geometrytrigonometry
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!
$$a = b - 2 \,c \cos \alpha $$
where $ \alpha = $ angle between the big side and one of equal sides length $c$.
Hint:
Let $b$ the longer basis $AD$. Than, given that the angle $\angle BAH$ is $45°$, the height of the trapezoid is $h= BH=AH=AD-BC=b-a$, and from the diagonal $AC=4a$ ( as hipothenuse of the triangle $ACD$ ) we have:
$$ AC^2=AD^2+CD^2 \qquad \iff \qquad b^2+(b-a)^2=16a^2 $$
solve this equation for $b>0$ and you have done.
The equation becomes: $$ b^2-2ab-15a^2=0 $$ with solutions : $$ b=\frac{a\pm\sqrt{a^2+30a^2}}{2} $$ and, for $b>0$: $$ b=\frac{a}{2}(1+\sqrt{31}) $$ so the area is: $$ A=\frac{1}{2}(b+a)h=\frac{1}{2}(b+a)(b-a)=\frac{1}{2}\left(\frac{a}{2}(1+\sqrt{31})+a \right)\left(\frac{a}{2}(1+\sqrt{31})-a \right)= $$ $$ =\frac{1}{2}\left(\frac{a^2}{4}(1+\sqrt{31})^2-a^2 \right)=\frac{1}{4}a^2(14+\sqrt{31}) $$
Best Answer
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.