Question 1058387
Treat the segment like a hypotenuse of a right triangle.  (y coordinate)/(hypotenuse length) is the sine of your theta;  


{{{8/sqrt((6^2)^2+(8^2)^2)}}}-----simplify this.


{{{8/sqrt(36+64)}}}


{{{8/sqrt(100)}}}


{{{8/10=highlight(4/5)}}}--------the {{{sine(theta)}}}