SOLUTION: Please help me find the sides of a, b, and c of the triangle that is a = 4 , b = 8, c = 10
So far I have {{{ 2^2 + 4^2 -2 *4*cos(c)=5^2 }}}
{{{ -16cos(c)=25-4-16 }}} {{{ -16cos(c
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Question 1077542: Please help me find the sides of a, b, and c of the triangle that is a = 4 , b = 8, c = 10
So far I have
Found 2 solutions by Alan3354, Boreal:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Please help me find the sides of a, b, and c of the triangle that is a = 4 , b = 8, c = 10
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You mean find the angles.
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So far I have
20 - 8cos(C) = 25
Not -16cos, -8
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**********
C =~ 128.68 degs
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I would use the Law of Sines to find a 2nd angle.
sin(C)/c = sin(A)/a
sin(A) = a*sin(C)/c = 4*0.7806/10 = 0.31225
A = 18.195 deg
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B = 180 - (A + C)
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
c^2=a^2+b^2-2ab cos C
10^2=4^2+8^2-2(32) cos C
100=80-64 cos C
20=-64 cos C
-5/16=cos C, as you got, which is 108.21 deg.
Now use Law of Sines
4/sin A=10/sin 108.21=10.527
sin A=4/10.527=22.33 degrees
B should be 49.46 deg
8/sin B=10.527; sin B=8/10.527=0.7600, which it is.
A=22.33 deg
B=49.46 deg
C=108.21 deg
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