Answer:
1) m∠B = 49°
2) m∠B = 26.5°
Step-by-step explanation:
In triangle ABC, we are given that m∠A = 29.7°, b = 41.5, and a = 27.2.
To find m∠B, we can use the Law of Sines:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Law of Sines}} \\\\\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\\\\\textsf{where:}\\\phantom{ww}\bullet \;\textsf{$A, B$ and $C$ are the angles.}\\\phantom{ww}\bullet\;\textsf{$a, b$ and $c$ are the sides opposite the angles.}\end{array}}[/tex]
Substitute the given values into the formula:
[tex]\dfrac{\sin 29.7^{\circ}}{27.2}=\dfrac{\sin B}{41.5}[/tex]
Solve for B:
[tex]\sin B=\dfrac{41.5\sin 29.7^{\circ}}{27.2}[/tex]
[tex]B=\sin^{-1}\left(\dfrac{41.5\sin 29.7^{\circ}}{27.2}\right)[/tex]
[tex]B=49.107465791...^{\circ}[/tex]
[tex]B=49^{\circ}\;\textsf{(nearest\;degree)}[/tex]
Therefore, the measure of angle B is 49° (rounded to the nearest degree).
[tex]\hrulefill[/tex]
In triangle ABC, we are given that C = 41°20′, b = 25.9, and c = 38.4.
To find m∠B, we can use the Law of Sines:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Law of Sines}} \\\\\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\\\\\textsf{where:}\\\phantom{ww}\bullet \;\textsf{$A, B$ and $C$ are the angles.}\\\phantom{ww}\bullet\;\textsf{$a, b$ and $c$ are the sides opposite the angles.}\end{array}}[/tex]
Begin by converting the measure of angle C into decimal degrees.
[tex]C=41^{\circ}20'=41^{\circ}+\left(\dfrac{20}{60}\right)^{\circ}=41.333...^{\circ}[/tex]
Now, substitute the given values into the formula:
[tex]\dfrac{\sin B}{25.9}=\dfrac{\sin 41.333...^{\circ}}{38.4}[/tex]
Solve for B:
[tex]\sin B=\dfrac{25.9\sin 41.333...^{\circ}}{38.4}[/tex]
[tex]B=\sin^{-1}\left(\dfrac{25.9\sin 41.333...^{\circ}}{38.4}\right)[/tex]
[tex]B=26.4522671...^{\circ}[/tex]
[tex]B=26.5^{\circ}\;\textsf{(nearest\;tenth)}[/tex]
Therefore, the measure of angle B is 26.5° (rounded to the nearest degree).
