Arowboat crosses a river with a velocity of 3.53 mi/h at an angle 62.5° north of west relative to the water. the river is 0.900 mi wide and carries an eastward current of 1.25 mi/h. how far upstream is the boat when it reaches the opposite shore?

It can be helpful to draw a diagram. (See the attached.)

There are a number of ways to find the resultant speed and direction. Here, we will make use of the Law of Cosines and the Law of Sines. We have two sides of the speed triangle and the angle between them, so we can use the Law of Cosines to find the resultant speed (s).
  s² = 1.25² +3.53² -2·1.25·3.53·cos(62.5°) ≈ 9.948468
  s ≈ √9.948468 ≈ 3.154119 . . . . miles per hour

Then the Law of Sines can be used to find the obtuse angle between the direction of the current and the direction of the boat in the current. That angle (α) will satisfy
  sin(α)/3.53 = sin(62.5°)/3.154119
  α = arcsin(3.53·sin(62.5°)/3.154119) ≈ 96.919153°

That is, the boat's path is 6.919° west of north. The tangent of this angle gives the ratio between the upstream distance traveled and the across-stream distance traveled (Tan = Opposite/Adjacent). We know the across-stream distance is 0.900 miles, so the upstream distance is
  upstream distance = tan(6.919°)·0.900 mi ≈ 0.1092 mi

  The boat is 0.1092 miles upstream when it reaches the opposite shore.

You must find the surface of the block therefore 98187 dived by 5628 = 583m squared