Question

identifying the ambiguous case
which triangle meets the criteria for the ambiguous case?
three triangle diagrams: first with sides 4, 6 and angle 66°; second with sides 3, 5 and angle 84°; third with side 7, angles 56°, 51°

Explanation:

Step1: Recall Ambiguous Case (SSA)

The ambiguous case (SSA) in triangles occurs when we know two sides and a non - included angle. For a triangle with sides \(a,b\) and non - included angle \(A\), we use the Law of Sines \(\frac{a}{\sin A}=\frac{b}{\sin B}\). The ambiguous case happens when \(a < b\) and \(A\) is acute, and \(a>b\sin A\) (two triangles), \(a = b\sin A\) (one right triangle), or \(a < b\sin A\) (no triangle).

Step2: Analyze each triangle

- **First triangle**: Sides \(4\) and \(6\), included angle \(66^{\circ}\)? No, wait, the angle \(66^{\circ}\) is between the sides? Wait, no, the first triangle has two sides \(4\) and \(6\) and the angle \(66^{\circ}\) between them? No, if it's SSA, the angle is non - included. Wait, let's re - check. The first triangle: sides \(4\) and \(6\), angle \(66^{\circ}\) at the vertex between them? No, if it's SSA, the angle is not between the two sides. Wait, maybe I misread. Wait, the first triangle: side \(4\), side \(6\), angle \(66^{\circ}\) opposite to one of them? Wait, no, let's check the angle - side configuration.
The ambiguous case (SSA) requires two sides and a non - included angle. Let's check each triangle:
- **First triangle**: Let's assume the sides are \(a = 4\), \(b = 6\), and angle \(A=66^{\circ}\) (non - included). Let's check the height \(h=b\sin A=6\sin66^{\circ}\approx6\times0.9135 = 5.481\). Since \(a = 4< h=5.481\)? No, wait \(a = 4\), \(b = 6\), angle \(A = 66^{\circ}\) (non - included). Wait, if \(a\) is opposite angle \(A\), and \(b\) is another side. Wait, maybe I made a mistake. Let's check the second triangle: sides \(3\) and \(5\), angle \(84^{\circ}\). The angle \(84^{\circ}\) is included? Because the two sides \(3\) and \(5\) meet at the \(84^{\circ}\) angle, so it's SAS, not SSA. The third triangle: angles \(56^{\circ}\) and \(51^{\circ}\), so the third angle is \(180-(56 + 51)=73^{\circ}\), and side \(7\). This is AAS or ASA, not SSA. Wait, the first triangle: sides \(4\) and \(6\), angle \(66^{\circ}\) (non - included). Let's calculate the height \(h\) corresponding to the side opposite the \(66^{\circ}\) angle? Wait, no. Let's recall: in SSA, we have two sides and a non - included angle. So for a triangle with side \(a\), side \(b\), and angle \(A\) (where \(A\) is not between \(a\) and \(b\)). Let's take the first triangle: suppose we have side \(a = 4\), side \(b = 6\), and angle \(A=66^{\circ}\) (non - included). The height \(h\) from the vertex opposite the angle \(A\) to the side \(b\) is \(h=a\sin A\)? Wait, no, let's use the Law of Sines. If we have \(a = 4\), \(b = 6\), \(\angle A=66^{\circ}\), then \(\frac{4}{\sin66^{\circ}}=\frac{6}{\sin B}\), so \(\sin B=\frac{6\sin66^{\circ}}{4}\approx\frac{6\times0.9135}{4}=\frac{5.481}{4} = 1.370\), which is impossible (since \(\sin B\leq1\)). Wait, maybe I got the sides wrong. Wait, maybe the first triangle has sides \(4\) and \(6\), and the angle \(66^{\circ}\) is non - included. Wait, no, maybe the first triangle is SSA. Wait, the ambiguous case (SSA) is also called the Law of Sines ambiguous case. Let's re - evaluate:

The second triangle: sides \(3\) and \(5\), angle \(84^{\circ}\) (included), so SAS, unique triangle.

The third triangle: two angles and a side, so AAS, unique triangle.

The first triangle: two sides (\(4\) and \(6\)) and a non - included angle (\(66^{\circ}\)). Let's check the height \(h\) of the triangle with respect to the side of length \(6\). The height \(h = 4\sin66^{\circ}\approx4\times0.9135 = 3.654\). Since \(h=3.654<6\) and \(4 < 6\), and \(4>h\) (because \(4>3.654\)), this satisfies the SSA ambiguous case (two possible triangles).

Answer:

The first triangle (with sides 4, 6 and angle \(66^{\circ}\))