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ritznca ritznca
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Posts: 52
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11 years ago
I have a problem that I keep coming up with an impossible answer! Here's the situation: a=162 b=185 and c=323. Everytime I try to use the law of cosines I get an error. Please help!
You can't use sin, cos, and tan because it is not a right triangle. I have used the law of cosines before on side-side-side triangles and it has worked
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ilovebballilovebball
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11 years ago
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wrote...
11 years ago
a^2 = b^2 + c^2 - 2(b)(c)CosA
CosA = (a^2 - b^2 - c^2)/[(-2)(b)(c)]
A = arccos{(a^2 - b^2 - c^2)/[(-2)(b)(c)]}
A = arccos{ (162^2 - 185^2 - 323^2)/[-2(185)(323)]
A = 19.99 deg   ANSWER

CosB = (b^2 - a^2 - c^2)/[(-2)(a)(c)]
B = arccos{(b^2 - a^2 - c^2)/[(-2)(a)(c)]
B = arccos{185^2 - 162^2 - 323^2)/[(-2)(162)(323)]
B = 22.98 deg    ANSWER

Cos C = (c^2 - a^2 - b^2)/[(-2)(a)(b)]
C = arccos{(c^2 - a^2 - b^2)/[(-2)(a)(b)]
C = arccos{(323^2 - 162^2 - 185^2)/[(-2)(162)(185)]
C = 137.03 deg    ANSWER

Checking:
A + B + C = 19.99 + 22.98 + 137.03 = 180 deg

Hope this helps.

teddy boy
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