Practical Applications - Trigonometry

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:

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.

First, let's incorporate the given information into a diagram. Start by labelling the plane's bearing of along with its velocity 330km. Next, draw a line segment to complete the triangle and determine the measures of the angles of the triangle. We can determine the angle , we constructed the diagram such that there is a right angle, and finally we can find the third angle by taking .

The question is asking us how far south and how far east the plane is from its starting point, so we need to now use trigonometry to determine the lengths of the missing sides of the triangle. We will call these sides s for the southward distance and e for the eastward distance.

km

km

Therefore the airplane is 139.46 km south of its starting point and 299.08 km east of its starting point.

To solve this problem, begin with a diagram, and label all known information. We know that we can label two angles as and one length as 240. Then, by deductive reasoning, we can label another angle as because .

Next, we need to use trigonometry to find the answers to each question we're being asked. To find the distance AB, set up

Next, we can find the distance BC. There are two ways to do this since we know two angles of the triangle, but either way you need to use the tangent function.

Finally, we are asked to find the bearing from Ship B to A. Be careful, because in the initial problem we are given the Bearing from Ship A to Ship B. While the angle will remain the same, the direction is different because we are starting at a different initial point (B instead of A). With B as your starting point, A is north and west. Therefore the bearing from Ship B to Ship A is .

Therefore the correct distance between Ships A and B is 640.67 miles, the distance between ships B and C is 594.02 miles, and the bearing from Ship B to Ship A is .

Begin by diagramming the given information; you'll see that the three ships create a right triangle. To solve the question, we need to find the unknown angles of the triangle, then frame our answers as the proper bearings.

First let's solve for the topmost angle in the diagram, which we'll call .

By alternate interior angles, the other diagrammed angle will also be equal to .

Now, put your cursor on Island Pioneer, and see if you need to move north or south, and then east or west to get to Sea Terraformer. You need to move north, then west. Therefore the bearing between Island Pioneer and Sea Terraformer is . Now put your cursor on Sea Terraformer and complete the same process to get to Island Pioneer; you'll go south, then east. Therefore the bearing from Sea Terraformer to Island Pioneer is .

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.