Based on the information provided, it appears that we are dealing with a triangle with vertices at K, M, and A, where the angle at M has measure 50 degrees, and the lengths of the sides opposite to M are 6x-5 and 17x-10. To find the measure of angle KMA, we need to use the law of cosines:
(6x-5)^2 + (17x-10)^2 - 2(6x-5)(17x-10)cos(KMA) = (side opposite to angle 50)^2
Simplifying the left-hand side:
36x^2 - 60x + 25 + 289x^2 - 340x + 100 - 204x^2 + 2040x - 1000 cos(KMA) = (6x+5)^2
Simplifying the right-hand side:
(6x+5)^2 = 36x^2 + 60x + 25
Substituting this into the previous equation and simplifying:
49x^2 - 80x + 75 - 1000 cos(KMA) = 0
Solving for cos(KMA):
cos(KMA) = (49x^2 - 80x + 75)/1000
To find the measure of angle KMA, we need to take the inverse cosine of this value:
KMA = cos^{-1}[(49x^2 - 80x + 75)/1000]
The value of x is not given, so we cannot find the exact measure of angle KMA. However, we can simplify the expression for cos(KMA) by factoring the numerator:
cos(KMA) = [(49x-15)(x-5)]/1000
This expression shows that cos(KMA) is positive when 15/49 < x < 5, and negative when x < 15/49 or x > 5. In either case, we can find the measure of angle KMA by taking the inverse cosine of the absolute value of cos(KMA):
KMA = cos^{-1}(|(49x^2 - 80x + 75)/1000|)
Note that the absolute value is necessary because the cosine function is only defined for values between -1 and 1, but the expression for cos(KMA) can have values outside this range.