Answer

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**Hint:**The rough figure that represents the given information is shown below.

We solve this problem by using the simple formula of speed that is

\[\text{Speed}=\dfrac{\text{Distance}}{\text{time}}\]

For finding the distance we use the tangent trigonometric ratio formula that is

\[\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}\]

By using this formula we calculate the distance travelled by bird in 2 sec to find the speed.

**Complete step-by-step answer:**

We are given that the bird is initially at a height of 80m at position B.

So, from the figure we can say that

\[\Rightarrow BD=80m\]

Let us assume that the bird moves to point C after 2 sec.

We are given that the bird maintains a constant height.

So, we can say that

\[\Rightarrow CE=BD=80m\]

We know that the tangent trigonometric ratio formula that is

\[\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}\]

Now, let us consider the triangle \[\Delta ABD\]

Now, by applying the tangent trigonometric ratio formula we get

\[\Rightarrow \tan {{45}^{\circ }}=\dfrac{DB}{DA}\]

We know that from the standard table of trigonometric ratios we have

\[\tan {{45}^{\circ }}=1\]

Now, by substituting the required values in above equation we get

\[\begin{align}

& \Rightarrow 1=\dfrac{80}{DA} \\

& \Rightarrow DA=80 \\

\end{align}\]

Now, let us consider the triangle \[\Delta ACE\]

Now, by applying the tangent trigonometric ratio formula we get

\[\Rightarrow \tan {{30}^{\circ }}=\dfrac{CE}{EA}\]

We know that from the standard table of trigonometric ratios we have

\[\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}\]

Now, by substituting the required values in above equation we get

\[\begin{align}

& \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{80}{EA} \\

& \Rightarrow EA=80\sqrt{3} \\

\end{align}\]

We know that from the figure the length ‘ED’ can be written as

\[\Rightarrow ED=EA-DA\]

Now, by substituting the required values in above equation we get

\[\begin{align}

& \Rightarrow ED=80\sqrt{3}-80 \\

& \Rightarrow ED=80\left( \sqrt{3}-1 \right) \\

\end{align}\]

We are given that the bird maintain a constant height so, we can say that

\[\Rightarrow BC=ED=80\left( \sqrt{3}-1 \right)\]

Let us assume that the speed of bird as \['v'\]

We are given that the bird reaches point C after 2 sec so, we can take the time as

\[\Rightarrow t=2\]

We know that the formula of speed that is

\[\text{Speed}=\dfrac{\text{Distance}}{\text{time}}\]

By using the above formula we get the speed of bird as

\[\Rightarrow v=\dfrac{BC}{t}\]

Now, by substituting the required values in above equation we get

\[\begin{align}

& \Rightarrow v=\dfrac{80\left( \sqrt{3}-1 \right)}{2} \\

& \Rightarrow v=40\left( \sqrt{3}-1 \right) \\

\end{align}\]

Therefore the speed of bird is \[40\left( \sqrt{3}-1 \right){}^{m}/{}_{s}\]

**Note:**Students may make mistakes for trigonometric ratio formula.

We have the tangent trigonometric ratio formula that is

\[\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}\]

This formula is applicable only when the triangle is right angled triangle. But, students may use this formula for all types of triangles which is a blunder mistake.

So, the formula significance needs to be taken care of.

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