Given field is in the shape of a triangle with side lengths 19 meters, 26 meters and 36 meters. Find the largest angle (measured in degrees) in any of the three corners of the field. Help me pls

Question

jdoe0001 · Answer

using the Law of Cosines $\bf a^2 = b^2+c^2-2bc(cos(a)) \implies
\cfrac{a^2-b^2-c^2}{-2bc} = cos(a)$

anonymous · Answer

yea thats what i used too

jdoe0001 · Answer

say for example, let's do the side opposite to the 36m in length, because that one with the longer side opposite to it, will be the obtuse angle

anonymous · Answer

yea

jdoe0001 · Answer

so you got it then :)

anonymous · Answer

i did the whole thing and kept getting a negative number, but ill try again

anonymous · Answer

i would help if you could go thru the whole thing just to be sure...

jdoe0001 · Answer

keep in mind that  $\bf -1 \le cosine \le 1$

so negative is fine, so long is $\bf \le -1$

anonymous · Answer

ok

jdoe0001 · Answer

$\bf \cfrac{36^2-19^2-26^2}{-2(19)(26)} = cos(a)$

anonymous · Answer

i could have put the bottom part in any order right?

jdoe0001 · Answer

yes, they're just factor, so yes

jdoe0001 · Answer

$ \bf \cfrac{36^2-19^2-26^2}{-2(19)(26)} = cos(a)\
cos^{-1}(\text{some value}) = cos^{-1}(cos(a))$

anonymous · Answer

i think i forgat the negative along the way dats why

anonymous · Answer

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