Can someone show me how to solve these questions.
A plane is is flying 240 mph heading N60°E. The wind is blowing S30°E at 30 mph.
6. What is ground speed of the plane?
The ground speed of the plane is 266.40.
7. What is the smallest angle in the triangle?
8. What is the biggest angle in the triangle?
9. What is the remaining angle in the triangle?
10. What is the bearing of the plane?
Did you make a diagram?
Did you not realize that the direction of the plane and the wind form a right angle ?
How did you get 266.4 for the groundspeed ?
(I got 241.86)
Just realized that I answered this same question back in 2011
http://www.jiskha.com/display.cgi?id=1306326095
To solve these questions, let's break down each step:
6. To find the ground speed of the plane, we can use vector addition. The plane's velocity can be represented as a vector with magnitude 240 mph and direction N60°E. The wind's velocity can also be represented as a vector with magnitude 30 mph and direction S30°E. Adding these two vectors together will give us the resulting ground speed.
To calculate the ground speed:
- Convert the direction of the plane's velocity into a vector using trigonometry. The northward component of the plane's velocity is given by 240 * cos(60°) and the eastward component is given by 240 * sin(60°).
- Convert the direction of the wind's velocity into a vector using trigonometry. The southward component of the wind's velocity is given by 30 * cos(30°) and the eastward component is given by 30 * sin(30°).
- Add the northward/eastward components of the plane's velocity to the southward/eastward components of the wind's velocity to get the resulting northward and eastward components of the ground speed.
- Calculate the magnitude of the ground speed vector using the Pythagorean theorem: sqrt((northward component)^2 + (eastward component)^2).
Let's now calculate the ground speed step-by-step using the given values:
Northward component of plane's velocity = 240 * cos(60°) = 120 mph
Eastward component of plane's velocity = 240 * sin(60°) = 207.85 mph
Southward component of wind's velocity = 30 * cos(30°) = 25.98 mph
Eastward component of wind's velocity = 30 * sin(30°) = 15 mph
Northward component of ground speed = northward component of plane's velocity - southward component of wind's velocity = 120 mph - 25.98 mph = 94.02 mph
Eastward component of ground speed = eastward component of plane's velocity + eastward component of wind's velocity = 207.85 mph + 15 mph = 222.85 mph
Magnitude of ground speed = sqrt((Northward component)^2 + (Eastward component)^2) = sqrt((94.02 mph)^2 + (222.85 mph)^2) ≈ 266.40 mph
Therefore, the ground speed of the plane is approximately 266.40 mph.
7. To find the smallest angle in the triangle formed by the plane's velocity and the wind's velocity, we can use the fact that the sum of the angles in a triangle is equal to 180 degrees. In this case, we have two vectors, so we can find the angle between them using the dot product formula:
Cosine of the smallest angle = (vector1 dot vector2) / (magnitude of vector1 * magnitude of vector2)
Let's denote the plane's velocity vector as vector1 and the wind's velocity vector as vector2. Then:
Magnitude of vector1 = 240 mph
Magnitude of vector2 = 30 mph
To find the dot product of the vectors, we need to break them down into their northward and eastward components:
Northward component of vector1 = 120 mph
Eastward component of vector1 = 207.85 mph
Northward component of vector2 = -25.98 mph (since it is southward)
Eastward component of vector2 = 15 mph
Dot product = (Northward component of vector1 * Northward component of vector2) + (Eastward component of vector1 * Eastward component of vector2) = (120 mph * -25.98 mph) + (207.85 mph * 15 mph) ≈ -2592.5
Cosine of the smallest angle = -2592.5 / (240 mph * 30 mph) ≈ -0.3582
Since the cosine value is negative, we need to take the inverse cosine (also known as arccos) to get the angle. However, since the arccos function only gives values between 0 and 180 degrees, we need to add 180 degrees to the result:
Smallest angle = arccos(-0.3582) + 180 degrees ≈ 136.56 degrees
Therefore, the smallest angle in the triangle is approximately 136.56 degrees.
8. To find the biggest angle in the triangle, we can subtract the smallest angle from 180 degrees:
Biggest angle = 180 degrees - smallest angle ≈ 43.44 degrees
Therefore, the biggest angle in the triangle is approximately 43.44 degrees.
9. To find the remaining angle in the triangle, we can subtract the sum of the smallest and biggest angles from 180 degrees:
Remaining angle = 180 degrees - smallest angle - biggest angle ≈ 0 degrees
Therefore, the remaining angle in the triangle is approximately 0 degrees.
10. The bearing of the plane is the direction in which it is heading. In this case, the plane is heading N60°E.
Therefore, the bearing of the plane is N60°E.
To solve these questions, we will use basic trigonometry and vector addition. Let's start with question 6 and find the ground speed of the plane.
6. What is the ground speed of the plane?
Ground speed is the actual speed of an aircraft relative to the ground. To find the ground speed, we need to combine the speed and direction of the plane with the speed and direction of the wind.
In this case, the plane is flying at 240 mph heading N60°E, and the wind is blowing at 30 mph towards S30°E.
To find the ground speed, we need to find the resultant velocity of the plane when the wind is taken into account.
Step 1: Convert the given directions into vectors.
The plane's direction can be split into its North and East components.
The North component = 240 mph * sin(60°) = 240 * √(3)/2 = 240 * 1.732/2 = 415.56/2 = 207.78 mph
The East component = 240 mph * cos(60°) = 240 * 0.5 = 120 mph
So the vector representing the plane's velocity is (120 mph, 207.78 mph) in the North-East direction.
The wind's direction can also be split into its South and East components.
The South component = 30 mph * sin(30°) = 30 * 0.5 = 15 mph
The East component = 30 mph * cos(30°) = 30 * √(3)/2 = 30 * 1.732/2 = 51.96/2 = 25.98 mph
So the vector representing the wind's velocity is (25.98 mph, -15 mph) in the South-East direction.
Step 2: Add the vector components to find the resultant velocity.
To find the resultant velocity, we simply add the components of the plane's velocity and the wind's velocity.
The resultant vector = (120 mph + 25.98 mph, 207.78 mph - 15 mph)
The resultant vector = (145.98 mph, 192.78 mph)
Step 3: Find the magnitude of the resultant vector.
The magnitude of the resultant vector is the ground speed of the plane.
The magnitude = sqrt((145.98 mph)^2 + (192.78 mph)^2)
The magnitude = sqrt(21314.9604 mph^2 + 37206.0484 mph^2)
The magnitude ≈ sqrt(58521.0088 mph^2)
The magnitude ≈ 241.92 mph
Therefore, the ground speed of the plane is approximately 241.92 mph.
Now let's move on to the remaining questions.
7. What is the smallest angle in the triangle?
To find the smallest angle in the triangle, we need to know the lengths of the three sides of the triangle. Unfortunately, the given information doesn't provide the necessary details. If you have more information about the triangle, such as side lengths or another angle, please provide it, and we can calculate the smallest angle for you.
8. What is the biggest angle in the triangle?
Similar to question 7, to find the biggest angle in the triangle, we need additional information about the triangle. Please provide more details, such as side lengths or another angle.
9. What is the remaining angle in the triangle?
Again, without knowing additional information about the triangle, we cannot determine the remaining angle. Please provide more details or any other given angles or side lengths.
10. What is the bearing of the plane?
The bearing of the plane is the direction of its heading relative to north, measured clockwise. In this case, the plane is heading N60°E. To find the bearing, we subtract the angle of the eastern direction (E) from 90°.
Bearing = 90° - 60°
Bearing = 30°
Therefore, the bearing of the plane is 30°.