Triangle ABC has vertices A(3,2), B(6,6) and C(0,14). List the angles in order from largest to smallest.

Answer 1

To find the angles in a triangle, we can use the Law of Cosines. Then we can compare the values to determine the largest and smallest angles.

First, we find the lengths of the sides of the triangle using the distance formula.

Side AB:
AB = √((6 - 3)^2 + (6 - 2)^2) = √(3^2 + 4^2) = √(9 + 16) = √25 = 5

Side BC:
BC = √((0 - 6)^2 + (14 - 6)^2) = √((-6)^2 + 8^2) = √(36 + 64) = √100 = 10

Side AC:
AC = √((0 - 3)^2 + (14 - 2)^2) = √((-3)^2 + 12^2) = √(9 + 144) = √153 ≈ 12.37

Now, we can use the Law of Cosines to find the angles.

Angle A:
cos(A) = (AB^2 + AC^2 - BC^2) / (2 * AB * AC)
cos(A) = (5^2 + 12.37^2 - 10^2) / (2 * 5 * 12.37)
cos(A) ≈ 0.968

Angle B:
cos(B) = (AB^2 + BC^2 - AC^2) / (2 * AB * BC)
cos(B) = (5^2 + 10^2 - 12.37^2) / (2 * 5 * 10)
cos(B) ≈ 0.871

Angle C:
cos(C) = (BC^2 + AC^2 - AB^2) / (2 * BC * AC)
cos(C) = (10^2 + 12.37^2 - 5^2) / (2 * 10 * 12.37)
cos(C) ≈ 0.758

Now we can determine the largest and smallest angles by comparing the values of cosines.

The largest angle is angle A, as cos(A) = 0.968.
The smallest angle is angle C, as cos(C) ≈ 0.758.

Therefore, the angles in the triangle ABC from largest to smallest are A, B, and C.