Ariver has a constant current of 3 km per hour. if a motorboat, capable of maintaining a constant speed of 20km per hour, tries to reach a point directly opposite the dock, what will his actual speed and direction be? if the river is 2 km wide, how long will it take to cross?

Construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have: 

theta = arcsin(3/20) = approx. 8.63° 

The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line. 

x = sqrt(20^2 - 3^2) 
= sqrt(400 - 9) 
= sqrt 391 

The boat's crossing time = 
0.5 km/(sqrt 391 km/hr) 
= (0.5/sqrt 391) hr 
= approx. 0.025 hr 
= approx. 91 seconds

Ithink the answer is d i hope it my friend message me if i’m wrong and i’ll change my answer and fix it for you