Use the given information to solve the triangle. If two solutions exists, find both.
A=110 degrees
a=125
b=100
a is greater than b so I think that there is only one triangle. I am not sure where to go from here.
To solve the triangle, we can use the Law of Sines and Law of Cosines.
1. Start by labeling the triangle with the given information:
A = 110 degrees (angle opposite side a)
a = 125 (length of side opposite angle A)
b = 100 (length of side opposite angle B)
2. Since a > b, there is only one possible triangle.
3. To find angle B, we can use the Law of Sines:
sin(B) / b = sin(A) / a
sin(B) / 100 = sin(110) / 125
Rearranging the equation:
sin(B) = (100 * sin(110)) / 125
sin(B) ≈ 0.5515
Using the arcsine function, we find:
B ≈ arcsin(0.5515)
B ≈ 33.18 degrees
4. Now, to find angle C, we know that the sum of all angles in a triangle is 180 degrees:
C = 180 - A - B
C = 180 - 110 - 33.18
C ≈ 36.82 degrees
So, the angles of the triangle are approximately:
A = 110 degrees
B ≈ 33.18 degrees
C ≈ 36.82 degrees
5. Finally, we can find the remaining side length using the Law of Cosines:
c^2 = a^2 + b^2 - 2ab*cos(C)
c^2 = 125^2 + 100^2 - 2(125)(100)*cos(36.82)
Solving the equation:
c ≈ √(125^2 + 100^2 - 2(125)(100)*cos(36.82))
c ≈ 85.48
So, the length of side c is approximately 85.48.
Therefore, the triangle is approximately:
A = 110 degrees
B ≈ 33.18 degrees
C ≈ 36.82 degrees
a = 125
b = 100
c ≈ 85.48