Inclined Planes and Air Navigation - Trigonometry
Card 1 of 28
Select the answer that correctly matches the following air navigation terms to their definitions.
Select the answer that correctly matches the following air navigation terms to their definitions.
Tap to reveal answer
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 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.
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In the following diagram, a blue box sits on an inclined plane. The box has weight W and exerts force 
against the inclined plane and force 
down the inclined plane. Which of the following correctly relates these vectors together?
In the following diagram, a blue box sits on an inclined plane. The box has weight W and exerts force 
Tap to reveal answer
The correct answer is 
because 
and 
are component vectors for the weight 
.
The correct answer is 
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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?
A 450 pound barrel rests on a 
Tap to reveal answer
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.
First, draw a diagram of the given information. We can label the angle of inclination as 
Using the above formulas, we get:
Next understand that the minimum force needed to prevent the barrel from rolling down the plane corresponds to 
← Didn't Know|Knew It →
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?
A 150 pound barrel rests on a 
Tap to reveal answer
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 
Because these two angles are equal, 
Using the above formulas, we get:
Next understand that the minimum force needed to prevent the barrel from rolling down the plane corresponds to 
← Didn't Know|Knew It →
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?
A 65 pound barrel rests on a 
Tap to reveal answer
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.
We can draw a diagram of the given information, including the weight of the box and the force 
Because these two angles are equal, 
Using the above formulas, we get:
Next understand that the minimum force needed to prevent the barrel from rolling down the plane corresponds to 
← Didn't Know|Knew It →
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.
The heading of an airplane is 
Tap to reveal answer
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 
.
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:
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 
We can find the course by subtraction:
Therefore the groundspeed is 222.77 miles per hour and the course is 
← Didn't Know|Knew It →
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.
The airspeed of an airplane is 210 kilometers per hour. There is a wind of 25 kilometers per hour from 
Tap to reveal answer
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 
.
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 
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 
Now to find the heading, subtract this from 
Therefore the heading is 
← Didn't Know|Knew It →
Select the answer that correctly matches the following air navigation terms to their definitions.
Select the answer that correctly matches the following air navigation terms to their definitions.
Tap to reveal answer
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 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.
← Didn't Know|Knew It →
In the following diagram, a blue box sits on an inclined plane. The box has weight W and exerts force against the inclined plane and force down the inclined plane. Which of the following correctly relates these vectors together?
In the following diagram, a blue box sits on an inclined plane. The box has weight W and exerts force
Tap to reveal answer
The correct answer is because and are component vectors for the weight .
The correct answer is
← Didn't Know|Knew It →
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?
A 450 pound barrel rests on a
Tap to reveal answer
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.
First, draw a diagram of the given information. We can label the angle of inclination as
Using the above formulas, we get:
Next understand that the minimum force needed to prevent the barrel from rolling down the plane corresponds to
← Didn't Know|Knew It →
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?
A 150 pound barrel rests on a
Tap to reveal answer
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
Because these two angles are equal,
Using the above formulas, we get:
Next understand that the minimum force needed to prevent the barrel from rolling down the plane corresponds to
← Didn't Know|Knew It →
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?
A 65 pound barrel rests on a
Tap to reveal answer
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.
We can draw a diagram of the given information, including the weight of the box and the force
Because these two angles are equal,
Using the above formulas, we get:
Next understand that the minimum force needed to prevent the barrel from rolling down the plane corresponds to
← Didn't Know|Knew It →
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.
The heading of an airplane is
Tap to reveal answer
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 .
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:
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
We can find the course by subtraction:
Therefore the groundspeed is 222.77 miles per hour and the course is
← Didn't Know|Knew It →
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.
The airspeed of an airplane is 210 kilometers per hour. There is a wind of 25 kilometers per hour from
Tap to reveal answer
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 .
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
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
Now to find the heading, subtract this from
Therefore the heading is
← Didn't Know|Knew It →
Select the answer that correctly matches the following air navigation terms to their definitions.
Select the answer that correctly matches the following air navigation terms to their definitions.
Tap to reveal answer
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 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.
← Didn't Know|Knew It →
In the following diagram, a blue box sits on an inclined plane. The box has weight W and exerts force against the inclined plane and force down the inclined plane. Which of the following correctly relates these vectors together?
In the following diagram, a blue box sits on an inclined plane. The box has weight W and exerts force
Tap to reveal answer
The correct answer is because and are component vectors for the weight .
The correct answer is
← Didn't Know|Knew It →
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?
A 450 pound barrel rests on a
Tap to reveal answer
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.
First, draw a diagram of the given information. We can label the angle of inclination as
Using the above formulas, we get:
Next understand that the minimum force needed to prevent the barrel from rolling down the plane corresponds to
← Didn't Know|Knew It →
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?
A 150 pound barrel rests on a
Tap to reveal answer
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
Because these two angles are equal,
Using the above formulas, we get:
Next understand that the minimum force needed to prevent the barrel from rolling down the plane corresponds to
← Didn't Know|Knew It →
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?
A 65 pound barrel rests on a
Tap to reveal answer
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.
We can draw a diagram of the given information, including the weight of the box and the force
Because these two angles are equal,
Using the above formulas, we get:
Next understand that the minimum force needed to prevent the barrel from rolling down the plane corresponds to
← Didn't Know|Knew It →
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.
The heading of an airplane is
Tap to reveal answer
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 .
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:
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
We can find the course by subtraction:
Therefore the groundspeed is 222.77 miles per hour and the course is
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