Question

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.

Answer 1

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
$$\text{Speed}={\displaystyle \frac{\text{Distance}}{\text{time}}}$$
For finding the distance we use the tangent trigonometric ratio formula that is
$$\tan \theta ={\displaystyle \frac{\text{opposite side}}{\text{adjacent side}}}$$
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
$$\tan \theta ={\displaystyle \frac{\text{opposite side}}{\text{adjacent side}}}$$
Now, let us consider the triangle $$\mathrm{\Delta }ABD$$
Now, by applying the tangent trigonometric ratio formula we get
$$\Rightarrow \tan {45}^{\circ }={\displaystyle \frac{DB}{DA}}$$
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
$$\begin{array}{rl}& \Rightarrow 1={\displaystyle \frac{80}{DA}}\\ & \Rightarrow DA=80\end{array}$$
Now, let us consider the triangle $$\mathrm{\Delta }ACE$$
Now, by applying the tangent trigonometric ratio formula we get
$$\Rightarrow \tan {30}^{\circ }={\displaystyle \frac{CE}{EA}}$$
We know that from the standard table of trigonometric ratios we have
$$\tan {30}^{\circ }={\displaystyle \frac{1}{\sqrt{3}}}$$
Now, by substituting the required values in above equation we get
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{80}{EA}}\\ & \Rightarrow EA=80\sqrt{3}\end{array}$$
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
$$\begin{array}{rl}& \Rightarrow ED=80\sqrt{3}-80\\ & \Rightarrow ED=80(\sqrt{3}-1)\end{array}$$
We are given that the bird maintain a constant height so, we can say that
$$\Rightarrow BC=ED=80(\sqrt{3}-1)$$
Let us assume that the speed of bird as $${}^{\prime }{v}^{\prime }$$
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
$$\text{Speed}={\displaystyle \frac{\text{Distance}}{\text{time}}}$$
By using the above formula we get the speed of bird as
$$\Rightarrow v={\displaystyle \frac{BC}{t}}$$
Now, by substituting the required values in above equation we get
$$\begin{array}{rl}& \Rightarrow v={\displaystyle \frac{80(\sqrt{3}-1)}{2}}\\ & \Rightarrow v=40(\sqrt{3}-1)\end{array}$$
Therefore the speed of bird is $$40(\sqrt{3}-1){}^m/{}_s$$

Note: Students may make mistakes for trigonometric ratio formula.
We have the tangent trigonometric ratio formula that is
$$\tan \theta ={\displaystyle \frac{\text{opposite side}}{\text{adjacent side}}}$$
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.