Respuesta :

superkoopasirf
[tex]\sin{(A - B)} \equiv \sin{A}\cos{B} - \cos{A}\sin{B} \\\\ \therefore \sin{(90 - x)}\\ = \sin{(90)}\cos{(x)} - \cos{(90)}\sin{(x)} \\ = (1)\cos{(x)} - (0)\sin{(x)} \\ = \cos{x} - 0 \\ = \cos{x} \\ = \frac{1}{3} [/tex]
Julik
[tex]90а= \frac{ \pi }{2} \\ sin(90а-x)=sin (\frac{ \pi }{2}-x)=cos (x) \\ cos(x)= \frac{1}{3} \\ sin(x)= \frac{1}{3} \approx0.33333 \\ x=19а[/tex]