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Ran into this problem recently: I have a scalene trapezoid with parallel bases $b_1$ and $b_2$, and legs $l_1$ and $l_2$. Both base side lengths are known, but only one leg is known. In addition, one angle $\alpha$ is known, but that angle is adjacent to the unknown leg. Since the angles of a trapezoid leg are supplemental, the other angle adjacent to the unknown leg, $\beta$, is also known. The length of the midpoint $m$ can also be determined.

My question is, can the trapezoid be determined with this information? Specifically, I'm looking for the diagonals $d_1$ and $d_2$ as well as the unknown leg side length. Can the parallel bases be used to solve this in a simple(r) way?

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  • $\begingroup$ This problem is equivalent to finding triangle by two sides $l_1$ and $|b_2-b_1|$ and angle $\alpha$ opposite to one of them (to $l_1$). This problem has two solutions in general, because equation for second leg is quadratic (cosine rule). $\endgroup$ Commented Jun 28, 2022 at 7:20

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Let the angle adjacent to the known leg be $\beta\in (0,\pi/2)$. Also let the angle $\alpha$ vary in the interval $(0,\pi)$. WLOG let the base adjacent to $\alpha,\beta$ have length $b_1$, and the known length be $l_1$ (adjacent to $\beta$). Dropping heights from the vertices on base $b_2$ to $b_1$ yields easily

$$h=l_1\sin\beta=l_2\sin\alpha\\ b_1=b_2+l_1\cos\beta+l_2\cos\alpha$$

Rearranging a bit we obtain

$$l_2=l_1\frac{\sin\beta}{\sin\alpha}\\ b_1-b_2=\frac{l_1}{\sin\alpha}\sin(\alpha+\beta)$$

For a given $\alpha,b_1, b_2, l_1$ we see that in principle we can express $\beta, l_2$ in terms of known quantities. By studying the behavior of the function $ f(x)=\sin(\alpha+x), x\in(0,\pi/2)$ we see that for any $\alpha\in (0,\pi)$ there is an interval in which $f$ is not one-to-one and hence the same set of known parameters can possibly parametrize two different trapezoids.

The casework here is a little tedious- I'll add details when I have more time.

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Imagine $b_1$, $l_1$, $b_2$, and $l_2$ as being long straight parts of a meccano set, with $b_1$ and $b_2$ joined to the ends of $l_1$ and $l_2$ joined to the other end of $b_1$. For now, put $b_1$ at the bottom and $b_2$ at the top, with $l_1$ the known leg on the left and $l_2$, the unknown length, at its correct angle on the right. Put $l_1$ and $b_2$ "flat", with $l_1$ lying on top of $b_1$ and $b_2$ continuing the line of $l_1$.

Now begin to to rotate $l_1$ anti-clockwise, so its end moves in an arc of a circle. Because $b_2$ is parallel to $b_1$ its right hand end must also follow a similar arc further to the right. You would need this arc to intersect the leg $l_2$. The arc may intersect the leg $l_2$ at zero, one, or two places. This means that in general this information does not determine the trapezium; there may be no trapezium matching the information, or there may be one or there may be two.

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