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 } \\)
\\( \bigcirc \\) a) \\( \angle b = 59 ^ { \circ } \\)
\\( \bigcirc \\) b) \\( \angle b = 53 ^ { \circ } \\)
\\( \bigcirc \\) c) \\( \angle b = 72 ^ { \circ } \\)
\\( \bigcirc \\) d) \\( \angle b = 61 ^ { \circ } \\)

Explanation:

Step1: Apply the Law of Sines

The Law of Sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}$. Substituting the given values: $\frac{3.7}{\sin 46^{\circ}}=\frac{4.9}{\sin B}$.

Step2: Solve for $\sin B$

Cross - multiply to get $3.7\sin B = 4.9\sin 46^{\circ}$. Then $\sin B=\frac{4.9\sin 46^{\circ}}{3.7}$.
Calculate $\sin 46^{\circ}\approx0.7193$. So $\sin B=\frac{4.9\times0.7193}{3.7}\approx\frac{3.5246}{3.7}\approx0.9526$.

Step3: Find $\angle B$

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

Answer:

C. $\angle B = 72^{\circ}$