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.

Part a would be letting z = AC I got z^2−2xzcos(θ)+(x2−y2)=0 but I don't get part b.
Pls help!

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7 years ago

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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.

Part a would be letting z = AC I got z^2−2xzcos(θ)+(x2−y2)=0 but I don't get part b.
Pls help!

Part a would be letting z = AC I got z^2−2xzcos(θ)+(x2−y2)=0 but I don't get part b. 
Pls help!