June 2016 Paper 1 Q16
16 Angle \(\theta\) is obtuse and \(\sin\theta = \dfrac{\sqrt{11}}{6}\)
Work out the value of \(\cos\theta\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| Sketch of right-angled triangle with \(\sqrt{11}\) on ‘opposite’ and 6 on hypotenuse | M1 | accept O = \(\sqrt{11}\) and H = 6 without a sketch for M1 |
| Adjacent = \(\sqrt{(6^2 - (\sqrt{11})^2)}\) | M1 | |
| Adjacent = 5 | A1 | |
| \(-\frac{5}{6}\) | A1 | |
| Alternative method 2 | ||
| (\(\sin^2\theta\) =) \(\dfrac{11}{36}\) | B1 | |
| their \(\frac{11}{36} + \cos^2\theta = 1\) | M1 | oe |
| \(\cos^2\theta = \frac{25}{36}\) | A1 | |
| \(-\frac{5}{6}\) | A1 | |
Additional guidance
An answer of \(\frac{5}{6}\) implies M1 M1 A1 A0 ... unless from clearly incorrect working.
In alt 2, their \(\frac{11}{36}\) must not involve a trig function
eg \(\sin^2\left(\frac{11}{36}\right) + \cos^2\theta = 1\) scores B1 M0 A0 A0