| Nov Math II Law of Cosine Quesiton
Guys you know that question which said there was an isosceles triangle with sides 5,10,10 and asked you to calculate the largest angle?
For that most people con CC say the answer was 29 something...but im fairly confident im right...
c^2 = a^2 + b^2 - 2abcosine(C)
5^2 = 10^2 + 10^2 - 2(10)(10)cosine(c)
25 - 200 = -2(100)cos(c)
-175/-200 = cos(c)
cos(c) = .875
Do inverse cos(.875) you get = 28.955 deg. which is...the SMALLEST angle. so
180 - 28.955 = 151.044
since its isosceles, divide this angle by 2 to get the largest angle
151.0444/2 = 75.522 degress. Which was an answer i belive. thats what i put. im pretty sure this is right.
and if the answer was ~29 or w/e it would have to be the largest angle, therefore since it is isoseceles 29 x 2 = 58 [[sum of two angles is 58]] 180 - 58 = 122. clearly a larger angle. so how can ~29 deg. be right?
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