Question

determine \\( \angle b \\) to the nearest degree for the triangle with the given information. \\( a = 3.7 \mathrm{~m}, b = 4.9 \mathrm{~m}, \angle a = 46^{\circ} \\) a) \\( \angle b = 53^{\circ} \\) b) \\( \angle b = 72^{\circ} \\) c) \\( \angle b = 61^{\circ} \\) d) \\( \angle b = 59^{\circ} \\)

Explanation:

Step1: Apply the sine law

The sine law states that \(\frac{a}{\sin A}=\frac{b}{\sin B}\). Substituting the given values \(a = 3.7\), \(b = 4.9\), and \(A=46^{\circ}\), we get \(\frac{3.7}{\sin46^{\circ}}=\frac{4.9}{\sin B}\).

Step2: Solve for \(\sin B\)

Cross - multiply to obtain \(3.7\sin B=4.9\sin46^{\circ}\). Then \(\sin B=\frac{4.9\sin46^{\circ}}{3.7}\).
Calculate \(\sin46^{\circ}\approx0.7193\). So \(\sin B=\frac{4.9\times0.7193}{3.7}\approx\frac{3.5246}{3.7}\approx0.9526\).

Step3: Find \(B\)

Take the inverse sine of \(0.9526\), \(B=\sin^{- 1}(0.9526)\approx72^{\circ}\).

Answer:

B. \(\angle B = 72^{\circ}\)