Question

Plssss help in need of answers soon

Answer 1

You have to apply the law of cosine
In a triangle ABC

The square of a side is equal to the sum of the square of the 2 other sides minus twice the cos of the angle opposed to the side AB

AB = 10. BC = 5 AND AC = 8
Let C the angle facing AB, B, facing AC and A facing BC

AB² = BC² + AC² - 2.BC.AC.cos² C. Now plug
10² = 8² + 5² - 2(8)(5).cos C
100 = 64 + 25 - 80 cos C
80cos C = -11; cos C =-11/80 = - 0.1375
cos C = - 0.1375 . To find the angle C, let's use the inverse function
coc C = -0.1375 and cos⁻¹ (-0.1375) = So C = 98°
Follow the same procedure for the other angle. And when you have got 2 angles, you will find the 3rd one
If you need further explanation, I will be on line for about 15 min max

Answer 2

My guess is that you're doing the Law of Cosines? You have everything you need for that except the angle theta, which is the thing you need to find. It's set up like this: (8)^2 = (10)^2 + (5)^2 -[2(10)(5)cos A] I used A instead of theta. Doing that math, you have: 64 = 100 + 25 -[ 100 cos A]; 64 = 125 - 100 cos A;
-61 = - 100 cos A; -61 / -100 = cos A; .61 = cos A. Now use your inverse function on your calculator to find cos^-1(.61) and that equals 52.4