In triangle ABC, side AB = 10, side AC = 13, and side BC = 16. What is the measure of the larges angle in degrees?
the larger angle Ø must be across from the larger side, so by the cosine law....
16^2 = 10^2+13^2 - 2(10)(13)cosØ
260cosØ = 13
cosØ = 13/260 = 1/20 = .05
Ø = 87.1°
To find the measure of the largest angle in triangle ABC, we can use the Law of Cosines. The Law of Cosines states that in any triangle with sides a, b, and c, and opposite angles A, B, and C, the following equation holds:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, we are given sides AB, AC, and BC. Let's assign A to angle A, B to angle B, and C to angle C. We are interested in finding angle C, which is the largest angle.
Plugging in the given values into the Law of Cosines equation, we have:
BC^2 = AB^2 + AC^2 - 2 * AB * AC * cos(C)
Substituting the known values, we get:
16^2 = 10^2 + 13^2 - 2 * 10 * 13 * cos(C)
Simplifying further, we have:
256 = 100 + 169 - 260 * cos(C)
Combine like terms:
-13 = -260 * cos(C)
Dividing by -260, we get:
cos(C) = -13 / -260
cos(C) = 1/20
To find the measure of angle C, we can take the inverse cosine (cos^-1) of 1/20 using a calculator:
C = cos^-1(1/20)
Calculating this, we find:
C ≈ 85.12 degrees
Therefore, the measure of the largest angle (angle C) in triangle ABC is approximately 85.12 degrees.