Question 1179470
Here's a diagram that explains the ambiguous case and the calculations for the two possible areas of Sam's triangular piece of land.

**Diagram:**

[asy]
unitsize(0.8 cm);

pair A, B, C, D;

A = (0,5);
B = (0,0);
C = (8,0);
D = (8,3);

draw(A--B--C--cycle);
draw(C--D);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, NE);
label("$24^\circ$", B + (0.8,0.2));
label("$180$", (A + C)/2, NE);
label("$340$", (C + D)/2, NE);
[/asy]

Let:

* AB be the first side (road)
* BC be the second side (wall)
* AC be the third side (fence)
* D be the point where the fence meets the wall

We are given:

* Angle ABC = 24°
* AC = 180 feet
* CD = 340 feet

**Ambiguity:**

The ambiguity arises because we are not given which side of the wall the fence lies on. This leads to two possible triangles:

* **Triangle ABC:** The fence lies on the same side of the wall as the road.
* **Triangle ABD:** The fence lies on the opposite side of the wall as the road.

**Calculations:**

**Triangle ABC:**

1. **Find angle ACB:**
   Using the Law of Sines in triangle ACD:
   ```
   sin(∠ACD) / AC = sin(∠CAD) / CD
   sin(∠ACD) / 180 = sin(24°) / 340
   sin(∠ACD) ≈ 0.214
   ∠ACD ≈ 12.4°
   ```
   Since ∠ACD and ∠ACB are supplementary angles:
   ```
   ∠ACB ≈ 180° - 12.4° = 167.6°
   ```

2. **Find angle BAC:**
   ```
   ∠BAC ≈ 180° - ∠ABC - ∠ACB 
   ∠BAC ≈ 180° - 24° - 167.6° = -11.6° 
   ```
   Since angles cannot be negative, this triangle is not possible.

**Triangle ABD:**

1. **Find angle ADB:**
   Using the Law of Sines in triangle ACD:
   ```
   sin(∠ADB) / AC = sin(∠CAD) / CD
   sin(∠ADB) / 180 = sin(24°) / 340
   sin(∠ADB) ≈ 0.214
   ∠ADB ≈ 12.4°
   ```

2. **Find angle BAD:**
   ```
   ∠BAD ≈ 180° - ∠ABD - ∠ADB 
   ∠BAD ≈ 180° - 24° - 12.4° = 143.6°
   ```

3. **Find AD:**
   Using the Law of Sines in triangle ABD:
   ```
   AD / sin(∠ABD) = BD / sin(∠BAD)
   AD / sin(24°) = 340 / sin(143.6°)
   AD ≈ 224.3 \text{ feet}
   ```

4. **Find AB:**
   Using the Law of Cosines in triangle ABD:
   ```
   AB^2 = AD^2 + BD^2 - 2 * AD * BD * cos(∠ADB)
   AB^2 ≈ 224.3^2 + 340^2 - 2 * 224.3 * 340 * cos(12.4°)
   AB ≈ 129.4 \text{ feet}
   ```

5. **Calculate the area:**
   Using Heron's formula with sides AB, BD, and AD:
   ```
   s = (AB + BD + AD) / 2 ≈ (129.4 + 340 + 224.3) / 2 ≈ 346.85
   Area ≈ √[s(s - AB)(s - BD)(s - AD)] 
   Area ≈ √[346.85(346.85 - 129.4)(346.85 - 340)(346.85 - 224.3)]
   Area ≈ 14595 \text{ square feet}
   ```

**Conclusion:**

The two possible lengths for the first side are approximately 129.4 feet. The area of the triangular piece of land is approximately 14595 square feet.