Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

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Example Questions

Example Question #1 : Inclined Planes And Air Navigation

A 150 pound barrel rests on a  inclined plane. What is the minimum force (ignoring friction) needed to keep the barrel from rolling down the incline? What is the force the barrel exerts against the inclined plane? Will the barrel stay in place or roll?

Possible Answers:

Minimum Force needed to prevent rolling: 146.72 lbs

Force barrel exerts against plane: 31.19 lbs

Will the barrel roll down the plane? Yes

Minimum Force needed to prevent rolling: 153.35 lbs

Force barrel exerts against plane: 721.46 lbs

Will the barrel roll down the plane? No

Minimum Force needed to prevent rolling: 31.19 lbs

Force barrel exerts against plane: 146.72 lbs

Will the barrel roll down the plane? No

Minimum Force needed to prevent rolling: 721.46 lbs

Force barrel exerts against plane: 153.35 lbs

Will the barrel roll down the plane? Yes

Correct answer:

Minimum Force needed to prevent rolling: 31.19 lbs

Force barrel exerts against plane: 146.72 lbs

Will the barrel roll down the plane? No

Explanation:

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.
Screen shot 2020 08 03 at 6.06.45 pm

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

Screen shot 2020 08 03 at 6.17.12 pm

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. 

Example Question #1 : Inclined Planes And Air Navigation

A 450 pound barrel rests on a  inclined plane. What is the minimum force (ignoring friction) needed to keep the barrel from rolling down the incline? What is the force the barrel exerts against the inclined plane? Will the barrel stay in place or roll?

Possible Answers:

Minimum Force needed to prevent rolling:  lbs

Force barrel exerts against plane:  lbs

Will the barrel roll down the plane? No

Minimum Force needed to prevent rolling:  lbs

Force barrel exerts against plane:  lbs

Will the barrel roll down the plane? No

Minimum Force needed to prevent rolling: lbs

Force barrel exerts against plane:  lbs

Will the barrel roll down the plane? Yes

Minimum Force needed to prevent rolling:  lbs

Force barrel exerts against plane:  lbs

Will the barrel roll down the plane? Yes

Correct answer:

Minimum Force needed to prevent rolling:  lbs

Force barrel exerts against plane:  lbs

Will the barrel roll down the plane? No

Explanation:

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).

Screen shot 2020 08 03 at 5.26.51 pm

Screen shot 2020 08 03 at 6.17.01 pm

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. 

Example Question #1 : Inclined Planes And Air Navigation

A 65 pound barrel rests on a  inclined plane. What is the minimum force (ignoring friction) needed to keep the barrel from rolling down the incline? What is the force the barrel exerts against the inclined plane? Will the barrel stay in place or roll?

Possible Answers:

Minimum Force needed to prevent rolling: 67.29 lbs

Force barrel exerts against plane: 251.14 lbs

Will the barrel roll down the plane? No

Minimum Force needed to prevent rolling: 16.82 lbs

Force barrel exerts against plane: 62.79 lbs

Will the barrel roll down the plane? No

Minimum Force needed to prevent rolling: 16.82 lbs

Force barrel exerts against plane: 62.79 lbs

Will the barrel roll down the plane? Yes

Minimum Force needed to prevent rolling: 251.14 lbs

Force barrel exerts against plane: 67.29 lbs

Will the barrel roll down the plane? Yes

Correct answer:

Minimum Force needed to prevent rolling: 16.82 lbs

Force barrel exerts against plane: 62.79 lbs

Will the barrel roll down the plane? No

Explanation:

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.

Screen shot 2020 08 03 at 6.07.48 pm

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

Screen shot 2020 08 03 at 6.17.20 pm

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. 

Example Question #2 : Inclined Planes And Air Navigation

The heading of an airplane is  and the airspeed is  miles per hour. Find the groundspeed and course if there is a wind of  miles per hour from . Refer to the figure below. 

 

Screen shot 2020 08 04 at 9.47.06 am

Possible Answers:

Groundspeed: 255 miles per hour

Course: 

Groundspeed: 222.77 miles per hour

Course: 

Groundspeed: 255 miles per hour

Course: 

Groundspeed: 222.77 miles per hour

Course: 

Groundspeed: 222.77 miles per hour

Course: 

Correct answer:

Groundspeed: 222.77 miles per hour

Course: 

Explanation:

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 .

Example Question #1 : Inclined Planes And Air Navigation

The airspeed of an airplane is 210 kilometers per hour. There is a wind of 25 kilometers per hour from . Find the heading and the groundspeed in order to track . Refer to the figure below.

 

Screen shot 2020 08 04 at 10.00.04 am

Possible Answers:

Groundspeed: 185 kmph

Heading: 

Groundspeed: 208.51 kmph

Heading: 

Groundspeed: 185 kmph

Heading: 

Groundspeed: 208.51 kmph

Heading: 

Correct answer:

Groundspeed: 208.51 kmph

Heading: 

Explanation:

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 .

Example Question #1 : Vectors

Determine the magnitude of vector A.

 

 

 

Possible Answers:

Correct answer:

Explanation:

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:

 

 

Example Question #1 : Vectors

Which of the following is the correct term for the sum of two vectors?

Possible Answers:

Magnitude

Resultant

Sum

Velocity

Correct answer:

Resultant

Explanation:

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.

Example Question #2 : Vectors

Determine the resultant of  and  .

Possible Answers:

Correct answer:

Explanation:

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.

 

 

Example Question #2 : Vectors

Consider the following graphs where  begins at the origin and ends at  and .  Which of the following depicts the correct resultant of these two vectors.

Screen shot 2020 08 27 at 1.29.58 pm

Possible Answers:

Screen shot 2020 08 27 at 1.32.49 pm

Screen shot 2020 08 27 at 1.32.16 pm

Screen shot 2020 08 27 at 1.31.51 pm

Screen shot 2020 08 27 at 1.33.16 pm

Correct answer:

Screen shot 2020 08 27 at 1.33.16 pm

Explanation:

To find the resultant we must sum the two vectors:

Now we must graph the resultant

Screen shot 2020 08 27 at 1.31.14 pm

Example Question #1 : Vectors

How many degrees above the x-axis is ?

Screen shot 2020 08 27 at 2.25.43 pm

Possible Answers:


Correct answer:

Explanation:

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.

Screen shot 2020 08 27 at 2.26.24 pm

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.

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