Question 73955
a^2 = b^2+c^2-2bc(CosA)
So, Cos(A)=[b^2+c^2-a^2]/2bc

------
Please help me round angle measures to the nearest degree and side measures to the nearest tenth for this problem: A triangular garden has dimensions 54 feet, 48 feet, and 62 feet. Find the angles at each corner of the garden. (Law of Cosines
---------------
Let "48 ft" be the side opposite angle A; then
CosA = [54^2 + 62^2 - 48^2]/(2*54*62) 
CosA = [0.6654719...
Take the cos^-1(A) to get A=48.28 degrees
------------
Let "54" be the side opposite angle B; then
CosB = [48^2+62^2-54^2]/(2*48*54] = 0.54301075...
Take the cos^-1(B) to get B=57.111 degrees
-----------
Then the 3 angle is 180-48.28-57.11=74.61 degrees
===============
Cheers,

Stan H.