The diagram shows a scale drawing of a lacrosse field. The diagram is 5

The measure of an angle is three times the measure of its supplementary angle. What is the measure of each angle? I will mark br

Korvikt [17]

Answer:

Step-by-step explanation: i wouldn’t really be able to help because you did say what the supplementary is but what ever the supplementary is you multiply it with 3

4 0

3 years ago

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

romanna [79]

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:

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

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:

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