Answer:
3° < x < 32.538°
Explanation:
The possible values of the angle marked with an x-expression may not be as obvious as they first seem. There is a somewhat complicated interaction between the angle value and the length of the side(s) marked 'a'.
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We can label the vertices of the figure A through D, clockwise from the bottom.
minimum x
The minimum value of angle ACB (marked 3x-9°) is 0°. That will be its measure when AB +BC = AC, or B lies on line AC. In this case, the value of x is ...
3x -9° = 0°
3x = 9° . . . . add 9°
x = 3° . . . . . divide by 3
maximum x
value of 'a'
The maximum value of angle ACB will be found where both triangles are isosceles: AC ≅ BC ≅ DC. The length of 'a' can be found for that case using the Law of Sines:
27/sin(93°) = a/sin(43.5°) . . . where (180° -93°)/2 = 43.5° is the base angle
a = 27×sin(43.5°)/sin(93°) ≈ 18.611
value of 3x -9°
Angle ACB will obey the same Law of Sines requirement. If z represents the base angle in isosceles triangle ABC, then we have ...
sin(z)/18.611 = sin(180-2z)/26
Recognizing that sin(180-2z) = sin(2z) and using the double-angle formula, we find ...
sin(z)/18.611 = 2sin(z)cos(z)/26
sin(z)(1/18.611 -cos(z)/13) = 0 . . . . . subtract the right side, factor
This has solutions that make the factors zero:
sin(z) = 0 ⇒ z = 0 . . . . . . an extraneous solution in this case
cos(z) = 13/18.611 ≈ 45.6925°
Then the value of x can be found from ...
3x -9° = 180° -2z ≈ 88.615°
x ≈ 3° +(88.615°/3) = 32.538°
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Additional comment
In the above, we have shown numerical values for the side lengths and angles in the maximum-x case. The various expressions can be combined to give ...
3(x -3°) = 180° -2(arccos(13/27×sin(93°)/sin(43.5°)))
x = 63° -2/3·arccos(13/27·sin(93°)/sin(43.5°))