Question

If $x$ is acute and $\sin (x+{10}^0)=\cos (3x-{68}^o)$, find $x$ in degrees

Answer 1

$\sin (x+{10}^o)=\cos (3x-{68}^o)$

$=\sin ({90}^o-3x+{68}^o)$

$=\sin ({158}^o-3x)$

$x+{10}^o=158-3x$

$4x=148$

$x={37}^o$