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# i think i got part a down $$z^2 - 2(x)(z)cos(\theta) + (x^2 - y^2)= 0$$ I need help with part b

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In most geometry courses, we learn that there's no such thing as "SSA Congruence". That is, if we have triangles $ABC$ and $DEF$ such that $AB = DE$, $BC = EF$, and $\angle A = \angle D$, then we cannot deduce that $ABC$ and $DEF$ are congruent. However, there are a few special cases in which SSA "works". That is, suppose $ABC$ is a triangle. Let $AB = x$, $BC = y$, and $\angle A = \theta$. For some values of $x$, $y$, and $\theta$, we can uniquely determine the third side, $AC$. (a) Use the Law of Cosines to derive a quadratic equation in $AC$. (b) Use the quadratic polynomial you found in part (a) in order to find conditions on $x, y,$ and $\theta$ which guarantee that the side $AC$ is uniquely determined.

Jul 17, 2020

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Aug 3, 2021