Original expresion
[tex]\sin (6A)=\cos (9A)[/tex]
Let's use the following relationship to determine the answer
[tex]\sin 6A=\cos (90-6A)[/tex]
Now in the original expression
[tex]\begin{gathered} \cos (90-6A)=\cos 9A \\ 90-6A=9A \\ 90=6A+9A \\ 15A=90 \\ A=\frac{90}{15} \\ A=6 \end{gathered}[/tex]
The answer would be A = 6
