Question

question 4 (mandatory) (1 point) 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 = 59^{\circ} \\) b) \\( \angle b = 61^{\circ} \\) c) \\( \angle b = 72^{\circ} \\) d) \\( \angle b = 53^{\circ} \\)

Explanation:

Step1: Apply the Law of Sines

The Law of Sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}$. We know $a = 3.7$, $b=4.9$, and $A = 46^{\circ}$. Plugging these values into the formula gives $\frac{3.7}{\sin46^{\circ}}=\frac{4.9}{\sin B}$.

Step2: Solve for $\sin B$

Cross - multiply to get $3.7\sin B=4.9\sin46^{\circ}$. 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 the angle $B$

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

Answer:

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