Answer:
c.68
Step-by-step explanation:
We are given that
\(sin(x-20)=cos 42^{\circ}\) Where 0 <x< 90
We have to solve for x
\(sin(x-20)=cos (90-48)\)
\(sin(x-20)=sin 48\)
\(cos (90-\theta)=sin\theta\)
Cancel sin on both sides then we get
\(x-20=48\)
\(x=48+20=68\)
Hence, the value of x=68
Answer:c.68
