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Apr 16, 2022

#1
+1

SSA Congruence is not a valid Congruence Theorem.

Apr 16, 2022
#2
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Yes, this is true, but I think the problem wants us to find the cases where "SSA congruence" would work.

MaxWong  Apr 17, 2022
#3
+9459
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(a) By Law of Cosines,

$$c^2 = a^2 + b^2 - 2ab\cos \theta\\ a^2 - (2b\cos \theta)a+(b^2 - c^2) = 0$$

Now, we use the quadratic formula to get (try to fill in the steps on your own):

$$a = {b\cos \theta \pm \sqrt{ c^2 - b^2 \sin^2 \theta}}$$

(b) The condition where there are no solutions for a corresponds to the case that the expression inside the square root is negative, i.e. $$b\sin \theta > c$$.

(c) The condition where there is 1 solution for a corresponds to the case that the square root equals 0 (since $$b\cos \theta \pm 0 = b\cos \theta$$), i.e., $$b\sin \theta = c$$.

Apr 17, 2022