[tex]\tan\theta=\dfrac78[/tex]
Recall that
[tex]1+\tan^2\theta=\sec^2\theta[/tex]
which means
[tex]\sec^2\theta=\dfrac{113}{64}\implies\cos^2\theta=\dfrac{64}{113}[/tex]
Given that [tex]0^\circ<\theta<90^\circ[/tex], we should expect [tex]\cos\theta>0[/tex], so we when we take the square root, we assume a positive sign. Then
[tex]\cos\theta=\sqrt{\dfrac{64}{113}}=\dfrac8{\sqrt{113}}[/tex]
