Question

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

Answer 1

We have,

$\sin (x+{10}^0)=\cos (3x-{68}^0)$

$\Rightarrow \sin (x+{10}^0)=sin[{90}^0-(3x-{68}^0)]$

$\Rightarrow x+{10}^0={90}^0-(3x-{68}^0)$

$\Rightarrow x+{10}^0={90}^0-3x+{68}^0$

$\Rightarrow x+3x={158}^0-{10}^0$

$\Rightarrow 4x={148}^0$

$\Rightarrow x=\frac{{148}^0}{4}$

$\Rightarrow x={37}^0$

Hence, this is the answer.