The perimeter of triangle ABC is 30 ft. AB = 3x+1, BC = 4x, and AC = 5x-7. List the angles of triangle ABC in order from smallest to largest. (Smallest on top, largest on bottom) (1 point)
= <A
= <B
= <C
In order to determine the angles of triangle ABC, we need to find the values of x first.
The perimeter of a triangle is the sum of the lengths of its sides. So,
AB + BC + AC = 30
(3x+1) + (4x) + (5x-7) = 30
12x - 6 = 30
12x = 36
x = 3
Now that we have found the value of x, we can substitute it back into the expressions for the lengths of AB, BC, and AC.
AB = 3x + 1 = 3(3) + 1 = 10
BC = 4x = 4(3) = 12
AC = 5x - 7 = 5(3) - 7 = 8
Now we can use the lengths of the sides to find the angles using the Law of Cosines.
Let's find angle A:
cos(A) = (BC^2 + AC^2 - AB^2) / (2 * BC * AC)
cos(A) = (12^2 + 8^2 - 10^2) / (2 * 12 * 8)
cos(A) = (144 + 64 - 100) / (2 * 12 * 8)
cos(A) = 108 / 192
cos(A) = 0.5625
A = arccos(0.5625)
A ≈ 55.51°
Let's find angle B:
cos(B) = (AB^2 + AC^2 - BC^2) / (2 * AB * AC)
cos(B) = (10^2 + 8^2 - 12^2) / (2 * 10 * 8)
cos(B) = (100 + 64 - 144) / (2 * 10 * 8)
cos(B) = 20 / 160
cos(B) = 0.125
B = arccos(0.125)
B ≈ 82.75°
Now that we know angle A and angle B, we can find angle C using the fact that the sum of the angles in a triangle is 180°.
C = 180° - A - B
C = 180° - 55.51° - 82.75°
C ≈ 41.74°
So, the angles of triangle ABC in order from smallest to largest are:
∠A ≈ 55.51°
∠B ≈ 82.75°
∠C ≈ 41.74°