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A triangle has sides of length 5, 8, and 10. What is the measure of its largest angle?
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Use the law of cosines
I did but can not seem to get the right answer
\[10^2=5^2+8^2-2(5)(8)\cos A\] \[100-89=80\cos A\] \[\frac{11}{80}=\cos A\]=.1375 \[\cos^{-1} (.1375)=82.1^{o}\]
That's what I got, but the back of the book said it was 97.9 @Mertsj
\[100-89=-80\cos A\]
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You do get what the book says. Mert just forgot the little negative sign
How is it the same as what the book says?
try using the calculator
arccos(-11/80)
I did so if I use the negative I will get the right answer?
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yes
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