Name the smallest angle of triangle ABC.
line AC is equal to 10
line CB is equal to 8
line BA is equal to 9
Answers
<A
<B
<C
Two angles are the same size and smaller than the thirds
To find the smallest angle in triangle ABC, we can use the Law of Cosines. The Law of Cosines relates the lengths of the sides of a triangle to the cosines of its angles. For a triangle with sides a, b, and c, and angles opposite these sides A, B, and C, respectively, we have:
c² = a² + b² - 2ab * cos(C)
We want to find the smallest angle, which is the angle opposite the shortest side. In triangle ABC, the shortest side is side BA which is 9 units long.
Now let's compare the other two sides:
- AC = 10 units
- CB = 8 units
Since CB < BA < AC, we can deduce that angle C is larger than angle B, which in turn must be larger than angle A (because the larger the side, the larger the opposite angle in any triangle). So, the smallest angle is opposite the shortest side, which is side BA. Therefore, the smallest angle is <C.
However, let's verify this with a calculation using the Law of Cosines:
For angle A using sides BC and AC:
cos(A) = (b² + c² - a²) / (2bc)
cos(A) = (8² + 10² - 9²) / (2 * 8 * 10)
cos(A) = (64 + 100 - 81) / 160
cos(A) = (83) / 160
cos(A) ≈ 0.51875
For angle B using sides AC and AB:
cos(B) = (a² + c² - b²) / (2ac)
cos(B) = (9² + 10² - 8²) / (2 * 9 * 10)
cos(B) = (81 + 100 - 64) / 180
cos(B) = (117) / 180
cos(B) ≈ 0.65
For angle C using sides AB and BC:
cos(C) = (a² + b² - c²) / (2ab)
cos(C) = (9² + 8² - 10²) / (2 * 9 * 8)
cos(C) = (81 + 64 - 100) / 144
cos(C) = (45) / 144
cos(C) ≈ 0.3125
Since cos(C) is the greatest value among the three cosine results, angle C is the smallest (because the bigger the cosine of an angle, the smaller the angle itself in the range 0 to 180 degrees). Therefore, <C is the smallest angle in triangle ABC.
The final answer is <C.