SOLUTION
We need to get the slopes of the lines A and B.
Slope of A considering the points (-4, 3) and (0, 0) which is at the origin, we have
[tex]\begin{gathered} m=\frac{0-3}{0-(-4)} \\ m=\frac{-3}{4} \\ m=-\frac{3}{4} \end{gathered}[/tex]Since line B is a horizontal line, the slope is 0.
angle between two slopes is given as
[tex]\begin{gathered} tan\theta=|\frac{m_2-m_1}{1+m_1m_2}| \\ where\text{ }\theta\text{ is the angle between them and } \\ m_1\text{ and m}_2\text{ are slopes of the line } \end{gathered}[/tex]So, we have
[tex]\begin{gathered} tan\theta=|\frac{m_2-m_1}{1+m_1m_2}| \\ tan\theta=|\frac{0-(-\frac{3}{4})}{1+0(-\frac{3}{4})}| \\ tan\theta=|\frac{\frac{3}{4}}{1}| \\ tan\theta=|\frac{3}{4}| \\ tan\theta=\frac{3}{4} \\ \theta=tan^{-1}\frac{3}{4} \\ \theta=36.86989 \\ \theta=36.87\degree \end{gathered}[/tex]Hence the angle between A and B is 36.87 degrees
Area of a sector is given as
[tex]A=\frac{\theta}{360\degree}\times\pi r^2[/tex]Note that the radius r is the length of line B, which is 5 units. So, the area becomes
[tex]\begin{gathered} A=\frac{\theta}{360\degree}\times\pi r^2 \\ A=\frac{36.87}{360}\times\pi\times5^2 \\ A=8.04376\text{ units}^2 \end{gathered}[/tex]Hence the answer is 8.04 square units. The last option is the answer