June 2017 Paper 1 Q16
16 Here is a triangle.
Not drawn accurately
\(\sin x^\circ = \dfrac{1}{\sqrt{12}}\)
Work out the value of \(y\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sin 120^\circ = \dfrac{\sqrt{3}}{2}\) | B1 | |
| \(\dfrac{y}{\sin 120^\circ} = \dfrac{6}{\sin x^\circ}\) or \(\dfrac{y}{\sin 120^\circ} = \dfrac{6}{1/\sqrt{12}}\) | M1 | oe |
| \(y = 6\sqrt{12} \times \sin 120^\circ\) or \(y = 12\sqrt{3} \times \sin 120^\circ\) | M1dep | dep on previous M mark earned |
| 18 | A1 |
Additional guidance
They might use a wrong value for \(\sin 120^\circ\) ... eg \(\sin 120^\circ = \frac{1}{2}\)
then write eg \(\dfrac{y}{1/2} = \dfrac{6}{1/\sqrt{12}}\) followed by \(y = 6\sqrt{12} \times \frac{1}{2}\), this scores M1 M1
because their method is correct ... they will already have lost the B1 mark and will be unable to score the A1 mark
Do not condone the use of 120 instead of \(\sin 120^\circ\)