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# A bird is sitting on a tree which is 80m high. The angle of elevation of a bird from a point on a ground is ${{45}^{\circ }}$ then the bird flies away from the point of observation horizontally and remains at a constant height. After 2 sec the angle of elevation from the point of observation becomes ${{30}^{\circ }}$. Find the speed of the flying bird.

Last updated date: 13th Jun 2024
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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.

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.