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.
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.
1 answer