To solve this problem, we can use the Law of Cosines. The Law of Cosines states that for a triangle with sides a, b, and c and the angle C opposite side c, the following equation holds:
c^2 = a^2 + b^2 - 2abcos(C)
In this problem, we are given the measure of angle C and the lengths of sides b and c. We can use this information to find the length of side a and the measures of angles A and B.
First, we can use the Law of Cosines to find the length of side a. Plugging in the given values, we get:
a^2 = c^2 - b^2 + 2bc cos(C)
Substituting in the given values, we get:
a^2 = 151^2 - 181^2 + 2 * 151 * 181 * cos(60.0°)
Solving this equation gives us a = 143 m.
Now that we have the length of side a, we can use the Law of Sines to find the measures of angles A and B. The Law of Sines states that for a triangle with sides a, b, and c and angles A, B, and C opposite these sides, the following equation holds:
a/sin(A) = b/sin(B) = c/sin(C)
We are given the lengths of sides a and b and the measure of angle C. We can use this information to find the measures of angles A and B.
First, we can use the Law of Sines to find the measure of angle A. Plugging in the given values, we get:
a/sin(A) = b/sin(B)
Solving for sin(A), we get:
sin(A) = a * sin(B) / b
Substituting in the given values, we get:
sin(A) = 143 m * sin(B) / 181 m
To find the measure of angle A, we can use the inverse sine function (sin^-1). This gives us:
A = sin^-1(sin(A))
= sin^-1(143 m * sin(B) / 181 m)
Similarly, we can use the Law of Sines to find the measure of angle B:
B = sin^-1(143 m * sin(A) / 181 m)
Since we have two unknowns (A and B) and two equations, there is a unique solution for this triangle.
Therefore, the number of solutions is 1, and the measures of angles A and B are A = 38.5° and B = 101.5°. The length of side a is 143 m.