Question

A bird is sitting on the top of a 100 m high tower. From a point on the ground, the angle of elevation o
bird is 45º. The bird flies away horizontally in such a way that it's height remain the same from the ground.
After 10 seconds, the angle of elevation of the bird from the same point is 30°. Find the speed of flying of the
bird. [Take -5 = 1.73] 1.3​

Answer 1

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Answer 2

The speed of the flying bird is 4.2264 m/s.

Explanation:

Given:

        First angle of elevation is $45^{o}$.

        Second angle of elevation is $30^{o}$

        Height of bird is 100 m

        Time taken is 10 seconds.

Let the base distance for $45^{o}$ elevation be 'y'. and the base distance for $30^{o}$ elevation  be 'x'.

So, Using trigonometry for calculation of bases,

                     ⇒   tan $45^{o}$ = $\frac{100}{y}$

                     ⇒  y  = 100 m

                     ⇒  tan $30^{o}$ = $\frac{100}{x}$

                     ⇒   x = $\frac{100}{\sqrt{3} }$

(y - x) is the distance for calculating speed which is 42.262 m.

So,                              ⇒  speed =  4.262 m/s