Math question  plz correct
posted by May .
Two airplanes leave an airport at the same time. One travels at 355km/h and the other at 450km/h. Two hrs later they are 800km apart. Find the angle between their courses.
a^2 = b^2 + c^2  2bc Cos A
800^2= 450^2 + 355^2  2(450)(355) Cos A
640000= 202 500 + 126 025  319 500 Cos A
640 000328 525 = 319 500 Cos A
311 475= 319 500 Cos A
311 475/ 319 500= Cos A
0.9749= Cos A
167.13 = A
Can you please explain to me what is wrong with this? I checked the back of the book and the answer is suppose to be 58.17
thanks~!

A quick look finds that you missed the fact that the time the airplanes are 800km apart is 2 hours not 1 hour. The legs should be 2*355km and 2*450km.

hi, so should I multiply 202 500 and 126 025 by 2?

Actually, you would have to multiply them by 4 because they are squared. You have the right idea.
For the legs use:
b=2*450km=900km
c=2*355km=710km
This would then give you
800^2=900^2 + 710^2  2(900)(710)cos(A)
Solve for A using the same method you did before. 
ohh, okay. thank you!

Glad to help!