Answer: 39) 1 40) 2
41) 1 42) 0
<u>Step-by-step explanation:</u>
39) ∠A = ? ∠B = ? ∠C = 129°
a = ? b = 15 c = 45
Use Law of Sines to find ∠B:

If ∠B = 15°, then ∠A = 180° - (15° + 129°) = 36°
If ∠B = 165°, then ∠A = 180° - (165° + 129°) = -114°
Since ∠A cannot be negative then ∠B ≠ 165°
∠A = 36° ∠B = 15° ∠C = 129° is the only valid solution.
40) ∠A = 16° ∠B = ? ∠C = ?
a = 15 b = ? c = 19
Use Law of Sines to find ∠C:

If ∠C = 20°, then ∠B = 180° - (16° + 20°) = 144°
If ∠C = 160°, then ∠B = 180° - (16° + 160°) = 4°
Both result with ∠B as a positive number so both are valid solutions.
Solution 1: ∠A = 16° ∠B = 144° ∠C = 20°
Solution 2: ∠A = 16° ∠B = 4° ∠C = 160°
41) ∠A = ? ∠B = 75° ∠C = ?
a = 7 b = 30 c = ?
Use Law of Sines to find ∠A:

If ∠A = 13°, then ∠C = 180° - (13° + 75°) = 92°
If ∠A = 167°, then ∠C = 180° - (167° + 75°) = -62°
Since ∠C cannot be negative then ∠A ≠ 167°
∠A = 13° ∠B = 75° ∠C = 92° is the only valid solution.
42) ∠A = ? ∠B = 119° ∠C = ?
a = 34 b = 34 c = ?
Use Law of Sines to find ∠A:

If ∠A = 61°, then ∠C = 180° - (61° + 119°) = 0°
If ∠A = 119°, then ∠C = 180° - (119° + 119°) = -58°
Since ∠C cannot be zero or negative then ∠A ≠ 61° and ∠A ≠ 119°
There are no valid solutions.