Answer:
[tex]\sin \left(B\right)=\frac{3\sqrt{3}}{8}[/tex]
Step-by-step explanation:
Given : In triangle ABC, c = 8, b = 6, and ∠C = 60°
We have to find the value of [tex]\sin B[/tex]
Consider the given triangle ABC,
LAWS OF SINE : States that for a given triangle AbC, with a, b, c and Side a faces angle A,side b faces angle B and side c faces angle C.
We have,
[tex]\frac{a}{\sin A}=\frac{b}{\sin B}= \frac{c}{\sin C}[/tex]
Consider the last two ratios, we have,
[tex]\frac{b}{\sin B}= \frac{c}{\sin C}[/tex]
Substitute, the value , we have,
c = 8, b = 6, and ∠C = 60°
[tex]\frac{6}{\sin B}= \frac{8}{\sin 60^{\circ}}[/tex]
Put [tex]\sin \left(60^{\circ \:}\right)=\frac{\sqrt{3}}{2}[/tex]
we have,
[tex]\frac{6}{\sin \left(B\right)}=\frac{8}{\frac{\sqrt{3}}{2}}[/tex]
Simplify, we have,
[tex]\sin \left(B\right)=\frac{3\sqrt{3}}{8}[/tex]
