In most geometry courses, we learn that there's no such thing as "SSA Congruence".

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.

I have solved part (a) and gotten $z^2-2xz\cos (\theta) + (x^2-y^2)=0$.

I don't quite get part (b). It is obvious $90^{\circ}$ is a solution but do I have to prove that this is right? Or do I need to prove that this is the only solution? Or is there other answers?

My work: All I know is that the discriminant would be $4x^2z^2\cos^2 \theta-4(-x^2+y^2)$

P.S. My answer for part(a) is correct, right?

In the generic case, there are two triangles (up to congruence). To see this, suppose we are given AB, BC, and the angle at A. Fix the segment AB and let BC rotate through a circle (centered at B). Form the line from A at the given angle from AB. That line can intersect the circle at two points.

True, it can be tangent to that circle but that's precisely the case in which the leg BC is perpendicular to AC.

Note: Some of the other answers here are incorrect. The discriminant does not have to be zero.

My solution:

From the cosine rule, we get $y^2=x^2+z^2-2xz\cos\theta$.

$\implies z^2-2xz\cos\theta+x^2-y^2=0$ ------------------- (1)

For real solutions, the discriminant has to be non-negative.

$4x^2\cos^2\theta-4(x^2-y^2)\ge 0$

$\implies x^2\cos^2\theta-(x^2-y^2)\ge 0$

$\implies y^2-x^2\sin^2\theta\ge 0$

$\implies (y+x\sin\theta)(y-x\sin\theta)\ge 0$

We know that $y+x\sin\theta>0$

So, $y-x\sin \theta\ge 0$

$\implies x\sin\theta\le y$

In addition, we want either:

1) repeated positive roots, in which case the discriminant will be $0$. So, $x\sin\theta=y$, in which case $z=x\cos\theta$. So, $\theta$ better be acute as well.

OR

2) only one root to be positive. Then the product of the roots will be $0$ or negative. So, from our quadratic equation [eq. (1)], we get $x^2-y^2\le 0$.

$\implies x^2\le y^2$

$\implies x\le y$

Note that in this case, $x\sin\theta\le x\le y\implies x\sin \theta\le y$, so the condition that the roots be real is also satisfied.

Thus, the condition we require is:

$\boxed{x\sin\theta=y\text{ and }\theta<90^0}$

OR

$\boxed{x\le y}$

In order for $AC = z$ to be uniquely determined, we want the discriminant to equal 0.

The discriminant is, in fact: $$4x^2 \cos^2(θ) - 4x^2 + 4y^2$$

Check your work for part (a).

We can simplify this to get

\begin{align*} 4x^2 \cos^2(θ) - 4x^2 + 4y^2 &= 4x^2 (\cos^2(θ)-1) + 4y^2 \\ &= -4x^2 \sin^2{θ} + 4y^2 = 0 \end{align*}

Simplifying gets:

$y^2 - x^2 \sin^2{θ} = (y-x \sin{θ})(y+x \sin{θ}) = 0$ Thus, the solution set is:

$y = x \sin{θ}; y = -x \sin\theta$