Basic trig question

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Jeff7

Jeff7

Lifer
Jan 4, 2001
41,596
19
81
I last had geometry and trig in 10th and 11th grade, back before 1999, and I don't remember some basics.

I've got a triangle, angles of 55, 60, and 65 degrees. The side opposite the 65deg angle is 5.4378".

Is there a way of finding the size of the other sides with that information? I'd think there would be, but I've got no idea how to do it anymore.

Sad thing is, I'm currently taking a Calc2 class, and getting a solid B in the course. But I can't remember basic trig.
 
Jeff7

Jeff7

Lifer
Jan 4, 2001
41,596
19
81
That equation requires knowing the dimensions of two sides though. I've only got the dimension for one side, but I know all three angles.
 
I

IdioticBuffoon

Senior member
Sep 11, 2005
327
0
0
Originally posted by: Jeff7
That equation requires knowing the dimensions of two sides though. I've only got the dimension for one side, but I know all three angles.
Click to expand...

Make two equations for two unknowns.
 
S

Soccer55

Golden Member
Jul 9, 2000
1,660
4
81
Originally posted by: Jeff7
That equation requires knowing the dimensions of two sides though. I've only got the dimension for one side, but I know all three angles.
Click to expand...

Try the Law of Sines instead:

a/sin(A) = b/sin(B) = c/sin(C)

where a,b,c are sides of the triangle and A,B,C are the angles of the triangle opposite the sides a,b,c respectively.

-Tom
 
Jeff7

Jeff7

Lifer
Jan 4, 2001
41,596
19
81
Excellent, thank you very much. Law of Sines it is.
I figured that there HAD to be a way, because 1 side remains constant, as do the 3 angles - the other sides can only take on a single value for each, and there had to be a way of figuring it.
Wish I'd known that yesterday. Would've saved me a LOT of time. Thanks a lot everyone for not predicting my needs yesterday. Sheesh.
 
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