Math  /  Trigonometry

QuestionNon-right Triangles: Law of Sines For the following exercises, assume α\alpha is opposite side a,βa, \beta is opposite side b\mathbf{b}, and γ\gamma is opposite side cc. Solve each triangle, if possible. Round each answer to the nearest tenth.
2. α=43.1,a=184.2,b=242.8\alpha=43.1^{\circ}, a=184.2, \mathbf{b}=242.8

Studdy Solution

STEP 1

What is this asking? We're given two sides of a triangle, a=184.2a = 184.2 and b=242.8b = 242.8, and an angle α=43.1\alpha = 43.1^\circ opposite to side aa.
We need to find the remaining angle β\beta opposite to side bb, the angle γ\gamma, and the side cc. Watch out! Remember, the Law of Sines can sometimes give two possible solutions for the angle, so we need to be careful and check for that!
Also, remember that the sum of the angles in a triangle is 180180^\circ.

STEP 2

1. Find angle β\beta
2. Find angle γ\gamma
3. Find side cc

STEP 3

We're given aa, bb, and α\alpha, so we can use the Law of Sines to find β\beta.
The Law of Sines says sin(α)a=sin(β)b\frac{\sin(\alpha)}{a} = \frac{\sin(\beta)}{b}.
Let's **plug in** our known values: sin(43.1)184.2=sin(β)242.8\frac{\sin(43.1^\circ)}{184.2} = \frac{\sin(\beta)}{242.8}.

STEP 4

To **isolate** sin(β)\sin(\beta), we **multiply** both sides by 242.8242.8: sin(β)=sin(43.1)184.2242.8\sin(\beta) = \frac{\sin(43.1^\circ)}{184.2} \cdot 242.8.

STEP 5

sin(β)0.6833184.2242.80.00371242.80.9006\sin(\beta) \approx \frac{0.6833}{184.2} \cdot 242.8 \approx 0.00371 \cdot 242.8 \approx 0.9006.

STEP 6

Now, we take the **inverse sine** to find β\beta: β=arcsin(0.9006)64.2\beta = \arcsin(0.9006) \approx 64.2^\circ.
Since sin(β)\sin(\beta) is positive, β\beta could also be in the second quadrant.
So, another possible value for β\beta is 18064.2=115.8180^\circ - 64.2^\circ = 115.8^\circ.
Let's consider both possibilities!

STEP 7

Since the sum of angles in a triangle is 180180^\circ, we have γ=180αβ18043.164.272.7\gamma = 180^\circ - \alpha - \beta \approx 180^\circ - 43.1^\circ - 64.2^\circ \approx 72.7^\circ.

STEP 8

Similarly, γ=180αβ18043.1115.821.1\gamma = 180^\circ - \alpha - \beta \approx 180^\circ - 43.1^\circ - 115.8^\circ \approx 21.1^\circ.

STEP 9

Using the Law of Sines again, sin(α)a=sin(γ)c\frac{\sin(\alpha)}{a} = \frac{\sin(\gamma)}{c}, so sin(43.1)184.2=sin(72.7)c\frac{\sin(43.1^\circ)}{184.2} = \frac{\sin(72.7^\circ)}{c}.

STEP 10

c=184.2sin(72.7)sin(43.1)184.20.95510.6833175.90.6833257.4c = \frac{184.2 \cdot \sin(72.7^\circ)}{\sin(43.1^\circ)} \approx \frac{184.2 \cdot 0.9551}{0.6833} \approx \frac{175.9}{0.6833} \approx 257.4.

STEP 11

Using the Law of Sines again, sin(43.1)184.2=sin(21.1)c\frac{\sin(43.1^\circ)}{184.2} = \frac{\sin(21.1^\circ)}{c}.

STEP 12

c=184.2sin(21.1)sin(43.1)184.20.35970.683366.260.683397.0c = \frac{184.2 \cdot \sin(21.1^\circ)}{\sin(43.1^\circ)} \approx \frac{184.2 \cdot 0.3597}{0.6833} \approx \frac{66.26}{0.6833} \approx 97.0.

STEP 13

We found two possible solutions for the triangle:
1. β64.2\beta \approx 64.2^\circ, γ72.7\gamma \approx 72.7^\circ, and c257.4c \approx 257.4
2. β115.8\beta \approx 115.8^\circ, γ21.1\gamma \approx 21.1^\circ, and c97.0c \approx 97.0

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