Answer: 2
Step-by-step explanation:
[tex]\text{If the angle is in Quadrant II and sin A }=\dfrac{4}{5}, \text{then cos A }=-\dfrac{3}{5}.\\use\ Pythagorean\ Theorem: x^2+4^2=5^2\implies x=3,\quad cos =\dfrac{x}{h}\\\\\\tan\bigg(\dfrac{A}{2}\bigg)=\dfrac{1-cosA}{sinA}\\\\\\.\qquad \qquad =\dfrac{1-(-\dfrac{3}{5})}{\dfrac{4}{5}}\\\\\\.\qquad \qquad =\dfrac{\dfrac{8}{5}}{\dfrac{4}{5}}\implies \dfrac{8}{5}\div\dfrac{4}{5}\implies \dfrac{8}{5}\times\dfrac{5}{4}=\dfrac{8}{4}=\large\boxed{2}[/tex]
