Answer:
∠B = 19°
Step-by-step explanation:
Given : In triangle ABC, if ∠A = 120°, a = 8, and b = 3
We have to find the measure of B that is ∠B
Consider the given triangle ABC,
Using Sine rule ,
For a triangle with measure of angle A, B and C and side a faces angle A,
side b faces angle B and side c faces angle C
[tex]\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}[/tex]
we have, a = 8 , b = 3 and ∠A = 120°
Consider first two ratios,
[tex]\frac{8}{\sin 120^{\circ}}=\frac{3}{\sin B}[/tex]
Solve for B, we have,
[tex]\sin \left(120^{\circ \:}\right)=\frac{\sqrt{3}}{2}[/tex]
[tex]\frac{8}{\frac{\sqrt{3}}{2}}=\frac{3}{\sin \left(B\right)}[/tex]
[tex]\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c[/tex]
[tex]8\sin \left(B\right)=\frac{\sqrt{3}}{2}\cdot \:3[/tex]
Simplify, we have,
[tex]\sin \left(B\right)=\frac{3\sqrt{3}}{16}[/tex]
Taking sine inverse both side, we have,
[tex]B=\sin^{-1}\left(\frac{3\sqrt{3}}{16}\right)[/tex]
We have, [tex]B=18.95^{\circ \:}[/tex]
Thus, ∠B = 19°
