Answer:
∠B = 25.4°
∠C = 114.6°
c = 4.2
Step-by-step explanation:
To solve the triangle given a = 3, b = 2 and A = 40°, 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, and solve for angle B:
[tex]\dfrac{\sin 40^{\circ}}{3}=\dfrac{\sin B}{2}\\\\\\ \sin B=\dfrac{2\sin 40^{\circ}}{3}\\\\\\ B=\sin^{-1}\left(\dfrac{2\sin 40^{\circ}}{3}\right)\\\\\\B=25.373993939...^{\circ}\\\\\\B=25.4^{\circ}\; \sf (nearest\;tenth)[/tex]
According to the Angle Sum Property of a Triangle, the sum of the interior angles of a triangle is always equal to 180°. Therefore:
[tex]A+B+C=180^{\circ}\\\\40^{\circ}+25.373993939...^{\circ}+C=180^{\circ}\\\\65.373993939...^{\circ}+C=180^{\circ}\\\\C=180^{\circ}-65.373993939...^{\circ}\\\\C=114.6260060608...^{\circ}\\\\C=114.6^{\circ}[/tex]
Now, use the Law of Sines again to find the measure of side c:
[tex]\dfrac{\sin 40^{\circ}}{3}=\dfrac{\sin 114.6260060608...^{\circ}}{c}\\\\\\ c=\dfrac{3\sin 114.6260060608...^{\circ}}{\sin 40^{\circ}}\\\\\\ c=4.2426785555...\\\\\\c=4.2\; \sf (nearest\;tenth)[/tex]
Therefore, the solutions are:
[tex]\Large\boxed{\boxed{\sf \angle B = 25.4^{\circ}\quad \angle C = 114.6^{\circ}\quad c = 4.2}}[/tex]
