Question

solve this .......☺️

Answer 1

Draw a perpendicular EC on AB

Horizontal component of u = u cos30

Horizontal component of 2u/3 = 2u/3 cosФ

Let both moving point will meet at point E in time T

Distance AC = speed × time = u cos30 × t

Distance CB = speed × times = 2u/ cosФ × t

Now,

in ΔAEC

tan30° = CE/ AC

CE = AC × tan30°

Similarly, in ΔECB

CE = CB × tanФ

Hence,

CB × tanФ = AC × tan30°

2ut/3 cosФ × tanФ = ut cos30° × tan30°

2u/3 × sinФ = u sin30°

sinФ = 3/4

Ф = sin⁻¹(3/4)