Answer: [tex]\frac{\sqrt{113}}{7}[/tex]
Step-by-step explanation:
As X is an acute angle, all 6 trigonometric functions with an argument of X are positive.
Using the identity [tex]1+\tan^{2} X=\sec^{2} X[/tex],
[tex]1+\left(\frac{8}{7} \right)^{2}=\sec^{2} X\\\\\sec^{2} X=\frac{113}{49}\\\\\therefore \sec X=\boxed{\frac{\sqrt{113}}{7}}[/tex]
