Question

19 multiple choice 1 point two triangles can be formed using the given measurements. solve both triangles. b = 54°, a = 14, b = 13 a = 29.4°, c = 96.6°, c = 16; a = 150.6°, c = 83.4°, c = 16 a = 60.6°, c = 65.4°, c = 14.6; a = 119.4°, c = 6.6°, c = 1.8 a = 60.6°, c = 65.4°, c = 11.6; a = 119.4°, c = 6.6°, c = 11.6 a = 29.4°, c = 96.6°, c = 10.6; a = 150.6°, c = 83.4°, c = 10.6

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). We know \(B = 54^{\circ}\), \(a = 14\), and \(b = 13\). So, \(\frac{14}{\sin A}=\frac{13}{\sin 54^{\circ}}\). Then \(\sin A=\frac{14\sin 54^{\circ}}{13}\).

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Since \(\sin A\approx0.8712\), \(A=\sin^{- 1}(0.8712)\approx60.6^{\circ}\) or \(A = 180^{\circ}-60.6^{\circ}=119.4^{\circ}\) (because \(\sin(A)=\sin(180 - A)\) in the range \(0^{\circ}

Step2: Find angle \(C\) for each case

Case 1: When \(A = 60.6^{\circ}\), using \(A + B+C=180^{\circ}\), then \(C=180^{\circ}-(A + B)=180^{\circ}-(60.6^{\circ}+54^{\circ})=65.4^{\circ}\).
Case 2: When \(A = 119.4^{\circ}\), then \(C=180^{\circ}-(119.4^{\circ}+54^{\circ})=6.6^{\circ}\).

Step3: Find side \(c\) for each case

Using the Law of Sines \(\frac{c}{\sin C}=\frac{b}{\sin B}\).
Case 1: When \(C = 65.4^{\circ}\), \(c=\frac{b\sin C}{\sin B}=\frac{13\sin65.4^{\circ}}{\sin54^{\circ}}\).

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Case 2: When \(C = 6.6^{\circ}\), \(c=\frac{13\sin6.6^{\circ}}{\sin54^{\circ}}\).

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Answer:

A = 60.6°, C = 65.4°, c = 14.6; A = 119.4°, C = 6.6°, c = 1.8