Higher June 2019 Paper 3 Q27
27 In this question, all lengths are in centimetres.
\(A\) is a point on a circle, centre \(O\).
\(B\) is a point on a different circle, centre \(O\).
\(AB = 20\)
Not drawn accurately
The equation of the larger circle is \(\quad x^2 + y^2 = 144\)
radius of smaller circle : radius of larger circle = 4 : 5
Work out the size of angle \(AOB\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sqrt{144}\) or 12 | B1 | radius of larger circle may be seen on diagram |
| \(\dfrac{4}{5} \times\) their 12 or 9.6 | M1 | their 12 must be a value may be seen on diagram |
| \((\cos AOB =)\) \(\dfrac{\text{their } 12^2 + \text{their } 9.6^2 - 20^2}{2 \times \text{their } 12 \times \text{their } 9.6}\) or \(\dfrac{144 + 92.16 - 400}{230.4}\) or \(-\dfrac{32}{45}\) or \(-0.71…\) | M1dep | oe |
| \(\cos^{-1}\) their \(-\dfrac{32}{45}\) | M1dep | dep on M2 |
| 135.(…) | A1 |
Additional guidance
| 144 \(\dfrac{4}{5} \times 144 = 115.2\) \((\cos AOB =)\ \dfrac{144^2 + 115.2^2 - 20^2}{2 \times 144 \times 115.2}\) | B0 M1 M1M0A0 |
| 12 seen, but a different value used for the radius of the larger circle cannot score B1M1 | |
| \(x + y = 12\) seen, but \(x = 6\) used to find radius \(OA = 4.8\) | B0M1 |