Question

Amit, standing on a horizontal plane, finds a bird flying at a distance of
200 m from him at an elevation of 30°. Deepak standing on the roof of a
50 m high building, finds the angle of elevation of the same bird to be 45°
Amit and Deepak are on opposite sides of the bird. Find the distance of
the bird from Deepak.​

Answer 1

Answer:

Step-by-step explanation:

Let Amit is standing at point A, bird at point B, deepak at point D

height of the bird = BC, let E be a point on BC corresponding to point D at 50 m height

given,

AB = 200

∠A = 30°

∠D = 45°

In triangle ABC

Sin 30 = BC/AB

=> 1/2 = BC/200

=> BC = 100 m

=> BE = BC- EC = 100 - 50 = 50 m

In triangle,BED,

sin 45 = BE/BD

=> 1/√2 = 50/BD

=> BD = 50√2 = 70.7 m

Answer 2

Let us assume that B is boy named Amit and bird is flying at point A. As per question we have given that distance between boy and bird is 200 m i.e. AB = 200 m.

Also, a boy named Deepak standing at a roof of height 50 m. (In figure DE = 50 m.)

Angle of elevation between bird and Amit is 30° and better Deepak and bird is 45°.

Make a point parallel to Deepak i.e. F point. Join FE. Also, FC = 50 m.

In ∆ABC

→ sin 30° = AC/BC

→ 1/2 = AC/200

→ 1/1 = AC/100

→ AC = 100 m

Now,

→ AC = AF + FC

→ 100 = AF + 50

→ 100 - 50 = AF

→ 50 = AF

Also, In ∆AFC

→ sin 45° = AF/AE

→ 1/√2 = 50/AE

Cross-multiply them

→ AE = 50√2 m

Therefore,

The distance of bird from Deepak is 50√2m.