The given expression is
[tex]\sin (\angle ABC+\mathring{60})[/tex]At first, we will use this rule to solve the question
[tex]\sin (A+B)=\sin A\cos B+\cos A\sin B[/tex]By using this rule with the given expression, then
[tex]\sin (\angle ABC+\mathring{60})=\sin \angle ABC\cos 60+\cos \angle ABC\sin 60[/tex]From the given triangle
[tex]\begin{gathered} \sin \angle ABC=\frac{AC}{AB} \\ \sin \angle ABC=\frac{4}{5} \\ \cos \angle ABC=\frac{BC}{AB} \\ \cos \angle ABC=\frac{3}{5} \end{gathered}[/tex]Since angle 60 is a special angle then
[tex]\begin{gathered} \sin 60=\frac{\sqrt[]{3}}{2} \\ \cos 60=\frac{1}{2} \end{gathered}[/tex]Let us substitute these values in the rule above
[tex]\sin (\angle ABC+60)=(\frac{4}{5})(\frac{1}{2})+(\frac{3}{5})(\frac{\sqrt[]{3}}{2})_{}[/tex]Simplify the right side
[tex]\sin (\angle ABC+60)=\frac{2}{5}+\frac{3}{10}\sqrt[]{3}[/tex]The correct answer is B
