check the picture below.
on the first quadrant, namely 0° < x < 90°, the sine of one acute angle, equals the cosine of the other acute angle, as you see in the example in the picture.
therefore
[tex]\bf sin(90^o-\theta )=cos(\theta )\qquad \boxed{\theta =42}\qquad sin(90^o-42^o )=cos(42^o )
\\\\\\
sin(48^o)=cos(42^o)\\\\
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sin(\stackrel{48^o}{x^o-20^o})=cos(42^o)
\\\\\\
x-20=48\implies x=48+20\implies x=68[/tex]
