A stationary person observes that rain is falling vertically down at `30 km//hr`. A cyclist is moving up on an inclined plane making an angle `30^(@)` with horizontal at `10 km//hr`. In which direction should the cyclist hold his umbrella to prevent himself from the rain ?

Options

A. At an angle `tan^(-1) ((3sqrt3)/(5))` with horizontal
B. At an angle `tan^(-1) ((3 sqrt3)/(5))` with inclined plane
C. At an angle `tan^(-1) ((sqrt3)/(7))` with inclined plane
D. At an angle `tan^(-1) ((3 sqrt3)/(7))` with vertical.

Answer

B

Explanation

`tan theta = (35)/(5 sqrt3) = (7)/(sqrt3)`
Angle of umbrella with inclined plane `= theta - 30^(@)`
`tan (theta - 30^(@)) = (tan theta - tan 30^(@))/(1 + tan theta tan 30^(@)) = (3 sqrt3)/(5)`
`theta - 30^(@) = tan^(-1) ((3sqrt3)/(5))`