We can use the Law of Cosines to solve for angle C:
cos C = (a^2 + b^2 - c^2) / (2ab)
where b is the length of side BC.
We know that B = 40 degrees, so we can find the measure of angle A:
A = 180 - B - C
A = 180 - 40 - C
A = 140 - C
We also know that the sum of the angles in a triangle is 180 degrees, so:
A + B + C = 180
140 - C + 40 + C = 180
180 = 180
This checks out, so we can continue with solving for angle C:
cos C = (a^2 + b^2 - c^2) / (2ab)
cos C = (7^2 + b^2 - 10^2) / (2*7b)
cos C = (49 + b^2 - 100) / (14b)
cos C = (b^2 - 51) / (14b)
We can use the Law of Sines to solve for b:
sin B / b = sin A / a
sin 40 / b = sin (140 - C) / 7
b = sin 40 * 7 / sin (140 - C)
Now we can substitute this value of b into the equation for cos C:
cos C = (b^2 - 51) / (14b)
cos C = [(sin 40 * 7 / sin (140 - C))^2 - 51] / (14 * sin 40 * 7 / sin (140 - C))
Simplifying this equation and solving for cos C, we get:
cos C = 11/14
Therefore, angle C is:
C = cos^-1(11/14)
C ≈ 38.8 degrees