Question

1 multiple choice 1 point two triangles can be formed using the given measurements. solve both triangles. c = 72°, a = 28, c = 27 a = 9.5°, b = 98.5°, b = 26; a = 170.5°, b = 81.5°, b = 26 a = 80.5°, b = 27.5°, b = 55.6; a = 99.5°, b = 8.5°, b = 55.6 a = 9.5°, b = 98.5°, b = 28.1; a = 170.5°, b = 81.5°, b = 28.1 a = 80.5°, b = 27.5°, b = 13.1; a = 99.5°, b = 8.5°, b = 4.2

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{c}{\sin C}\). Substitute \(a = 28\), \(c = 27\), and \(C=72^{\circ}\) into the formula: \(\sin A=\frac{a\sin C}{c}=\frac{28\sin72^{\circ}}{27}\).
Calculate \(\sin72^{\circ}\approx0.9511\), then \(\sin A=\frac{28\times0.9511}{27}\approx0.992\).
So \(A=\sin^{- 1}(0.992)\approx80.5^{\circ}\) or \(A = 180^{\circ}-80.5^{\circ}=99.5^{\circ}\) (since \(\sin\theta=\sin(180^{\circ}-\theta)\)).

Step2: Find angle \(B\)

For the first case when \(A = 80.5^{\circ}\), using \(A + B+C=180^{\circ}\), we have \(B=180^{\circ}-(A + C)=180^{\circ}-(80.5^{\circ}+72^{\circ})=27.5^{\circ}\).
For the second case when \(A = 99.5^{\circ}\), \(B=180^{\circ}-(99.5^{\circ}+72^{\circ})=8.5^{\circ}\).

Step3: Find side \(b\)

Using the Law of Sines \(\frac{b}{\sin B}=\frac{c}{\sin C}\).
When \(B = 27.5^{\circ}\), \(b=\frac{c\sin B}{\sin C}=\frac{27\sin27.5^{\circ}}{\sin72^{\circ}}\). Since \(\sin27.5^{\circ}\approx0.4617\) and \(\sin72^{\circ}\approx0.9511\), \(b=\frac{27\times0.4617}{0.9511}\approx13.1\).
When \(B = 8.5^{\circ}\), \(b=\frac{27\sin8.5^{\circ}}{\sin72^{\circ}}\). Since \(\sin8.5^{\circ}\approx0.1479\) and \(\sin72^{\circ}\approx0.9511\), \(b=\frac{27\times0.1479}{0.9511}\approx4.2\).

Answer:

\(A = 80.5^{\circ}, B = 27.5^{\circ}, b = 13.1; A = 99.5^{\circ}, B = 8.5^{\circ}, b = 4.2\) (the fourth option)