Practical Applications - Trigonometry

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 .

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 31.19 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 .

We can use the pythagorean theorem to solve this problem. Using as our hypotenuse, we can drop a vertical vector perpendicular to the x-axis. We will call this and it is 4 units in length. We can also extend a vector from the origin that connects to . We will call this and it is 3 units in length.

Using the pythagorean theorem:

When summing two vectors, you have both an x and y component and you sum these separately leaving you with a coordinate as your answer. This coordinate is called a resultant.

When determining the resultant of two vectors, you are finding the sum of two vectors. To do this you must add the x component and the y component separately.

To find the resultant we must sum the two vectors:

Now we must graph the resultant

First, we must understand what we are solving for. We are solving for the angle that is formed by and the x-axis. To do this, we can extend a vector from the origin which stops directly under the end of . We will call this new vector and it will be 7 units long. We will also extend a vector upwards that is perpendicular to the x-axis. We will call this and it will be 3 units long.

Now we can use the relationship that where is the adjacent side and is the opposite side.

And so is 23.2 degrees above the x-axis.

When finding the difference of two vectors, you must subtract the x and y components separately.

To find the difference of two vectors we must consider the x and y components separately.

And then we must correctly graph this vector

When finding the magnitude of the vector, you use either the Pythagorean Theorem by forming a right triangle with the vector in question or you can use the distance formula. This is much more clear considering the distance vector that the magnitude of the vector is in fact the length of the vector.

The correct image depicting an aeronautical bearing of is

This image begins at north, and moves clockwise from it.

Three of the given incorrect answers depict , , . The fourth incorrect answer does not represent a standard bearing convention as it is neither an acute angle, nor in the clockwise direction. That incorrect answer looks like:

The bearing of a point B from a point A in a horizontal plane is defined as the acute angle made by the ray drawn from A through B with the north-south line through A. The bearing is read from the north or south line toward the east or west. Bearing is typically only represented in degrees (or degrees and minutes) rather than radians. To find , start in the south direction, then move towards the west:

The other incorrect answer choices provided depict , , and .

This question and its answer choices give you a few clues to work with. First, we need to identify the bearing angle being shown. The options in the answer choices are either , , or . Because the angle begins in the south direction and moves towards the west, the correct bearing is . That means only two of the answer choices could be correct. We now need to understand how the miles per hour corresponds to the problem. Notice that there is no answer choice that has the bearing of and velocity of miles per hour. Rather, we need to choose between miles per hour for hours or miles per hour for hours. Because miles per hour for hours corresponds to (whereas the other option corresponds to only ), the correct answer is "A motorboat traveling at miles per hour for hours."

First, let's set up a diagram using the given information. This looks like this:

Next, let's convert this info into a triangle so that we can use trigonometry to solve the problem. We need to calculate that the ship going miles per hour for hours will have traveled miles.

Now we can use trigonometry to determine the missing sides, s and e.

Therefore the ship has travelled 181.87 miles south and 105 miles east.