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How do I determine the length of the shorter base of a trapezoid from the longer base length, height, and only two angles?

An example would be 24" longer base, with 45 deg angles at both ends with only 1" in height. Both upper angles would be 135 deg.

What would the length of the shorter base be?

How do you solve for it?

Thanks!

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Suppose the longer base length is $b$ and the $2$ angles are $\alpha$ and $\beta$ with height being $h$

The shorter base length is

$$b-h\cot \alpha-h\cot \beta = b-h(\cot \alpha+ \cot \beta)$$

To see this, notice that the height and the base form perpendicular angle.

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If you decompose the figure into a rectangle and two right triangles, and note the triangles are isosceles, you'll see the length you seek is a longer base minus the height doubled.

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One way to think about it is this: the longer base and the angles determine a triangle. The height determines where the similar triangle is cut off to give the trapezium. Thus, if we let the base you want be $b,$ and the other $B,$ then it follows that $$\frac bB=\frac aA.$$

You can determine $A,$ the height of the main triangle, from the given information. Then $a,$ the height of the cut-off triangle, is given by $a=A-h,$ with $h$ being the height of the trapezium, also given.

The work is in determining $A.$ One way is to first calculate some other side of the large triangle (use the cosine rule, for example -- remember all three angles are known, and one side). Then you now have two sides and an included angle. This allows you to determine $A$ as a scaled sine of the included angle.

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