Question

State the number of possible triangles that can be formed using the given measurements.

Answer 1

Answer:  39) 1              40) 2

                41) 1              42) 0

Step-by-step explanation:

39)     ∠A = ?        ∠B = ?       ∠C = 129°

            a = ?          b = 15         c = 45

Use Law of Sines to find ∠B:

$\dfrac{\sin B}{b}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin B}{15}=\dfrac{\sin 129}{45}\rightarrow \quad \angle B=15^o\quad or \quad \angle B=165^o$

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:

$\dfrac{\sin A}{a}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin 16}{15}=\dfrac{\sin C}{19}\rightarrow \quad \angle C=20^o\quad or \quad \angle C=160^o$

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:

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{7}=\dfrac{\sin 75}{30}\rightarrow \quad \angle A=13^o\quad or \quad \angle A=167^o$

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:

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{34}=\dfrac{\sin 119}{34}\rightarrow \quad \angle A=61^o\quad or \quad \angle A=119^o$

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.