To find the conditions on x, y, and theta which guarantee that the side AC is uniquely determined, we need to consider the quadratic equation derived in part (a), which is:
z^2 − 2xz cos(theta) + (x^2 − y^2) = 0
Let's analyze the quadratic equation to determine when it has a unique solution for z (AC).
For a quadratic equation to have a unique solution, the discriminant (b^2 - 4ac) should be equal to zero. In this case, our equation is:
(-2xz cos(theta))^2 - 4(1)((x^2 - y^2)) = 0
Simplifying this equation, we get:
4x^2z^2 cos^2(theta) - 4(x^2 - y^2) = 0
Dividing both sides by 4, we have:
x^2z^2 cos^2(theta) - (x^2 - y^2) = 0
Rearranging the terms, we get:
x^2z^2 cos^2(theta) = x^2 - y^2
Now, we can solve for z^2:
z^2 = (x^2 - y^2) / (x^2 cos^2(theta))
Since z (AC) represents the length of a side, it must be positive. Therefore, the condition for AC to be uniquely determined is:
(x^2 - y^2) / (x^2 cos^2(theta)) > 0
To establish the condition more explicitly, we consider the following cases:
1. x ≠ 0: If x ≠ 0, then it cancels out, simplifying the condition to:
(x^2 - y^2) cos^2(theta) > 0
Since x and y are real numbers, (x^2 - y^2) will be positive or negative. In order for the product to be positive, both factors must have the same sign. Hence, we have two sub-cases:
a) (x^2 - y^2) > 0 and cos^2(theta) > 0
This implies x^2 > y^2 and cos^2(theta) > 0. Thus, the condition is:
x > y and theta is any real number.
b) (x^2 - y^2) < 0 and cos^2(theta) < 0
This implies x^2 < y^2 and cos^2(theta) < 0. However, cos^2(theta) cannot be negative since it is a square. Hence, this sub-case is not valid.
2. x = 0: If x = 0, the original quadratic equation becomes:
-2yz cos(theta) = x^2 - y^2
Since x = 0, y ≠ 0 for a non-degenerate triangle. Dividing both sides by -2ycos(theta), we get:
z = (y^2 - x^2) / (2ycos(theta))
In this case, the side AC can be uniquely determined with the condition that y ≠ 0 and cos(theta) ≠ 0.
To summarize:
AC is uniquely determined when the following conditions are met:
1. Case 1: x ≠ 0
a) x > y and theta is any real number.
2. Case 2: x = 0
a) y ≠ 0 and cos(theta) ≠ 0
These conditions ensure that the quadratic equation derived in part (a) has a unique solution for z (AC).