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 = 72^{circ} \\) b) \\( \angle b = 61^{circ} \\) c) \\( \angle b = 59^{circ} \\) d) \\( \angle b = 53^{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: $B = \sin^{-1}(0.9526)\approx72^{\circ}$.

Answer:

a) $\angle B = 72^{\circ}$