Dune buggies X and Y are 500m apart, and the bearing of Y from X is 215degres. X travels 200m due east and Y travels 100m due north. How far apart the dune buggies now?

Question

According to text book, answer is 577m. Would you please post the steps?

Someone posted answer for this question on your site. However answer is not correct. (https://www.wyzant.com/resources/answers/748512/dune-buggies-x-and-y-are-500m-apart-and-the-bearing-of-y-from-x-is-215degre)

Answer 1

Let X = the point (0,0), the origin

then Y is the point (500cos215, 500sin215)

after Y goes north 100, its new point is (500cos215, 100+500sin215)

the new X is (200,0)

calculate the distance by taking the square root of the sum of squares of the x and y coordinates

Use the Pythagorean Theorem.

But when I do this, I get closer to 639, not 577

cos215 = -cos35 -.82

500sin35 =-410

from -410 to 200 = 610

the new XY distance is at least 610