Get all angles of an isosceles trapezoid

Question

Okay so this is the problem: "One of the acute base of an isosceles trapezoid is 3 less than 3 times a certain number, while the other acute angle is 24 more than twice the same number. Find the measures of all of the angles of the trapezoid." I think the missing "certain number" is 27 but I'm not sure. My brain said that it should be like this:

= (3 - 3x) + (24 + 2x)

= 3 - 3x + 24 + 2x

= -3x + 2x + 3 + 24

= -x + 27

x = 27

But I'm not good with math so I'm not sure. Please help in explaining it to me.

Answer 1

The first angle should be 3x - 3. You have the expression for the second angle written correctly.

The base angles are congruent in isosceles trapezoids so

3x - 3 =24 + 2x

I would subtract 2x from each side first

x - 3 = 24

Then add 3 to each side

x = 27. The work you showed doesn't really make sense mathematically because your equation doesn't equal anything.

We have found x, but they want the measure of all the angles.

So for the acute angles, we substitute 27 in for x in either expression

24 + 2(27) = 24 + 54 = 78.

The acute angles and obtuse angles inside the trapezoid are supplementary so they add up to 180.

180 - 78 = 102.

The acute angles measure 78 degrees and the obtuse angles measure 102 degrees.