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Let us start the solution to the above question by letting the height of the bird from the ground be x meters.

Now, let us draw the situation of the above question by drawing the diagram of the situation given in the question.

According to the above figure, Amit is at point A and Deepak is at point C. We have to find EC.

In $ \Delta ADE $ , we know that $ \angle EAD=30{}^\circ $ . Also, we know that $ \sin x=\dfrac{perpendicular}{hypotenuse} $ .

$ \sin \left( \angle EAD \right)=\dfrac{ED}{EA}=\dfrac{x}{200} $

$ \Rightarrow \sin 30{}^\circ =\dfrac{x}{200} $

Now, we know that $ \sin 30{}^\circ =\dfrac{1}{2} $ . If we put this in the above equation, we get

$ \dfrac{1}{2}=\dfrac{x}{200} $

$ \Rightarrow x=100 $

Now let us move to triangle EFC. We know that $ \angle ECF=45{}^\circ $ and EF=(x-50)=100-50=50m. So, if we again use the definition of sine function, we get

$ \sin \left( \angle ECF \right)=\dfrac{EF}{EC} $

$ \Rightarrow \sin 45{}^\circ =\dfrac{50}{EC} $

Now, we know that $ \sin 45{}^\circ =\dfrac{1}{\sqrt{2}} $ .

$ \dfrac{1}{\sqrt{2}}=\dfrac{50}{EC} $

$ EC=50\sqrt{2}m $

Therefore, the distance of the bird from Deepak is $ 50\sqrt{2}m $ .