Answer:
6 < x < 23.206
Explanation:
To properly answer this question, we need to make the assumption that angle DAC is non-negative and that angle BCA is acute.
The maximum value of the angle DAC can be shown to occur when points B, C, and D are on a circle centered at A*. When that is the case, the sine of half of angle DAC is equal to 16/22 times the sine of half of angle BAC. That is, ...
(2x -12)/2 = arcsin(16/22×sin(24°))
x ≈ 23.206°
Of course, the minimum value of angle DAC is 0°, so the minimum value of x is ...
2x -12 = 0
x -6 = 0 . . . . . divide by 2
x = 6 . . . . . . . add 6
Then the range of values of x will be ...
6 < x < 23.206
_____
* One way to do this is to make use of the law of cosines:
22² = AB² + AC² -2·AB·AC·cos(48°)
16² = AD² + AC² -2·AD·AC·cos(2x-12)
The trick is to maximize x while satisfying the constraints that all of the lengths are positive. This will happen when AB=AC=AD, in which case the equations be come ...
22² = 2·AB²·(1-cos(48°))
16² = 2·AB²·(1 -cos(2x-12))
The value of AB drops out of the ratio of these equations, and the result for x is as above.