Question

solve each triangle abc that exists.
b = 34°54 a = 38.6 b = 31.1
select the correct choice below and, if necessary, fill in the answer boxes within the choice.
oa. there is only one possible solution for the triangle.
the measurements for the remaining angles a and c and side c are as follows.
a =
c =
c =
(simplify your answer. round to the nearest (round to the nearest tenth
degree as needed. round to the nearest minute as needed.)
ob. there are two possible solutions for the triangle.
the measurements for the solution with the longer side c are as follows.
a₁ =
c₁ =
c₁ =
(simplify your answer. round to the nearest (round to the nearest tenth
degree as needed. round to the nearest minute as needed.)
the measurements for the solution with the shorter side c are as follows.
a₂ =
c₂ =
c₂ =
(simplify your answer. round to the nearest (round to the nearest tenth
degree as needed. round to the nearest minute as needed.)
oc. there are no possible solutions for this triangle.

Explanation:

Step1: Convert \(B = 34^{\circ}54'\) to decimal degrees

Since \(1^{\circ}=60'\), then \(54'=\frac{54}{60}=0.9^{\circ}\), so \(B = 34.9^{\circ}\)

Step2: Use the Law of Sines \(\frac{a}{\sin A}=\frac{b}{\sin B}\) to find \(\sin A\)

\(\sin A=\frac{a\sin B}{b}\)
Substitute \(a = 38.6\), \(b = 31.1\), \(B = 34.9^{\circ}\)
\(\sin A=\frac{38.6\times\sin(34.9^{\circ})}{31.1}\)
\(\sin(34.9^{\circ})\approx0.572\)
\(\sin A=\frac{38.6\times0.572}{31.1}\approx\frac{22.0792}{31.1}\approx0.7099\)
\(A=\sin^{- 1}(0.7099)\approx45.2^{\circ}\) or \(A = 180^{\circ}-45.2^{\circ}=134.8^{\circ}\)

Step3: Check for the first - case (\(A_1\approx45.2^{\circ}\))

Convert \(A_1 = 45.2^{\circ}\) to degrees and minutes: \(0.2^{\circ}\times60 = 12'\), so \(A_1=45^{\circ}12'\)
\(C_1=180^{\circ}-A_1 - B\)
\(C_1=180^{\circ}-45^{\circ}12'-34^{\circ}54'\)
\(C_1 = 99^{\circ}54'\) (since \(180^{\circ}=179^{\circ}60'\), \(179^{\circ}60'-45^{\circ}12'-34^{\circ}54'=(179 - 45-34)^{\circ}(60 - 12 - 54)'=99^{\circ}(- 6)'=99^{\circ}54'\))
Use the Law of Sines \(\frac{c_1}{\sin C_1}=\frac{b}{\sin B}\)
\(c_1=\frac{b\sin C_1}{\sin B}\)
\(\sin C_1=\sin(99^{\circ}54')=\sin(99.9^{\circ})\approx0.987\), \(\sin B=\sin(34.9^{\circ})\approx0.572\)
\(c_1=\frac{31.1\times0.987}{0.572}\approx\frac{30.6957}{0.572}\approx53.7\)

Step4: Check for the second - case (\(A_2 = 134.8^{\circ}\))

Convert \(A_2 = 134.8^{\circ}\) to degrees and minutes: \(0.8^{\circ}\times60 = 48'\), so \(A_2=134^{\circ}48'\)
\(C_2=180^{\circ}-A_2 - B\)
\(C_2=180^{\circ}-134^{\circ}48'-34^{\circ}54'\)
\(C_2=180^{\circ}-(134^{\circ}48'+34^{\circ}54')=180^{\circ}-169^{\circ}42'=10^{\circ}18'\)
Use the Law of Sines \(\frac{c_2}{\sin C_2}=\frac{b}{\sin B}\)
\(\sin C_2=\sin(10^{\circ}18')=\sin(10.3^{\circ})\approx0.179\)
\(c_2=\frac{31.1\times0.179}{0.572}\approx\frac{5.5669}{0.572}\approx9.7\)

Answer:

B. There are two possible solutions for the triangle.
The measurements for the solution with the longer side \(c\) are as follows.
\(A_1 = 45^{\circ}12'\), \(C_1 = 99^{\circ}54'\), \(c_1\approx53.7\)
The measurements for the solution with the shorter side \(c\) are as follows.
\(A_2 = 134^{\circ}48'\), \(C_2 = 10^{\circ}18'\), \(c_2\approx9.7\)