ABC,

`\tan45^(\circ)=(AB)/(BC)\\\Rightarrow1=(1000√(3))/(BC)\\\Rightarrow\ BC=1000√(3)`

In triangle DEC,

`\tan30^(\circ)=(DE)/(EC)\\\Rightarrow(1)/(√(3))=(1000√(3))/(EC)\\\Rightarrow\ EC=3*1000\\\Rightarrow\ EC=3000`

Now, EB=
`3000-1000√(3)=1000(3-√(3))`

Speed of eagle =
`=(EB)/(time)=(1000√(3))/(15)`

In km/hr

Speed of eagle =
`(1000√(3))/(15)*(3600)/(1000)=240√(3)km/h`