Question

1 multiple choice 1 point use the law of sines. find b to the nearest tenth. 62.0 42.6 27.9 99.2

Explanation:

Step1: Find angle \(C\)

The sum of angles in a triangle is \(180^{\circ}\). So \(C = 180^{\circ}-(54^{\circ}+32^{\circ})=94^{\circ}\).

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Here, we know \(c = 65\), \(C = 94^{\circ}\), and \(A=32^{\circ}\). We want to find \(b\). Using \(\frac{b}{\sin B}=\frac{c}{\sin C}\), substituting \(B = 54^{\circ}\), \(c = 65\), \(C = 94^{\circ}\). So \(b=\frac{65\times\sin(54^{\circ})}{\sin(94^{\circ})}\).
We know that \(\sin(54^{\circ})\approx0.809\) and \(\sin(94^{\circ})\approx0.997\). Then \(b=\frac{65\times0.809}{0.997}=\frac{52.585}{0.997}\approx52.8\) (This seems wrong, let's use the correct pair: using \(\frac{c}{\sin C}=\frac{b}{\sin B}\) is wrong, should use \(\frac{c}{\sin C}=\frac{a}{\sin A}\), no, wait the side - angle pairs: side \(BC = 65\) (opposite \(A = 32^{\circ}\)), side \(AC=b\) (opposite \(B = 54^{\circ}\)). So \(\frac{b}{\sin B}=\frac{BC}{\sin A}\).
So \(b=\frac{65\times\sin(54^{\circ})}{\sin(32^{\circ})}\). Since \(\sin(54^{\circ})\approx0.809\), \(\sin(32^{\circ})\approx0.5299\). Then \(b=\frac{65\times0.809}{0.5299}=\frac{52.585}{0.5299}\approx99.2\).

Answer:

\(99.2\)