To solve triangle ABC, we can use the Law of Cosines, which states that for any triangle with sides a, b, and c and opposite angles A, B, and C, respectively:
c^2 = a^2 + b^2 - 2ab*cos(C)
We are given B, a, and b, so we can solve for c as follows:
c^2 = 25^2 + 41^2 - 2(25)(41)cos(64)
c^2 = 625 + 1681 - 2135cos(64)
c^2 = 1829 - 2135*cos(64)
c^2 = 311.90
Taking the square root of both sides, we get:
c ≈ 17.7
So the length of side c is approximately 17.7 units.
To find the measures of angles A and C, we can use the Law of Sines, which states that for any triangle with sides a, b, and c and opposite angles A, B, and C, respectively:
a/sin(A) = b/sin(B) = c/sin(C)
We know a, b, and c, and we just solved for c, so we can use the Law of Sines to solve for angles A and C:
a/sin(A) = c/sin(C)
sin(A) = asin(C)/c
A = sin^{-1}(asin(C)/c)
A = sin^{-1}(25*sin(C)/17.7)
Similarly,
b/sin(B) = c/sin(C)
sin(B) = bsin(C)/c
B = sin^{-1}(bsin(C)/c)
B = sin^{-1}(41*sin(C)/17.7)
To find angle C, we can use the fact that the sum of the angles in a triangle is 180 degrees:
C = 180 - A - B
Using a calculator, we get:
A ≈ 41.6 degrees
B ≈ 74.1 degrees
C ≈ 64.3 degrees
Therefore, the measures of the angles in triangle ABC are approximately:
A ≈ 41.6 degrees
B = 64 degrees
C ≈ 64.3 degrees
And the lengths of the sides are approximately:
a = 25
b = 41
c ≈ 17.7