Answer:
[tex]\sin (b^\circ)=\dfrac{3}{5}[/tex].
Step-by-step explanation:
It is given that,
[tex]\tan (b^\circ)=\dfrac{3}{4}[/tex]
[tex]\cos (b^\circ)=\dfrac{4}{5}[/tex]
If a figure is dilated, then the image is similar to the figure. It means the corresponding angles of figure and image are congruent.
So, the value of sin(b°) after dilation is equal to the value of sin(b°) before dilation.
We know that,
[tex]\dfrac{\sin \theta}{\cos \theta}=\tan \theta[/tex]
[tex]\dfrac{\sin (b^\circ)}{\cos (b^\circ)}=\tan (b^\circ)[/tex]
[tex]\sin (b^\circ)=\tan (b^\circ)\times \cos (b^\circ)[/tex]
[tex]\sin (b^\circ)=\dfrac{3}{4}\times \dfrac{4}{5}[/tex]
[tex]\sin (b^\circ)=\dfrac{3}{5}[/tex]
Therefore, [tex]\sin (b^\circ)=\dfrac{3}{5}[/tex].
