HELP ASAP PLEASE!!!

cos 63°  = sin x

After switching sides:

\sin \left(x\right)=\cos \left(63^{\circ \:}\right)

Using \cos \left(x\right)=\sin \left(90^{\circ \:}-x\right)

We have:

\sin \left(x\right)=\sin \left(90^{\circ \:}-63^{\circ \:}\right)

So x = 180^{\circ \:}-27^{\circ \:}+360^{\circ \:}n,\:x=360^{\circ \:}n+27^{\circ \:}\(180^{\circ \:}-27^{\circ \:}+360^{\circ \:}n,\:x=360^{\circ \:}n+27^{\circ \:}\)

Oops, the LaTeX didn't show up :(

I'n just going to tell you the answer so I don't have to retype the whole thing :)

x = 180° - 27° + 360° n, x = 360° n + 27°

We jhave the identity....

cos (A)  = sin (90- A)

So

cos (63)  = sin (90- 63)

cos (63)  = sin (27).....so....x  = 27  (degrees)