Question

man wishes to swim across a river 0.5 km wide.
can swim at the rate of 2 km/h in still water
1 the river flows at the rate of 1 km/h. The angle
made by the directon (w.r.t. the flow of the river)
long which he should swim so as to reach a point
exactly opposite his starting point, should be :​

Answer 1

Answer:

Angle = 120°

Explanation:

Refer to the attached image to better understand the case.

Given;-

              Vm = 2 km/h [where, Vm is the velocity of man]

              Vr = 1 km/hr [where, Vr is the velocity of river]

               D = 0.5 km [where, D is the width of the river]

[Method 1]:

Now, from the question, its clear that we need to find out the minimum path. We know that, for minimum path;-

             Vm sin∅ = Vr

Now,

                    sin ∅ = Vr/ Vm

                    sin ∅ = 1/2 [given, Vr =1 and Vm =2]

                          ∅ = 30°

Now, Let angle made by direction be α. Now;-

                       α = ∅ + 90°

                       α = 30° + 90°

                       α = 120°

[Method 2]:

Let us consider the following;

                      Vm = Hypotenuse

            Vm cos∅ = Perpendicular

            Vm sin∅ = Adjacent

By, Pythagoras Theorem, we have;-

           (Hypotenuse)² = (Perpendicular)² + (Adjacent)²

Now, Vm = 2 km/h, Vm cos∅= 1 , Vm sin∅ = ?

                           (2)² = (1)² + (?)²   (Let ? be x)

                           (x)² = 4 - 1

                            x = √3

Now, here,  tan ∅ = Vm/Vr

                   tan ∅ = 1/√3

                          ∅ = 30°

Then, Let the angle made by direction be β. Now;-

                         β = ∅ + 90°

                         β = 30° + 90°

                         β = 120°

Hope it helps ;-))