A triangle and its sides
There are two sides which relate as 3:8 and they form the angle 60 degrees. Find the nearest sides of the angle. The third side is 21 cm.
I tried to use cosine theoreme and I assumed that the first side is a=3*x/11 and the second side is b=8*x/11 cm.
Cosine theoreme says that c^2=a^2+b^2-2ab*cos(gamma).
It's all right now.
check your formulas
if this can help, there is a little mistake in your cosin formula : it is c^2=a^2+b^2-2ab*cos(gamma) instead of c^2=a^2+b^2-2ab-cos(gamma)
(between the ab and the cos(gamma) there is multiplication and not subtraction).
If you want more info, it would be good if you can post a figure because I did not understand all your problem. You say : they formed (why in the past, don't they in your actual problem) the angle 60 degrees. Find the nearest sides of the angle. The third angle is 21 cm (an angle in cm ?).
Yeah, the formula was wrong. I got the right answer now.