Math  /  Trigonometry

QuestionSolve each triangle ABCA B C that exists. A=40.5a=8.8mb=10.7mA=40.5^{\circ} \quad a=8.8 m \quad b=10.7 m
Select the correct choice below and, if necessary, fill in the answer boxes within the choice. A. There is only one possible solution for the triangle.
The measurements for the remaining angles B and C and side C are as follows. B=\mathrm{B}=\square^{\circ} C=\mathrm{C}= \square c=c= \square (Round to the nearest (Round to the nearest (Round to the nearest tenth tenth as needed.) tenth as needed.) as needed.) B. There are two possible solutions for the triangle. The measurements for the solution with the longer side c are as follows. B1=B_{1}= \square (Round to the nearest C1=\mathrm{C}_{1}= \square c1=c_{1}= \square tenth as needed.) (Round to the nearest (Round to the nearest tenth The measurements for the tenth as needed.) as needed.) B2=B_{2}= \square C2=\mathrm{C}_{2}= \square (Round to the nearest (Round to the nearest tenth as needed.) tenth as needed.) c2=c_{2}=\square (Round to the nearest tenth as needed.) C. There are no possible solutions for this triangle.

Studdy Solution

STEP 1

What is this asking? Given an angle A and sides aa and bb of a triangle, we need to find the remaining angle B, angle C, and side cc, and figure out if there's one, two, or no solutions! Watch out! Remember the ambiguous case of the sine rule!
When we're given two sides and an angle opposite one of them (SSA), there might be two possible triangles, one, or even none!

STEP 2

1. Apply the sine rule to find angle B.
2. Determine the number of solutions.
3. Calculate the remaining values for each solution.

STEP 3

We know the sine rule states that sin(A)a=sin(B)b\frac{\sin(A)}{a} = \frac{\sin(B)}{b}.
We have A=40.5A = 40.5^{\circ}, a=8.8ma = 8.8 m, and b=10.7mb = 10.7 m.
Let's plug these values in!

STEP 4

sin(40.5)8.8=sin(B)10.7\frac{\sin(40.5^{\circ})}{8.8} = \frac{\sin(B)}{10.7} Multiplying both sides by **10.7** to isolate sin(B)\sin(B), we get: sin(B)=10.7sin(40.5)8.8\sin(B) = \frac{10.7 \cdot \sin(40.5^{\circ})}{8.8} sin(B)10.70.64948.87.08.80.7886\sin(B) \approx \frac{10.7 \cdot 0.6494}{8.8} \approx \frac{7.0}{8.8} \approx 0.7886

STEP 5

Taking the inverse sine of both sides gives us B=arcsin(0.7886)52.1B = \arcsin(0.7886) \approx 52.1^{\circ}.
This is our **principal value** for angle B.

STEP 6

Since sin(B)\sin(B) was positive, there *might* be two possible solutions for BB.
The other possible value for BB (let's call it B2B_2) is the supplement of the first one: B2=180B18052.1=127.9B_2 = 180^{\circ} - B \approx 180^{\circ} - 52.1^{\circ} = 127.9^{\circ}.

STEP 7

We need to check if both BB and B2B_2 lead to valid triangles.
For a triangle to exist, the sum of its angles must be 180180^{\circ}.
With A=40.5A = 40.5^{\circ} and B52.1B \approx 52.1^{\circ}, we have A+B40.5+52.1=92.6<180A + B \approx 40.5^{\circ} + 52.1^{\circ} = 92.6^{\circ} < 180^{\circ}, so this solution is valid.
With A=40.5A = 40.5^{\circ} and B2127.9B_2 \approx 127.9^{\circ}, we have A+B240.5+127.9=168.4<180A + B_2 \approx 40.5^{\circ} + 127.9^{\circ} = 168.4^{\circ} < 180^{\circ}, so this solution is also valid.
We have **two possible solutions**!

STEP 8

C1=180AB18040.552.1=87.4C_1 = 180^{\circ} - A - B \approx 180^{\circ} - 40.5^{\circ} - 52.1^{\circ} = 87.4^{\circ}. Now, using the sine rule again: c1sin(C1)=asin(A)\frac{c_1}{\sin(C_1)} = \frac{a}{\sin(A)}, so c1=asin(C1)sin(A)8.8sin(87.4)sin(40.5)8.80.9990.649413.5mc_1 = \frac{a \cdot \sin(C_1)}{\sin(A)} \approx \frac{8.8 \cdot \sin(87.4^{\circ})}{\sin(40.5^{\circ})} \approx \frac{8.8 \cdot 0.999}{0.6494} \approx 13.5 m.

STEP 9

C2=180AB218040.5127.9=11.6C_2 = 180^{\circ} - A - B_2 \approx 180^{\circ} - 40.5^{\circ} - 127.9^{\circ} = 11.6^{\circ}. Using the sine rule: c2=asin(C2)sin(A)8.8sin(11.6)sin(40.5)8.80.2010.64942.7mc_2 = \frac{a \cdot \sin(C_2)}{\sin(A)} \approx \frac{8.8 \cdot \sin(11.6^{\circ})}{\sin(40.5^{\circ})} \approx \frac{8.8 \cdot 0.201}{0.6494} \approx 2.7 m.

STEP 10

There are two possible solutions!
Solution 1: B152.1B_1 \approx 52.1^{\circ}, C187.4C_1 \approx 87.4^{\circ}, c113.5mc_1 \approx 13.5 m
Solution 2: B2127.9B_2 \approx 127.9^{\circ}, C211.6C_2 \approx 11.6^{\circ}, c22.7mc_2 \approx 2.7 m

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