Guys i really need geometry hw 21 and 22

These can be a bit of a mind-twister. A fairly straightforward approach is to define a variable (x) to be the angle, then use 90-x for the complement and 180-x for the supplement. Write the relations given in the text, solve for x, then calculate the measure of the angle asked for by the problem statement.

You don't have to do it that way. In 21, you can work directly with "the complement of the angle." You know that an angle and its complement make 90°, and that an angle and its supplement make 180°. This means the supplement is 90° more than the complement—a fact you can use here.

The supplement of an angle is 4 times the complement of the angle.

The complement of the angle.

Let x represent the complement of the angle (what the question asks for). Then the supplement of the angle is x+90.

The supplement is 4 × the complement

... x + 90 = 4x

... 90 = 3x . . . . . . . subtract x

... 30 = x

The complement of the angle is 30°.

Check

The angle is 60°; its complement is 30°, and its supplement is 120°. 120° is 4 times 30°, as required.

5 times the complement of an angle less 2 times the supplement is 40°

the supplement of the angle

We can do this as we did in the previous problem. We are asked to find the supplement of the angle. Let x represent the supplement of the angle. As discussed above, then the complement of the angle is x-90°.

... 5 × (the complement) - 2 × (the supplement) = 40

... 5(x-90) -2x = 40 . . . . . . . . . filling in the variable expressions for complement and supplement

... 5x -450 -2x = 40 . . . . . . . . eliminate parentheses

... 3x = 490 . . . . . . . . . . . . . . . collect terms, add 450

... x = 163 1/3

The supplement of the angle is 163 1/3°.

Check

The angle is 16 2/3°. Its complement is 73 1/3°.

5 times the complement is 366 2/3°. 2 times the supplement is 326 2/3°. The latter subtracted from the former is 366 2/3 - 326 2/3 = 40, as required.

step-by-step explanation:

i think you should think about the question and not look online.