For a triangle ABC if C=106.2, a=8.3 and b=11.78, what is c?
It is given that for the triangle ABC, the angle C is 106.2 degrees, side a = 8.3 and b = 11.78.
We can use the law of cosines to find c.
According to law of cosines: c^2 = a^2 + b^2 - 2ab*cos C
=> c^2 = 8.3^2 + 11.78^2 - 2*8.3*11.78*cos 106.2
=> c^2 = 68.89 + 138.76 + 54.55
=> c^2 = 262.2
=> c = sqrt(262.2)
we don't consider the negative value -16.19 as length is only positive.
The length of the side c = 16.19
Since we know two lengths of the sides of triangle ABC and the angle enclosed by them, then the length of the side c could be evaluated using the law of cosines.
c^2 = a^2 + b^2 - 2a*b*cos C
We'll replace a and b by its values:
c^2 = (8.3)^2 + (11.78)^2 - 2*8.3*11.78*cos 106.2
Since the angle 106.2 is in the 2nd quadrant, the value of the function cosine is negative.
cos 106.2 = -0.27
c^2 = 68.89 + 138.76 + 54.55
c^2 = 262.20
c = sqrt 262.20 => c = 16.19
We'll keep only the positive value since the length of a side cannot be negative.
Therefore, the requested length of the side c is c = 16.19 units.