Name the smallest angle of triangle ABC,
segment AC is 9, segment CB is 10, and segment AB is 8.
<C
<A
Two angles are the same size and smaller
than the third
<B
Thank you :) Your help is appreciated!
For any triangle, the smallest angle is always opposite the smallest side, and the largest angle is always opposite the largest side.
So look at your diagram to find the correct answer.
Thank you
its angle ACB
Name the smallest angle in this triangle
To determine the smallest angle of triangle ABC, we need to compare the three angles and find the one that is the smallest.
To find the angles, we can use the Law of Cosines, which states that in a triangle with sides of lengths a, b, and c, and the angle opposite side c is denoted as angle C, the following relationship holds:
c^2 = a^2 + b^2 - 2ab * cos(C)
Using the given side lengths, we can calculate the cosines of angles A, B, and C.
Given:
AC = 9
CB = 10
AB = 8
We have three sides and one angle, so let's find angle A first, which is opposite side AB:
cos(A) = (AB^2 + AC^2 - CB^2) / (2 * AB * AC)
= (8^2 + 9^2 - 10^2) / (2 * 8 * 9)
= (64 + 81 - 100) / (144)
= 45 / 144
= 0.3125
Next, let's find angle B, which is opposite side CB:
cos(B) = (CB^2 + AB^2 - AC^2) / (2 * CB * AB)
= (10^2 + 8^2 - 9^2) / (2 * 10 * 8)
= (100 + 64 - 81) / (160)
= 83 / 160
= 0.51875
Finally, let's find angle C, which is opposite side AC:
cos(C) = (AC^2 + CB^2 - AB^2) / (2 * AC * CB)
= (9^2 + 10^2 - 8^2) / (2 * 9 * 10)
= (81 + 100 - 64) / (180)
= 117 / 180
= 0.65
Now that we have the cosines of all three angles (A, B, and C), we can find the angles themselves by taking the inverse cosine (arccos) of each value.
The smallest angle will be the one with the smallest cosine value, as the cosine function is decreasing on the interval [0, π].
So, comparing the cosines we found:
cos(A) ≈ 0.3125
cos(B) ≈ 0.51875
cos(C) ≈ 0.65
The cosine value for angle A is the smallest, indicating that angle A is the smallest angle of triangle ABC.
Therefore, the smallest angle of triangle ABC is <A.