Moderate -0.8 This is a straightforward application of Pythagoras to find the opposite side from the given adjacent/hypotenuse ratio, then computing tan = opp/adj. It requires only basic trigonometric definitions and one calculation step, making it easier than average but not trivial since exact form (surds) is required.
M1 for use of \(\sin^2 \theta + \left(\frac{1}{3}\right)^2 = 1\)
M1 for \(\sin\theta = \frac{\sqrt{8}}{3}\) (ignore \(\pm\))
Answer
Marks
Guidance
Diag.: hypot = 3, one side = 1
M1
3rd side \(\sqrt{8}\)
M1
3
Question 8:
$\sqrt{8}$ or $2\sqrt{2}$ not $\pm\sqrt{8}$ | 3
M1 for use of $\sin^2 \theta + \left(\frac{1}{3}\right)^2 = 1$
M1 for $\sin\theta = \frac{\sqrt{8}}{3}$ (ignore $\pm$)
Diag.: hypot = 3, one side = 1 | M1
3rd side $\sqrt{8}$ | M1 | 3