It's the one about triangle xyz. ignore my work. i have to find all of the angles.

Given all 3 sides you need the Law of Cosines to find the first angle.
Let's say a = 22
b = 21
and c = 29

cos(A) = b^2 + c^2 -a^2 / 2bc
cos(A) = 21*21 + 29*29 -22*22 / 2*21*29
cos(A) = 441 + 841 -484 / 2*21*29
cos(A) = 798 / 1,218
cos(A) = 0.6551724138
angle A = 49.067 Degrees

For the second angle, we can use the easier Law of Sines.
sin (B) = (b * sin(A)) / a
sin (B) = (21 * sin (49.067)) / 22
sin (B) = (21 * 0.75548) / 22
sin (B) = 0.72114
Angle B = 46.149 Degrees

Angle C is really easy to solve.
Angle C = 180 -49.067 -46.149
Angle C = 84.784

25 is obviously the correct answer

X= 49; x = 4step 1: foilstep 2: equal both to 0step 3: find x individually