Answer:
sin θ [tex]=\frac{3}{5}[/tex]
It is given that , Angle θ lies in the second quadrant.
In the right triangle, Perpendicular =3 units, and Hypotenuse = 5 Units
As, sinθ is positive in second Quadrant also.That is , sin(π-θ )=sin θ
Also, [tex]sin(\frac{\pi}{2}-\Theta)=cos\Theta[/tex]
[tex]sin (\Theta) = 3/5\\\\ (\Theta) =sin^{-1} \frac{3}{5} {\text{or}} (\Theta) =\pi -sin^{-1} \frac{3}{5}\\\\cos(\frac{\pi}{2}-\Theta)= \frac{3}{5}\\\\ \frac{\pi}{2}-\Theta=cos^{-1}[\frac{3}{5}]\\\\ \Theta= \frac{\pi}{2}-cos^{-1}[\frac{3}{5}][/tex]
So, [tex]\pi-\Theta= \frac{\pi}{2}+cos^{-1}\frac{3}{5}, {\text{or}}\pi -sin^{-1} \frac{3}{5}[/tex]
are values of theta which lies in second quadrant.
