\(\cos^2 \theta + \frac{1}{8} = \sin^2 \theta\)..........(i) from trigometric ratios for an acute angle, where \(cos^{2} \theta + \sin^2 \theta = 1\) ........(ii) Substitute for equation (i) in (i) = \(\cos^2 \theta + \frac{1}{8} = 1 - \cos^2 \theta \) = \(\cos^2 \theta + \cos^2 \theta = 1 - \frac{1}{8}\) \(2\cos^2 \theta = \frac{7}{8}\) \(\cos^2 \theta = \frac{7}{2 \times 8}\) \(\frac{7}{16} = \cos \theta\) \(\sqrt{\frac{7}{16}}\) = \(\frac{\sqrt{7}}{4}\) but cos \(\theta\) = \(\frac{\text{adj}}{\text{hyp}}\) \(opp^2 = hyp^2 - adj^2\) \(opp^2 = 4^2 - (\sqrt{7})^{2}\) = 16 - 7 opp = \(\sqrt{9}\) = 3 \(\tan \theta = \frac{\text{opp}}{\text{adj}}\) = \(\frac{3}{\sqrt{7}}\) \(\frac{3}{\sqrt{7}}\) x \(\frac{\sqrt{7}}{\sqrt{7}}\) = \(\frac{3\sqrt{7}}{7}\)