Course and Ground Speed (Navigation problem)

AI Thread Summary
To solve the navigation problem, the plane's course and ground speed must be determined by analyzing the vectors involved. The plane flies at 120 degrees with an airspeed of 300 mph, while the wind affects its trajectory at 40 mph from north to south. Decomposing the vectors into components is crucial for visualizing the problem and applying the law of cosines to find the third side of the triangle formed by the plane's direction and the wind's vector. After calculating the third side, the law of sines can be used to determine the remaining angles. Ultimately, the final velocity of the airplane is the resultant of the two vectors combined.
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Homework Statement



A plane flies in the direction 120 degrees with an air speed of 300mph. The wind is blowing north to south at 40mph. Find the course and ground speed of the plane.

Homework Equations





The Attempt at a Solution



I attempted this problem, but i still cannot figure out how we get the second angle?
 
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Have you drawn a picture?
Do you know how to decompose vectors into their components? If you don't know how to work with vectors, your picture should help you get the largest angle in the triangle formed by the plane's direction vector and the wind's vector. You know two of the sides of the triangle and the angle between them, so you can use the law of cosines to find the third side of the triangle. After you have found it, you can use the law of sines to find one of the remaining angles in the triangle.
 
Draw the vectors as suggested. The final velocity of the airplane is the sum of the two vectors.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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