June 2017 Paper 2 Q8

8 The perimeter of a triangular flower bed, \(ABC\), is marked out using 27 metres of rope.

Not drawn accurately

Work out the size of angle \(BAC\). [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
(\(BC\) =) 12	B1	Allow as two 6s labelled on \(BC\) after perpendicular drawn from \(A\)
their \(12^2 = 7^2 + 8^2\) \(- 2 \times 7 \times 8 \times \cos A\) or \(144 = 49 + 64 - 112\cos A\) or \(144 = 113 - 112\cos A\) or \(\dfrac{7^2 + 8^2 - \text{their } 12^2}{2 \times 7 \times 8}\) or \(\dfrac{49 + 64 - 144}{112}\) or \(-\dfrac{31}{112}\) or [\(-0.277\), \(-0.27\)] or \(-0.28\)	M1	oe Do not allow if their 12 comes from use of Pythagoras’ theorem ie (\(BC\) =) \(\sqrt{7^2 + 8^2}\) or \(\sqrt{113}\) or 10.6(…) is B0M0
\(\cos^{-1}\left(\dfrac{7^2 + 8^2 - \text{their } 12^2}{2 \times 7 \times 8}\right)\)	M1dep	oe May be implied by final answer
[106, 106.1]	A1ft	Only ft B0M2
Alternative method 2
(\(BC\) =) 12	B1	Allow as two 6s labelled on \(BC\) after perpendicular drawn from \(A\)
(angle \(ABC\) =) \(\cos^{-1}\left(\dfrac{7^2 + \text{their } 12^2 - 8^2}{2 \times 7 \times \text{their } 12}\right)\) or 39.8… and \(\sin A = \dfrac{\sin \text{their } 39.8}{8} \times\) their 12 or \(\sin A = 0.96\ldots\)	M1	oe eg works out angle \(ACB\) (= 34.09(…) or 34.1) and uses sine rule Do not allow if their 12 comes from use of Pythagoras’ theorem ie (\(BC\) =) \(\sqrt{7^2 + 8^2}\) or \(\sqrt{113}\) or 10.6(…) is B0M0
\(180 - \sin^{-1}\)(their 0.96…)	M1dep	oe May be implied by final answer
[106, 106.1]	A1ft	Only ft B0M2

\(\cos^{-1}\) or \(\cos^{-1}\)ans does not score M1dep unless recovered

For the M1dep must have correct rearrangement but allow arithmetic errors

Answer outside range is A0
eg 106.2(…) from \(\cos^{-1}(-0.28)\)