Question

2. A bird is sitting on the top of a tree, which is 80 m high. The angle of elevation of bird, from a
point on the ground is 45°. The bird flies 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.



plz tell​

Answer 1

Answer:

Step-by-step explanation:

Let P be the position of a bird at the height of 80 m with the angle of elevation 45∘ from A.

Let after 2 secounds, it reaches, at Q from where its elevation angle is 30∘.

Now, in right ΔPBA,

tan45∘=PBAB⇒1=80AB

⇒AB=80m...(1)

In right ΔQCA,

tan30∘=QCAC⇒13–√=80AC

⇒AC=803–√m...(2)

∴BC=AC−AB

=803–√−80=80(3–√−1)

=80(1.732−1)=80×0.732=58.56m

Now, speed of bird =Distancetime=58.562m/sec=29.28m/sec.