Law of cosines: a2 = b2 + c2 – 2bccos(a) find the measure of j, the smallest angle in a triangle with sides measuring 11, 13, and 19. round to the nearest whole degree.

Given the sides 11,13,19, to calculate for the smallest angle using cosine rule, i.e;
a^2=b^2+c^2-2bc cos (A)

we shall proceed as follows;

The smallest angle is always opposite to the shortest side; Since 11 is our shortest side, the smallest angle is the angle opposite to this side. Thus;
let a=11 and our smallest  angle be A
thus
11^2=13^2+19^2-2*13*19 Cos A
121=169+361-494 Cos A
121=530-494 Cos A
-409=-494 Cos A
dividing through by -494 we get;
Cos A=0.8279

thus
A=Cos(-1) 0.8279
A=34.113°≈34°

The answer is A=34°

Idon’t even know the answer hehehehe