Question

36. A bird is sitting on the top of a tree, which is 90 m high. The angle of elevation of the bird, from
a point on the ground is 45°. The bird fries away from the point of observation horizontally and
remains at a constant height. After 2 seconds, the angle of elevation of the bird from the point
of observation becomes 30°. Find the speed of flying of the bird.​

Answer 1

Answer:

very good question

Try it yourself. lt's Very easy

Answer 2

Given:

let AB be in the tree where A is the top of the tree and B is its base

STEP BY STEP EXPLANATION:

let the fixed point in the ground be O

So AOB = ${45}^ {0}$

in ∆ AOB, AB = height of tree = 80 m

OB = AB = 80 m

Now let the bird fly away xm from point a to point C

Draw CD perpendicular to the ground

So CD = AB = 80 m and DB = a = xm

Give elevation is ${30}^{0}$

, COD = ${30}^{0}$

in ∆ COD,

tan ${30}^{0}$

$\frac{cd}{od} \: = \: \frac{cd}{ob \: + \: bd} \: = \frac{80}{80 \: + \: x} \\ \\ </p><p></p><p>= \frac{1}{ \sqrt{3} } = \: \frac{80}{80 \: + \: x}$

so x = 80 $( \sqrt{3} \: - \: 1)$ = 58.4

ANSWER:

∴ Now the bird flies 58.4 m in 2 secs

So the speed of the bird is 29.2 m / secs

Hence verified !