<p>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?</p>
<p>For that most people con CC say the answer was 29 something...but im fairly confident im right...</p>
<p>c^2 = a^2 + b^2 - 2abcosine(C)</p>
<p>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.</p>
<p>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?</p>
<p>why used law of cosines and complicated trigs -_-. Just divide the base into 2. (5/2). Then take cos-1(2.5/10). the angle should be 75.5. Double that, then 180-the result, it should give you the answer without wasting much of your time D:</p>
<p>If there are 2 sides of the same length (aka isosceles) and the third is SMALLER than the two, then the angle for that third side is smaller than the other two. An equilateral triangle is 60-60-60. If 1 angle is smaller than the other two (which will be the same since they are across of same sides) then it has to be less than 60 degrees. No other way is possible, unless you change the very definition of what constitutes a triangle.</p>