According to the given problem,
[tex]\cos (2x)^{\circ}=\sin (8x+5)^{\circ}[/tex]
Consider the formula,
[tex]\sin (90-\theta)=\cos \theta[/tex]
Apply the formula,
[tex]\sin (90-2x)=\sin (8x+5)[/tex]
Comparing both sides,
[tex]\begin{gathered} 90-2x=8x+5 \\ 8x+2x=90-5 \\ 10x=85 \\ x=\frac{85}{10} \\ x=8.5 \end{gathered}[/tex]
Obtain the value of the two angles,
[tex]\begin{gathered} 2x=2(8.5)=17 \\ 8x+5=8(8.5)+5=73 \end{gathered}[/tex]
It is evident that the smaller angle is 17 degrees, and the larger angle is 73 degrees.
Thus, the required value of the smaller acute angle of the triangle is 17 degrees.
