Question

question 10 (5 points) determine whether \\( \triangle a b c \\) should be solved by using the law of sines or the law of cosines. then solve the triangle. \\( a = 35 ^ { \circ }, b = 13, a = 11 \\) law of sines; \\( b \approx 42.7 ^ { \circ }, c \approx 102.3 ^ { \circ }, c \approx 18.7 \\) law of sines; \\( b \approx 102.3 ^ { \circ }, c \approx 42.7 ^ { \circ }, c \approx 18.7 \\) law of cosines; \\( b \approx 106.2 ^ { \circ }, c \approx 38.8 ^ { \circ }, c \approx 18.7 \\) law of cosines; \\( b \approx 38.8 ^ { \circ }, c \approx 106.2 ^ { \circ }, c \approx 18.7 \\)

Explanation:

Step1: Determine the law to use

We are given two sides \(a = 11\), \(b=13\) and an angle \(A = 35^{\circ}\). This is the SSA (side - side - angle) case. The Law of Sines is used for the SSA case. The Law of Sines formula is \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\).

Step2: Find angle \(B\)

Using \(\frac{a}{\sin A}=\frac{b}{\sin B}\), we substitute \(a = 11\), \(A = 35^{\circ}\), \(b = 13\).
\(\sin B=\frac{b\sin A}{a}=\frac{13\sin35^{\circ}}{11}\)
\(\sin35^{\circ}\approx0.5736\), so \(\sin B=\frac{13\times0.5736}{11}=\frac{7.4568}{11}\approx0.6779\)
\(B=\sin^{- 1}(0.6779)\approx42.7^{\circ}\) or \(B = 180^{\circ}-42.7^{\circ}=137.3^{\circ}\). But if \(B = 137.3^{\circ}\), then \(A + B=35^{\circ}+137.3^{\circ}=172.3^{\circ}\), and \(C=180-(A + B)=7.7^{\circ}\). Using the Law of Sines \(\frac{c}{\sin C}=\frac{a}{\sin A}\), \(c=\frac{a\sin C}{\sin A}\approx\frac{11\times\sin7.7^{\circ}}{\sin35^{\circ}}\approx2.5\) (not in options). So we take \(B\approx42.7^{\circ}\)

Step3: Find angle \(C\)

Since \(A + B + C=180^{\circ}\), \(C=180-(A + B)\)
\(C=180-(35 + 42.7)=102.3^{\circ}\)

Step4: Find side \(c\)

Using \(\frac{a}{\sin A}=\frac{c}{\sin C}\), \(c=\frac{a\sin C}{\sin A}\)
\(\sin C=\sin102.3^{\circ}\approx0.9763\), \(\sin A=\sin35^{\circ}\approx0.5736\)
\(c=\frac{11\times0.9763}{0.5736}\approx18.7\)

Answer:

Law of Sines; \(B\approx42.7^{\circ}\), \(C\approx102.3^{\circ}\), \(c\approx18.7\)