Answer: [tex]240\sqrt{3}km/h[/tex]
Step-by-step 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,
[tex]\tan45^{\circ}=\frac{AB}{BC}\\\Rightarrow1=\frac{1000\sqrt{3}}{BC}\\\Rightarrow\ BC=1000\sqrt{3}[/tex]
In triangle DEC,
[tex]\tan30^{\circ}=\frac{DE}{EC}\\\Rightarrow\frac{1}{\sqrt{3}}=\frac{1000\sqrt{3}}{EC}\\\Rightarrow\ EC=3\times1000\\\Rightarrow\ EC=3000[/tex]
Now, EB= [tex]3000-1000\sqrt{3}=1000(3-\sqrt{3})[/tex]
Speed of eagle =[tex]=\frac{EB}{time}=\frac{1000\sqrt{3}}{15}[/tex]
In km/hr
Speed of eagle =[tex]\frac{1000\sqrt{3}}{15}\times\frac{3600}{1000}=240\sqrt{3}km/h[/tex]
