An airplane left an airport and flew east 174 miles. Then it turned northward flying along a bearing of N22E. When it was 254 miles from the airport, there was an engine problem and it turned back to take the shortest route possible back to the airport. Find theta, the angle through which the plane turned
Is the angle 39.43?
Your answer is right
Your answer is not correct.
I see a triangle with sides 174 and 254 and the angle between those sides as 112º.
Let the third side be x. (The return path)
By Cosine Law:
x^2 = 174^2 + 254^2 - 2(174)(254)cos112º
x = 357.637
Let the angle at the turn be ß
By Sine Law:
sin ß/174 = sin 112/357.637
sin ß = .4510998
ß = 26.8º
So the plane must turn an angle of (180-26.8) or 153.2º
To find the angle through which the plane turned, we can use trigonometry.
Given that the plane initially flew east for 174 miles and then turned north along a bearing of N22E, we can determine the distance the plane traveled northward using the sine function.
Sin(theta) = Opposite / Hypotenuse
Sin(theta) = Distance North / Distance Travelled
Let's assume the angle through which the plane turned is theta.
Distance North = Distance Travelled * Sin(theta)
Distance North = 254 miles * Sin(theta)
Since the plane turned back to take the shortest route back to the airport, the distance it traveled northward is equal to the distance it originally traveled eastward.
Distance North = 174 miles
So, we have the equation:
174 miles = 254 miles * Sin(theta)
Now we can solve for theta.
Dividing both sides of the equation by 254 miles:
Sin(theta) = 174 miles / 254 miles
Sin(theta) ≈ 0.685
Taking the inverse sine of both sides to find theta:
theta = arcsin(0.685)
theta ≈ 42.35 degrees
Therefore, the angle through which the plane turned is approximately 42.35 degrees, not 39.43 degrees.
To solve the problem, we can use trigonometry and vector addition.
First, let's visualize the problem. The airplane initially flies east for 174 miles, as shown in the diagram below:
```
----------------------------> East (E)
|
| 174 miles
|
Airplane
|
|
```
Then, it turns northward along a bearing of N22E. The bearing N22E means it is 22 degrees east of north. So, the new direction of the airplane is:
```
|
|
N22E |
|
|
|
----------------------------> East (E)
|
|
|
Airplane
```
The airplane continues flying in this new direction for 254 miles, reaching a point (254 miles away from the starting point) before it encounters an engine problem and turns back towards the airport.
To find the angle through which the plane turned, we need to find the angle between the initial east direction and the final direction when it turns back towards the airport.
Let's break down the vectors involved:
Vector A: Initial direction (east) with length 174 miles.
Vector B: Final direction towards the airport (shortest route) with length 254 miles.
We can find the angle between these two vectors using the dot product formula:
θ = acos((A dot B) / (|A| * |B|))
First, let's find the dot product of A and B:
A dot B = (174 * 254)
Next, calculate the magnitude (length) of vectors A and B:
|A| = 174
|B| = 254
Now, substitute the values into the formula:
θ = acos((174 * 254) / (174 * 254))
Evaluating the expression:
θ = acos(1)
As the dot product of A and B is equal to the product of their magnitudes, the angle θ can be determined as:
θ = acos(1)
θ = 0 degrees
Therefore, the angle through which the plane turned is 0 degrees, indicating that the plane did not turn at all when returning to the airport.
So, the angle is not 39.43 degrees. Instead, it is 0 degrees.