Name the smallest angle of Triangle ABC. Image shows a triangle. The bottom left angle is a. The bottom right angle is B. The very top angle is c. Line AB =9. Line BC=8. Line AC=10
To find the smallest angle of Triangle ABC, we can use the Law of Cosines. This law states that in any triangle, the square of one side length is equal to the sum of the squares of the other two side lengths minus twice the product of the two other side lengths and the cosine of the included angle.
Let's label the smallest angle of Triangle ABC as angle A. Using the Law of Cosines, we have:
(AC)^2 = (AB)^2 + (BC)^2 - 2(AB)(BC)cos A
Substituting the given values:
10^2 = 9^2 + 8^2 - 2(9)(8)cos A
100 = 81 + 64 - 144cos A
100 = 145 - 144cos A
144cos A = 145 - 100
144cos A = 45
cos A = 45/144
cos A ≈ 0.3125
To find angle A, we can take the inverse cosine (arccos) of 0.3125.
A ≈ arccos(0.3125)
A ≈ 71.07 degrees
Therefore, the smallest angle of Triangle ABC (angle A) is approximately 71.07 degrees.