Answered By Rohtaz S

As mentioned that the boat cruising speed is 12km/hr. The straight has a 3km/hr speed towards west, which means that the pilot will have to negate this speed to travel in a straight line in the North

(Georgetown to Medan)

Lets assume the speed of the boat towards south = X km/hr

Required speed of the boat towards east = 3 km/hr

By Using Pythagoras theorem 

X²+3²=12²

^{Solve the equation and we get}

X = 11.6189km/hr

Angle = Tan^-1(3/11.6189) = 14.48^o

So, the rider has to set course at angle 14.48^o from the North direction towards East to enable him to travel straight North.

Now, lets calculate time taken:

^{Given, Distance between towns is 100KM}

^{Time taken to reach Medan = Distance/Speed = (100Km)/(11.6189km/hr)= 8.60Hours}

Answered By Rohtaz S

Correction - X is the speed of boat towards north

Answered By Majid B

The velocity of the boat relative to the Earth:  $\overline{v}_{BE}$v_B‍E 

The velocity of the water relative to the Earth:  $\overline{v}_{WE}$v_WE 

The velocity of the boat relative to the Water:   $\overline{v}_{BW}$v_BW 

 $\overline{v}_{BE}=\overline{v}_{BW}+\overline{v}_{WE}$v_BE=v_BW+v_WE     

 $v_{BE}=\sqrt{v_{BW}^2-v^2_{WE}}=\sqrt{12^2-3^2}=\sqrt{135}=11.62$v_BE=√v_BW²−v²_WE=√12²−3²=√135=11.62 km/h

 $\theta=\cos^{-1}\left(\frac{v_{WE}}{v_{BW}}\right)=\cos^{-1}\left(\frac{3}{12}\right)=75.52^{\circ}$‍θ=cos⁻¹(v_WEv_BW )=cos⁻¹(312 )=75.52^? 

  $\overline{v}_{BW}=12$v_BW=12 km/h @  $75.52^{\circ}$75.52^? N of E

 $t=\frac{d}{v_{BE}}=\frac{100}{11.62}=8.61$t=dv_BE =10011.62 =8.61 h

Answered By Sosimo H

Let use the coordinates system, where x=3 Km /h and y=12 Km/h , Now we need to calculate the angle formed by those two sides . In this case we use the tangent .

  $\tan\theta=12\div3=4$tanθ=12÷3=4  ;   $\theta=\tan^{-1}4=76$θ=tan⁻¹4=76 

Therefore the pilot should take the course of 76 degrees  in order to compensate for the current.

The time that the sailing take is : t=distance/ velocity = 100 Km /12 Km/h = 8.33 Hours 

 $^{\circ}$^?