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.

Question

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!

Answer 1

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).