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what is the law of cosine
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like pythagoras but not for right triangles \[a^2=b^2+c^2-2b c\ cos(A)\]
there's a diagram at the wikipedia site
i still can get this question thoough in tirangle ABC, a=8 b=9 and m of angle C = 135 what is the area of ABC
Area is base times height divided by 2, which can be found without the law of cosines, the height would be the sine of 135 times 8 if the base is 9
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sin(135) = sin(45) = sqrt(2)/2 Area = 9*8*(sqrt(2)/2)/2
thank you that was an answer
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