an airplane flies 200 km due west from city A and then 350 km in the direction of 32.0 degrees north of west from city B to city C.
a) in a straight-line distance, how far is city c from city a?
b) relative to city a, what direction is city c?
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To solve this problem, we can break it down into two components: the westward distance and the northward distance traveled by the airplane. Let's go step by step:
a) To find the straight-line distance between City A and City C, we can use the Pythagorean theorem. The Pythagorean theorem states that for a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
In this case, the westward distance is 200 km, and the northward distance is calculated by taking the sine of the angle and multiplying it by 350 km (sine = opposite/hypotenuse). So, the northward distance would be sin(32°) × 350 km.
Using the Pythagorean theorem, the straight-line distance between City A and City C can be calculated as:
Distance^2 = (200 km)^2 + (sin(32°) × 350 km)^2
Solving this equation will give us the value of the straight-line distance between City A and City C.
b) To determine the direction of City C relative to City A, we can use trigonometry. Specifically, we will use the tangent function to find the angle. The angle can be calculated as:
Angle = arctan((sin(32°) × 350 km) / 200 km)
This angle will give us the direction of City C with respect to City A.
By following these steps, we can find the answers to both parts of the problem.