A plane leaves an airport and flies South 58 Degrees East for 2 hours, then changes course to South 43 Degrees West and flies for 3 hours. How far away is the plane from the airport if it keeps a constant speed of 500km/h? Draw a diagram and solve to the nearest km.
I got 1636.32 km, I used cosine with angle 79 and the two side lengths 1000 and 1500. Correct answer? If not how do I do it?
Looks good to me.
To solve this problem, we can use the Law of Cosines to find the distance between the starting point (airport) and the current position of the plane.
First, let's draw a diagram to visualize the situation:
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Airport
Now, let's label the sides and angles of the triangle formed:
Side a: Distance from the airport to the final position of the plane (what we need to find)
Side b: Distance covered in the first part of the flight (2 hours at 500 km/h)
Side c: Distance covered in the second part of the flight (3 hours at 500 km/h)
Angle A: 58 degrees
Angle B: 180 - 58 - 43 = 79 degrees
Angle C: 43 degrees
Using the Law of Cosines, we have:
a² = b² + c² - 2bc * cos(A)
Plugging in the known values:
a² = (2 * 500)² + (3 * 500)² - 2 * (2 * 500) * (3 * 500) * cos(58)
a² = 250000 + 450000 - 600000 * cos(58)
a² = 700000 - 600000 * cos(58)
Calculating this expression, we get:
a² ≈ 338314.03
To find the value of a, we take the square root:
a ≈ √(338314.03) ≈ 581.89 km
Therefore, the plane is approximately 581.89 km away from the airport.
It seems that your calculation of 1636.32 km is incorrect. Please recheck your approach or calculations to find out where the mistake might have occurred.