Point B has coordinates (3,-4) and lies on the circle. Draw the perpendiculars from point B to the x-axis and y-axis. Denote the points of intersection with x-axis A and with y-axis C. Consider the right triangle ABO (O is the origin), by tha conditions data: AB=4 and AO=3, then by Pythagorean theorem:
[tex]BO^2=AO^2+AB^2 \\ BO^2=3^2+4^2 \\ BO^2=9+16 \\ BO^2=25 \\ BO=5[/tex].
{Note, that BO is a radius of circle and it wasn't necessarily to use Pythagorean theorem to find BO}
The sine of the angle BOA is
[tex]\sin \angle BOA= \dfrac{AB}{BO} = \dfrac{4}{5} =0.8[/tex]
Since point B is placed in the IV quadrant, the sine of the angle that isĀ drawn in a standard position with its terminal ray will be
[tex]\sin \theta=-0.8[/tex] .