A boy runs on a horizontal road with a speed of 4 m/s while it is raining.
He sees that the rain is making an angle `theta` with the vertical while running from west to east.
However , when he runs from east to west, the angle is `alpha`.
The rain is pouring down at an angle `45^@` with the vertical normal and at a speed of 8 m/s as shown in the figure. the ratio `(tan theta) /(tan alpha)` is

Options

A. `(1-sqrt2)^2`
B. `(1+sqrt2)^2`
C. `(1+sqrt2)`
D. `(sqrt2-1)`

Answer

B

Explanation

A relative motion between rain and boy is shown in the figure below.

Given , speed of the boy on horizontal road is v= 4m/s
The boy is runs from east to west , the angle is = ` alpha` the rain is pouring down at an angle = `45^@`
Now, `tan A= (4sin 45^@)/(8-4 cos 45^@) = (4(1/sqrt2))/(8-4(1/sqrt2)) =1/(2sqrt2-1)`
Similarly , `tan B = (4 sin (90^@ +45^@))/(8-4 cos (90^@+45^@)) =1/ (2sqrt2+1)`
As from figure,
`tan theta = tan (A+45^@) =(tanA +tan 45^@)/(1- tan A tan 45^@)`
`rArr tan theta = ((1+2sqrt2-1)/(2sqrt2-1))/((2sqrt2-1-1)/(2sqrt2-1))=(2sqrt2)/(2sqrt2-2) =sqrt2/(sqrt2-1)`
Similarly, `tan alpha = tan (beta -45^@) = (-2sqrt2)/(2+2sqrt2) =(-sqrt2)/((1+sqrt2))`
Hence, `(tan theta)/(tan alpha) =((1+sqrt2)/(1-sqrt2))xx((1+sqrt2)/(1+sqrt2))= (1+sqrt2)^2`
So, the correct option is b