Given: GA = OL Prove: GO = AL
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<h3>28.</h3>
The path of any fair ball will divide the 90° angle between foul lines into complementary angles. The angle with the first-base foul line is ...
... 90° -27° = 63°
<h3>30.</h3>The attached picture shows the two angles "constructed" using a protractor. It is not possible to construct a 97° angle with compass and straight-edge
<h3>32.</h3>Let x represent the measure of the complement. Then the angle that is 6° less than the meaure of the complement is x-6°. The angle and its complement total 90°, so we have ...
... (x-6°) + (x) = 90°
... 2x - 6° = 90° . . . . . simplify
... 2x = 96° . . . . . . . . .add 6°
... x = 48° . . . . . . . . . . divide by 2
... (x-6°) = 48°-6° = 42°
The angles are 42° and 48°.
<h3>34.</h3>Let x represent the measure of the angle. Then its supplement is 180°-x, and half that is (180°-x)/2. The angle is 3° more than this, so we have
... x = (180° -x)/2 +3°
... x -3° = (180° -x)/2 . . . . subtract 3°
... 2x -6° = 180° -x . . . . . .multiply by 2
... 3x -6° = 180° . . . . . . . .add x
... x - 2° = 60° . . . . . . . . . divide by 3 to get x by itself
... x = 62° . . . . . . . . . . . . add 2°
... The supplementary angle is 180° -62° = 118°, half of which is 59°.
One angle is 62°; half the measure of its supplement is 59°; it supplement is 118°.