Math  /  Geometry

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In the following exercise, two sides and an angle are given. First determine whether the information results in no triangle, one triangle, or two triangles. Solve the resulting triangle. a=23.2,b=25.4, and A=45.6a=23.2, b=25.4 \text {, and } A=45.6^{\circ}
Determine the value of sinB\boldsymbol{\operatorname { s i n }} \mathrm{B}. sinB=\sin B= \square (Round to four decimal places as needed.)
How many and what type of triangle does the given information produce? Use the rounded value of sinB\sin B. One or Two Oblique Triangles One Right Triangle No Triangle Time Remaining: 01:18:25 Next

Studdy Solution

STEP 1

What is this asking? We're given two sides of a triangle (aa and bb) and an angle (AA), and we need to figure out how many triangles can be formed with these measurements and solve for the missing parts. Watch out! Remember, the sine rule can sometimes give two possible solutions, so we need to be careful and check both!

STEP 2

1. Calculate sin(B)
2. Determine the number of triangles
3. Solve the triangle(s)

STEP 3

We're given aa, bb, and AA, and we want to find BB.
The *Law of Sines* connects these pieces!
It says: sinAa=sinBb \frac{\sin A}{a} = \frac{\sin B}{b}

STEP 4

Let's plug in what we know: a=23.2a = \textbf{23.2}, b=25.4b = \textbf{25.4}, and A=45.6A = \textbf{45.6}^{\circ}. sin45.623.2=sinB25.4 \frac{\sin 45.6^{\circ}}{23.2} = \frac{\sin B}{25.4}

STEP 5

To get sinB\sin B by itself, we'll multiply both sides of the equation by 25.425.4: sinB=25.4sin45.623.2 \sin B = \frac{25.4 \cdot \sin 45.6^{\circ}}{23.2}

STEP 6

Now, let's crunch the numbers! sinB25.40.714423.218.1437623.20.7821 \sin B \approx \frac{25.4 \cdot \textbf{0.7144}}{23.2} \approx \frac{\textbf{18.14376}}{23.2} \approx \textbf{0.7821} So, sinB0.7821\sin B \approx \textbf{0.7821}.

STEP 7

Since sinB\sin B is between 0 and 1 (specifically, sinB0.7821\sin B \approx \textbf{0.7821}), there *could* be one or two possible triangles.

STEP 8

We can find a possible value for angle BB by taking the inverse sine (arcsin) of our result: B=arcsin(0.7821)51.46 B = \arcsin(0.7821) \approx \textbf{51.46}^{\circ}

STEP 9

Since sin(x)=sin(180x)\sin(x) = \sin(180^{\circ} - x), there might be another possible angle BB: B2=18051.46=128.54 B_2 = 180^{\circ} - 51.46^{\circ} = \textbf{128.54}^{\circ}

STEP 10

Let's check if this second angle works.
We know A=45.6A = 45.6^{\circ}.
If we add this to B2B_2, we get 45.6+128.54=174.1445.6^{\circ} + 128.54^{\circ} = \textbf{174.14}^{\circ}.
Since this is less than 180180^{\circ}, it's possible to have a second triangle!

STEP 11

We have A=45.6A = \textbf{45.6}^{\circ}, B=51.46B = \textbf{51.46}^{\circ}, and we can find CC since the angles in a triangle add up to 180180^{\circ}: C=18045.651.46=82.94 C = 180^{\circ} - 45.6^{\circ} - 51.46^{\circ} = \textbf{82.94}^{\circ} We can use the Law of Sines again to find cc: sin45.623.2=sin82.94c \frac{\sin 45.6^{\circ}}{23.2} = \frac{\sin 82.94^{\circ}}{c} c=23.2sin82.94sin45.623.20.99290.714432.33 c = \frac{23.2 \cdot \sin 82.94^{\circ}}{\sin 45.6^{\circ}} \approx \frac{23.2 \cdot \textbf{0.9929}}{\textbf{0.7144}} \approx \textbf{32.33}

STEP 12

We have A=45.6A = \textbf{45.6}^{\circ}, B2=128.54B_2 = \textbf{128.54}^{\circ}, so: C2=18045.6128.54=5.86 C_2 = 180^{\circ} - 45.6^{\circ} - 128.54^{\circ} = \textbf{5.86}^{\circ} Using the Law of Sines: sin45.623.2=sin5.86c2 \frac{\sin 45.6^{\circ}}{23.2} = \frac{\sin 5.86^{\circ}}{c_2} c2=23.2sin5.86sin45.623.20.10220.71443.31 c_2 = \frac{23.2 \cdot \sin 5.86^{\circ}}{\sin 45.6^{\circ}} \approx \frac{23.2 \cdot \textbf{0.1022}}{\textbf{0.7144}} \approx \textbf{3.31}

STEP 13

sinB0.7821\sin B \approx \textbf{0.7821}.
The given information produces *two oblique triangles*.
Triangle 1: B51.46B \approx 51.46^{\circ}, C82.94C \approx 82.94^{\circ}, and c32.33c \approx 32.33.
Triangle 2: B2128.54B_2 \approx 128.54^{\circ}, C25.86C_2 \approx 5.86^{\circ}, and c23.31c_2 \approx 3.31.

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