Triangle A B C has vertices A (3,2), B (6,6), and C (0,14). List the angles in order from largest to smallest.
To find the largest angle, we need to find the side opposite to it, by calculating the lengths of the sides. We can use the distance formula.
The distance between points A and B is equal to the length of side AB:
AB = √[(6 - 3)^2 + (6 - 2)^2]
AB = √[3^2 + 4^2]
AB = √[9 + 16]
AB = √25
AB = 5
The distance between points B and C is equal to the length of side BC:
BC = √[(0 - 6)^2 + (14 - 6)^2]
BC = √[-6^2 + 8^2]
BC = √[36 + 64]
BC = √100
BC = 10
The distance between points A and C is equal to the length of side AC:
AC = √[(0 - 3)^2 + (14 - 2)^2]
AC = √[-3^2 + 12^2]
AC = √[9 + 144]
AC = √153
Now, we can use the Law of Cosines to find the largest angle:
cos(A) = (AB^2 + AC^2 - BC^2) / (2AB * AC)
cos(A) = (5^2 + √153^2 - 10^2) / (2 * 5 * √153)
cos(A) = (25 + 153 - 100) / (10 * √153)
cos(A) = 78 / (10 * √153)
cos(A) ≈ 0.404
Since the cosine of an angle is positive in the first and fourth quadrants, we know that angle A is acute.
Now, let's calculate the remaining angles:
cos(B) = (AC^2 + BC^2 - AB^2) / (2AC * BC)
cos(B) = (√153^2 + 10^2 - 5^2) / (2 * √153 * 10)
cos(B) = (153 + 100 - 25) / (20 * √153)
cos(B) = 228 / (20 * √153)
cos(B) ≈ 0.316
cos(C) = (AB^2 + BC^2 - AC^2) / (2AB * BC)
cos(C) = (5^2 + 10^2 - √153^2) / (2 * 5 * 10)
cos(C) = (25 + 100 - 153) / (100)
cos(C) = -28 / 100
cos(C) = -0.28
Now, we can find the values of the angles:
A = arccos(0.404)
A ≈ 66.42°
B = arccos(0.316)
B ≈ 71.73°
C = arccos(-0.28)
C ≈ 104.62°
Therefore, the angles in order from largest to smallest are: C, B, and A.