The heading of an airplane is the direction in which the airplane is pointed. The heading is measured clockwise from the north and expressed in degrees. 

The airspeed is the speed of the airplane in still air.

The course of an airplane is the direction in which it moves relative to the ground. The course is measured clockwise from the north.

The groundspeed is the speed of the airplane relative to the ground.

The drift angle is the positive difference between the heading and the course. 

You may use vectors to represent airspeed and heading, direction and speed of wind, or groundspeed and course. The groundspeed vector is the resultant of the airspeed vector and the wind vector.

The correct answer is  because  and  are component vectors for the weight .

We can draw a diagram of the given information, including the weight of the box and the force  against the inclined plane and the force  pushing down on the inclined plane. Additionally, we can label the angle of inclination as , but furthermore, in this type of problem, the angle formed between  and  is equal to that same measure, so that has also been labelled  also.

Because these two angles are equal,  and  (see second diagram).

Using the above formulas, we get:

 lbs

 lbs

Next understand that the minimum force needed to prevent the barrel from rolling down the plane corresponds to , so the minimum force to prevent the barrel from rolling is 31.19 lbs. The force against the inclined plan is  lbs. Finally, because , the barrel will not roll down the inclined plane. 

First, draw a diagram of the given information. We can label the angle of inclination as , but furthermore, in this type of problem, the angle formed between  and  is equal to that same measure, so that has also been labelled . Because these two angles are equal,  and  (see second diagram).

Using the above formulas, we get:

 lbs

 lbs

Next understand that the minimum force needed to prevent the barrel from rolling down the plane corresponds to , so the minimum force to prevent the barrel from rolling is  lbs. The force against the inclined plan is  lbs. Finally, because , the barrel will not roll down the inclined plane. 

We can draw a diagram of the given information, including the weight of the box and the force  against the inclined plane and the force  pushing down on the inclined plane. Additionally, we can label the angle of inclination as , but furthermore, in this type of problem, the angle formed between  and  is equal to that same measure, so that has also been labelled  also.

Because these two angles are equal,  and  (see second diagram).

Using the above formulas, we get:

 lbs

 lbs

Next understand that the minimum force needed to prevent the barrel from rolling down the plane corresponds to , so the minimum force to prevent the barrel from rolling is 16.82 lbs. The force against the inclined plan is  lbs. Finally, because , the barrel will not roll down the inclined plane. 

To work through an air navigation problem, we must first understand how these forces act on one another. In these problems, we have three vectors: one representing airspeed, one representing wind, and one representing groundspeed. The groundspeed vector is the resultant of the airspeed and wind vectors, as you can see in the above diagram. These three vectors create a triangle. 

To find the groundspeed, simply use the Pythagorean Theorem:

 miles per hour.

To find the course, we need to find the smallest angle of the triangle (the angle in between the blue and purple vectors), then subtract this from the given angle of . While we don't know that angle , we can use the tangent function to find it, as below:

We can find the course by subtraction:

Therefore the groundspeed is 222.77 miles per hour and the course is .

First, let's break down how this figure was constructed. In any air navigation problem, we have three vectors that create a triangle: groundspeed, airspeed, and wind. We've drawn the groundspeed vector due north, along ON. Next, the wind vector is drawn  off from the groundspeed vector, and finally, the airspeed vector closes the triangle. At this point, you can label all known quantities. 

To begin solving the problem, we can start by finding the groundspeed. This requires only Pythagorean Theorem to find:

Therefore the groundspeed is 208.51 kilometers per hour.

Next, let's find the heading such that the plane tracks . Since we can't solve for the heading directly, we need to find , then subtract it from .

Now to find the heading, subtract this from .

.

Therefore the heading is .