Solve for $x$. I would be able to solve this with law of cosines (with a LOT of work!) but the students that this problem was presented are not familiar with law of cosines. Is there something I am missing? My approach would be to calculate the hypotenuse of the right triangle by using Pythagorean theorem. Therefore I would know everything needed to use law of cosines. If needed $y = (3x-4)\tan36°$. I think there might a simpler solution for this but what is it?
2026-03-25 17:22:28.1774459348
Solve for $x$ from this quadrilateral without using law of cosines.
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1
In the diagram, take the right triangle and swap the position of the acute angles. You then have a large right triangle as the complement of $36^{\circ}$ is the supplement of $126^{\circ}$. Moreover, by arranging the side lengths as such, the $y$ variable is immediately eliminated.
Here's a diagram illustrating what we have now.
Solving for $x$ using the Pythagorean theorem:
\begin{align} (4x)^2+(3x-4)^2&=(5x-2)^2\\ 16x^2+9x^2-24x+16&=25x^2-20x+4\\ -4x&=-12\\ x&=3 \end{align}
Therefore, $\boxed{x=3}$.