To solve these, you make use of the relationships you know between complementary and supplementary angles: complementary angles sum to 90°; supplementary angles sum to 180°.
12. The right-angle indicator in the corner shows you the sum of the angles is 90°.
... (15x -2)° +(7x +4)° = 90°
... 22x = 88 . . . . . . . . . . . . . divide by °, collect terms, subtract 2
... x = 4 . . . . . . . . . . . . . . . . divide by the coefficient of x
Now, we need to find the two angles:
- ∠BAC = (15·4-2)° = 58°
- ∠CAD = (7·4+4)° = 32°
13. Same deal. The angles add up to 90°.
... (x -10)° +(4x -10)° = 90°
... 5x = 110 . . . . . . . . . . . . . divide by degrees, add 20, collect terms
... x = 22 . . . . . . . . . . . . . . divide by the coefficient of x
Now, we need to find the two angles:
- ∠UVW = (22 -10)° = 12°
- ∠XYZ = (4·22 -10)° = 78°
14. These supplementary angles add to 180°. The sum is different, but the method of solution is the same.
... (3x +17)° + ((1/2)x -5)° = 180°
... (3 1/2)x = 168 . . . . . . . . . . . . divide by °, subtract 12, collect terms
... x = 168/3.5 = 48 . . . . . . . . . . divide by the coefficient of x
Now, we need to find the two angles:
- ∠EFG = (3·48 +17)° = 161°
- ∠LMN = ((1/2)·48 -5)° = 19°
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Note that the answers for each of these problems are easy enough to check. Add the angle values and make sure the sum is 90° or 180° as the problem requires.