Question

classify each set of conditions by the possible number of triangles, if any, that can be created with those conditions. drag each set of conditions to the table. more than one triangle can be formed exactly one triangle can be formed no triangles can be formed m∠a = 35°, m∠b = 45°, ab = 3 m∠a = 30°, m∠b = 30°, m∠c = 60° ab = 5, bc = 8, ca = 3 m∠a = 50°, m∠b = 80°, bc = 5 m∠c = 70°, ca = 5, ab = 7

Answer:

To solve this, we analyze each set of conditions using triangle formation rules (e.g., angle - sum property, triangle inequality, ASA, SSA criteria):

1. Condition: \(m\angle A = 35^{\circ},m\angle B=45^{\circ},AB = 3\)

• Step 1: Find \(m\angle C\) using the angle - sum property of a triangle (\(m\angle A+m\angle B + m\angle C=180^{\circ}\)).

\(m\angle C=180-(35 + 45)=100^{\circ}\)

• Step 2: We have two angles and the included side (ASA - Angle - Side - Angle: \(\angle A\), \(AB\), \(\angle B\)). By the ASA congruence criterion, there is exactly one triangle that can be formed.

2. Condition: \(m\angle A = 90^{\circ},m\angle B = 30^{\circ},m\angle C=60^{\circ}\)

• Step 1: The sum of angles \(90 + 30+60 = 180^{\circ}\), which satisfies the angle - sum property. But we are not given any side lengths. However, for a given set of angle measures (AAA - Angle - Angle - Angle), there are infinitely many similar triangles (more than one triangle) with different side lengths that can be formed.

3. Condition: \(AB = 5,BC = 8,CA=3\)

• Step 1: Check the triangle inequality theorem, which states that for a triangle with side lengths \(a\), \(b\), \(c\), \(a + b>c\), \(a + c>b\) and \(b + c>a\).

Here, \(AB+CA=5 + 3=8\), and \(BC = 8\). But the triangle inequality requires \(a + b>c\) (strictly greater than, not equal to). So, no triangle can be formed.

4. Condition: \(m\angle A=50^{\circ},m\angle B = 80^{\circ},BC = 5\)

• Step 1: Find \(m\angle C\) using the angle - sum property: \(m\angle C=180-(50 + 80)=50^{\circ}\)
• Step 2: We have two angles and a non - included side (AAS - Angle - Angle - Side: \(\angle A\), \(\angle B\), \(BC\)). By the AAS congruence criterion, there is exactly one triangle that can be formed.

5. Condition: \(m\angle C = 70^{\circ},CA = 5,AB=7\)

• Step 1: We can use the Law of Sines: \(\frac{AB}{\sin C}=\frac{CA}{\sin B}\)

Substitute \(AB = 7\), \(CA = 5\), \(m\angle C=70^{\circ}\)
\(\sin B=\frac{CA\sin C}{AB}=\frac{5\sin70^{\circ}}{7}\)
\(\sin70^{\circ}\approx0.9397\), so \(\sin B=\frac{5\times0.9397}{7}\approx\frac{4.6985}{7}\approx0.6712\)

• Step 2: Since \(\sin B\approx0.6712\), \(B\) can be an acute angle (\(B_1\approx42.1^{\circ}\)) or an obtuse angle (\(B_2 = 180 - 42.1=137.9^{\circ}\)). We need to check if both angles are valid.

For \(B_2 = 137.9^{\circ}\), \(m\angle A=180-(70 + 137.9)=- 27.9^{\circ}\) (invalid, as angle measure cannot be negative). But for \(B_1\approx42.1^{\circ}\), \(m\angle A=180-(70 + 42.1)=67.9^{\circ}\) (valid). Wait, actually, when we use the Law of Sines in the SSA (Side - Side - Angle) case, if \(AB>CA\) ( \(7>5\)), then there is exactly one triangle. Wait, let's re - check.
The Law of Sines: \(\frac{AB}{\sin C}=\frac{CA}{\sin B}\)
\(\sin B=\frac{CA\sin C}{AB}=\frac{5\sin70^{\circ}}{7}\approx0.671\)
Since \(AB = 7\) and \(CA = 5\), and \(AB>CA\), the angle opposite \(AB\) is \(C = 70^{\circ}\), and the angle opposite \(CA\) is \(B\). Since \(AB>CA\), \(C>B\) (because in a triangle, the larger side is opposite the larger angle). Since \(\sin B\approx0.671\), \(B\approx42.1^{\circ}\) (and \(42.1^{\circ}<70^{\circ}\)), so there is exactly one triangle? Wait, no, if \(a>b\) in SSA (\(a = AB\), \(b = CA\)), then there is exactly one triangle. Wait, maybe I made a mistake earlier. Let's correct:
If we have \(AB = c = 7\), \(CA = b = 5\), \(\angle C = 70^{\circ}\)
By the Law of Sines, \(\frac{c}{\sin C}=\frac{b}{\sin B}\)
\(\sin B=\frac{b\sin C}{c}=\frac{5\sin70^{\circ}}{7}\approx0.671\)
Since \(c>b\), \(\angle C>\angle B\), so \(\angle B\) must be acute. So there is exactly one triangle? Wait, no, the initial thought about two triangles is when \(b>a\sin C\) and \(b < a\). Let's calculate \(a\sin C\) (here \(a\) is the side opposite \(\angle A\), wait, maybe I mixed up the notation. Let's use standard notation: in \(\triangle ABC\), \(a = BC\), \(b = AC\), \(c = AB\), \(\angle A\) opposite \(a\), \(\angle B\) opposite \(b\), \(\angle C\) opposite \(c\).
So \(c = 7\), \(b = 5\), \(\angle C = 70^{\circ}\)
We know that \(b\sin C=5\sin70^{\circ}\approx4.698\)
Since \(b\sin C\approx4.698 < b = 5 < c = 7\), there are two possible triangles. Wait, the formula for SSA: if \(b\sin C < a < b\) (wait, no, in our case \(a\) is unknown, \(b = 5\), \(c = 7\), \(\angle C = 70^{\circ}\)). The correct rule is: for SSA (side \(c\), side \(b\), angle \(C\)), if \(b\sin C < c\) and \(b < c\), then there are two triangles? No, if \(b < c\), then \(\angle B<\angle C\), so \(\angle B\) is acute, and there is exactly one triangle. Wait, I think I confused the notation. Let's use the first condition's approach. Since we have two sides and a non - included angle (SSA), and \(AB = 7\), \(CA = 5\), \(\angle C = 70^{\circ}\), and \(AB>CA\), the angle opposite \(AB\) is \(\angle C\), and the angle opposite \(CA\) is \(\angle B\). Since \(AB>CA\), \(\angle C>\angle B\), so \(\angle B\) must be acute, so there is exactly one triangle? No, actually, when \(c>b\) in SSA (\(c = AB\), \(b = CA\), \(\angle C\)), then there is exactly one triangle. So this set of conditions forms exactly one triangle? Wait, no, let's take an example. Suppose we have \(c = 7\), \(b = 5\), \(\angle C = 70^{\circ}\). We can draw \(AB = 7\), then at point \(C\), we have an angle of \(70^{\circ}\), and \(CA = 5\). Since \(AB>CA\), there is only one possible position for point \(B\). So exactly one triangle.

Now, let's classify each:

• \(m\angle A = 35^{\circ},m\angle B = 45^{\circ},AB = 3\): Exactly one triangle.
• \(m\angle A = 90^{\circ},m\angle B = 30^{\circ},m\angle C = 60^{\circ}\): More than one triangle (AAA, similar triangles).
• \(AB = 5,BC = 8,CA = 3\): No triangles (violates triangle inequality).
• \(m\angle A = 50^{\circ},m\angle B = 80^{\circ},BC = 5\): Exactly one triangle (AAS).
• \(m\angle C = 70^{\circ},CA = 5,AB = 7\): Exactly one triangle (SSA with \(c>b\), so one triangle)

Final Classification Table:

Condition	Number of Triangles
\(m\angle A = 90^{\circ},m\angle B = 30^{\circ},m\angle C = 60^{\circ}\)	More than one triangle
\(AB = 5,BC = 8,CA = 3\)	No triangles
\(m\angle A = 50^{\circ},m\angle B = 80^{\circ},BC = 5\)	Exactly one triangle
\(m\angle C = 70^{\circ},CA = 5,AB = 7\)	Exactly one triangle

If we were to fill the table in the problem:

• More than one triangle can be formed: \(m\angle A = 90^{\circ},m\angle B = 30^{\circ},m\angle C = 60^{\circ}\)
• Exactly one triangle can be formed: \(m\angle A = 35^{\circ},m\angle B = 45^{\circ},AB = 3\); \(m\angle A = 50^{\circ},m\angle B = 80^{\circ},BC = 5\); \(m\angle C = 70^{\circ},CA = 5,AB = 7\)
• No triangles can be formed: \(AB = 5,BC = 8,CA = 3\)