Math / Trigonometry

QuestionUsing the Law of Sines to solve the all possible triangles if $\angle A=119^{\circ}, a=30, b=19$ . If no answer exists, enter DNE for all answers. $\angle B$ is $\square$ degrees; $\angle C$ is $\square$ degrees; $c=\square ;$

Studdy Solution

STEP 1

What is this asking? We're given a triangle with an angle of $119$ degrees and two sides with lengths $30$ and $19$ .
We need to find the other angle and side using the Law of Sines! Watch out! Remember, the Law of Sines can sometimes give two possible solutions for a triangle, so we need to be extra careful and check for that!

STEP 2

1. Find angle B
2. Find angle C
3. Find side c

STEP 3

Let's start with the Law of Sines, which says $\frac{\sin(\angle A)}{a} = \frac{\sin(\angle B)}{b}$ .
We know $\angle A = 119^{\circ}$ , $a = 30$ , and $b = 19$ , so let's plug those values in: $\frac{\sin(119^{\circ})}{30} = \frac{\sin(\angle B)}{19}$ .

STEP 4

Now, we want to isolate $\sin(\angle B)$ , so we multiply both sides by $19$ : $19 \cdot \frac{\sin(119^{\circ})}{30} = \sin(\angle B)$ .

STEP 5

Let's calculate that value: $\sin(\angle B) \approx 19 \cdot \frac{0.8746}{30} \approx 0.5548$ .

STEP 6

To find $\angle B$ , we take the inverse sine (arcsin) of both sides: $\angle B = \arcsin(0.5548) \approx 33.7^{\circ}$ .

STEP 7

We know that the sum of the angles in a triangle is always $180^{\circ}$ .
We already know $\angle A = 119^{\circ}$ and $\angle B \approx 33.7^{\circ}$ , so we can find $\angle C$ : $\angle C = 180^{\circ} - 119^{\circ} - 33.7^{\circ} \approx 27.3^{\circ}$ .

STEP 8

Back to the Law of Sines!
This time, we'll use $\frac{\sin(\angle A)}{a} = \frac{\sin(\angle C)}{c}$ .
We plug in our known values: $\frac{\sin(119^{\circ})}{30} = \frac{\sin(27.3^{\circ})}{c}$ .

STEP 9

To solve for $c$ , we can cross-multiply: $c \cdot \sin(119^{\circ}) = 30 \cdot \sin(27.3^{\circ})$ .

STEP 10

Now, we divide both sides by $\sin(119^{\circ})$ to isolate $c$ : $c = \frac{30 \cdot \sin(27.3^{\circ})}{\sin(119^{\circ})}$ .

STEP 11

Let's calculate that: $c \approx \frac{30 \cdot 0.4586}{0.8746} \approx 15.7$ .

STEP 12

$\angle B \approx 33.7^{\circ}$ ; $\angle C \approx 27.3^{\circ}$ ; $c \approx 15.7$ .

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QuestionUsing the Law of Sines to solve the all possible triangles if $\angle A=119^{\circ}, a=30, b=19$ . If no answer exists, enter DNE for all answers. $\angle B$ is $\square$ degrees; $\angle C$ is $\square$ degrees; $c=\square ;$

Studdy Solution

STEP 1

STEP 2

1. Find angle B
2. Find angle C
3. Find side c

STEP 3

Let's start with the Law of Sines, which says $\frac{\sin(\angle A)}{a} = \frac{\sin(\angle B)}{b}$ .
We know $\angle A = 119^{\circ}$ , $a = 30$ , and $b = 19$ , so let's plug those values in: $\frac{\sin(119^{\circ})}{30} = \frac{\sin(\angle B)}{19}$ .

STEP 4

Now, we want to isolate $\sin(\angle B)$ , so we multiply both sides by $19$ : $19 \cdot \frac{\sin(119^{\circ})}{30} = \sin(\angle B)$ .

STEP 5

Let's calculate that value: $\sin(\angle B) \approx 19 \cdot \frac{0.8746}{30} \approx 0.5548$ .

STEP 6

To find $\angle B$ , we take the inverse sine (arcsin) of both sides: $\angle B = \arcsin(0.5548) \approx 33.7^{\circ}$ .

STEP 7

We know that the sum of the angles in a triangle is always $180^{\circ}$ .
We already know $\angle A = 119^{\circ}$ and $\angle B \approx 33.7^{\circ}$ , so we can find $\angle C$ : $\angle C = 180^{\circ} - 119^{\circ} - 33.7^{\circ} \approx 27.3^{\circ}$ .

STEP 8

Back to the Law of Sines!
This time, we'll use $\frac{\sin(\angle A)}{a} = \frac{\sin(\angle C)}{c}$ .
We plug in our known values: $\frac{\sin(119^{\circ})}{30} = \frac{\sin(27.3^{\circ})}{c}$ .

STEP 9

To solve for $c$ , we can cross-multiply: $c \cdot \sin(119^{\circ}) = 30 \cdot \sin(27.3^{\circ})$ .

STEP 10

Now, we divide both sides by $\sin(119^{\circ})$ to isolate $c$ : $c = \frac{30 \cdot \sin(27.3^{\circ})}{\sin(119^{\circ})}$ .

STEP 11

Let's calculate that: $c \approx \frac{30 \cdot 0.4586}{0.8746} \approx 15.7$ .

STEP 12

$\angle B \approx 33.7^{\circ}$ ; $\angle C \approx 27.3^{\circ}$ ; $c \approx 15.7$ .

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Studdy Solution

STEP 1

STEP 2

1. Find angle B2. Find angle C3. Find side c

STEP 3

STEP 4

Now, we want to **isolate** sin⁡(∠B)\sin(\angle B)sin(∠B), so we **multiply** both sides by 191919: 19⋅sin⁡(119∘)30=sin⁡(∠B)19 \cdot \frac{\sin(119^{\circ})}{30} = \sin(\angle B)19⋅30sin(119∘)​=sin(∠B).

STEP 5

Let's **calculate** that value: sin⁡(∠B)≈19⋅0.874630≈0.5548\sin(\angle B) \approx 19 \cdot \frac{0.8746}{30} \approx 0.5548sin(∠B)≈19⋅300.8746​≈0.5548.

STEP 6

To find ∠B\angle B∠B, we take the **inverse sine** (arcsin) of both sides: ∠B=arcsin⁡(0.5548)≈33.7∘\angle B = \arcsin(0.5548) \approx 33.7^{\circ}∠B=arcsin(0.5548)≈33.7∘.

STEP 7

STEP 8

STEP 9

To **solve for** ccc, we can **cross-multiply**: c⋅sin⁡(119∘)=30⋅sin⁡(27.3∘)c \cdot \sin(119^{\circ}) = 30 \cdot \sin(27.3^{\circ})c⋅sin(119∘)=30⋅sin(27.3∘).

STEP 10

Now, we **divide** both sides by sin⁡(119∘)\sin(119^{\circ})sin(119∘) to **isolate** ccc: c=30⋅sin⁡(27.3∘)sin⁡(119∘)c = \frac{30 \cdot \sin(27.3^{\circ})}{\sin(119^{\circ})}c=sin(119∘)30⋅sin(27.3∘)​.

STEP 11

Let's **calculate** that: c≈30⋅0.45860.8746≈15.7c \approx \frac{30 \cdot 0.4586}{0.8746} \approx 15.7c≈0.874630⋅0.4586​≈15.7.

STEP 12

∠B≈33.7∘\angle B \approx 33.7^{\circ}∠B≈33.7∘; ∠C≈27.3∘\angle C \approx 27.3^{\circ}∠C≈27.3∘; c≈15.7c \approx 15.7c≈15.7.

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1. Find angle B
2. Find angle C
3. Find side c

Now, we want to isolate $\sin(\angle B)$ , so we multiply both sides by $19$ : $19 \cdot \frac{\sin(119^{\circ})}{30} = \sin(\angle B)$ .

Let's calculate that value: $\sin(\angle B) \approx 19 \cdot \frac{0.8746}{30} \approx 0.5548$ .

To find $\angle B$ , we take the inverse sine (arcsin) of both sides: $\angle B = \arcsin(0.5548) \approx 33.7^{\circ}$ .

To solve for $c$ , we can cross-multiply: $c \cdot \sin(119^{\circ}) = 30 \cdot \sin(27.3^{\circ})$ .

Now, we divide both sides by $\sin(119^{\circ})$ to isolate $c$ : $c = \frac{30 \cdot \sin(27.3^{\circ})}{\sin(119^{\circ})}$ .

Let's calculate that: $c \approx \frac{30 \cdot 0.4586}{0.8746} \approx 15.7$ .

$\angle B \approx 33.7^{\circ}$ ; $\angle C \approx 27.3^{\circ}$ ; $c \approx 15.7$ .