This is trig stuff and im completely stumped. The sides of a triangle are: a=36 b=77 and c=85. i know i use the laws of cosine to find the angle across from the largest side which is c. i need to know all the angles, A, B, and C. Please help. I tryed pluggin it into a calc using law of cosines and i get a zero..whats this about?
2006-11-29 13:15:39 · 6 answers · asked by Allison B 2 in Science & Mathematics ➔ Mathematics
c^2 = a^2+ b^2 - 2abcosC
C = 90
A = 25
B = 65
2006-11-29 13:22:39 · answer #1 · answered by Rajkiran 3 · 0⤊ 0⤋
The laws of cosine apply to all angles not just to the one with the largest opposite side.
sqaure of one side = sum of squares of the other 2 sides - 2*product of the other tow sides* cosine of the angle between the other two sides. No restriction on longest side
You probably made a mistake in calc. Please recheck.
2006-11-29 21:21:21 · answer #2 · answered by nn 3 · 0⤊ 0⤋
You got cosC = 0, not C=0. That means C is 90 degrees, a right angle.
2006-11-29 21:21:13 · answer #3 · answered by bictor717 3 · 0⤊ 0⤋
This is a right triangle, so
C = 90 degrees
sin A = a/c
cos B = a/c
2006-11-29 21:26:09 · answer #4 · answered by Anonymous · 0⤊ 0⤋
we have a^2 + b^2 = c^2
so this is a right triangle
C = 90
sin A = a/c = 0.423 => A ~ 25
sin B = b/c = 0.906 => B ~ 65
2006-11-29 21:22:27 · answer #5 · answered by James Chan 4 · 0⤊ 0⤋
the law of cosines should net you the right answers. However this might help:
http://hyperphysics.phy-astr.gsu.edu/Hbase/lcos.html
2006-11-29 21:19:16 · answer #6 · answered by Poetic Disaster 1 · 0⤊ 0⤋