Posted by **May** on Tuesday, May 27, 2008 at 7:51pm.

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

640 000-328 525 = -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~!

- Math question - plz correct -
**Quidditch**, Tuesday, May 27, 2008 at 8:07pm
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.

- Math question - plz correct -
**May**, Tuesday, May 27, 2008 at 8:27pm
hi, so should I multiply 202 500 and 126 025 by 2?

- Math question - plz correct -
**Quidditch**, Tuesday, May 27, 2008 at 8:39pm
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.

- Math question - plz correct -
**May**, Tuesday, May 27, 2008 at 9:00pm
ohh, okay. thank you!

- Math question - plz correct -
**Quidditch**, Tuesday, May 27, 2008 at 9:06pm
Glad to help!

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