question_answer

A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is \[45{}^\circ \]. The bird flies away horizontally in such a way that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is \[30{}^\circ \]. Find the speed of flying of the bird. (Take \[\sqrt{3}=1.732\])

Answer:

Let B be the initial position of bird sitting on top of tree of length 80 m.

After 2 sec, the position of bird becomes C.

Let the distance travel by bird from B to C is x m.

Now, in \[\Delta \text{ }ABO\]

\[\tan \,45{}^\circ =\frac{AB}{AO}=\frac{80}{y}\]

\[y=80\,m\]                                           ?(i)

And, in \[\Delta \,DCO\]

\[\tan \,30{}^\circ =\frac{CD}{DO}=\frac{80}{x+y}\]

\[\frac{1}{\sqrt{3}}=\frac{80}{x+80}\]                     [Using eq. (i)]

\[x+80=80\sqrt{3}\]

\[x=80\left( \sqrt{3}-1 \right)=80\times 0.732\]

\[\therefore x=58.56\,m\]

Hence, speed of flying of the bird \[=\frac{58.56}{2}\]

\[\left( Speed=\frac{Dis\tan ce}{Time} \right)\]

\[=29.28\,m/s\]