Foundation November 2021 Paper 2 Q24
24 Here is triangle \(ABC\).
Not drawn accurately
(a) Assume that angle \(ACB = 90^\circ\)
Work out the length \(AB\). [3 marks]
(b) The actual length \(AB\) is greater than the answer to part (a).
What does this mean about angle \(ACB\)?
Tick one box. [1 mark]
- It is \(90^\circ\)
- It is less than \(90^\circ\)
- It is more than \(90^\circ\)
- It could be any of the above.
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(15^2\) or 225 and \(7^2\) or 49 or 274 | M1 | |
| \(\sqrt{7^2 + 15^2}\) or \(\sqrt{49 + 225}\) | M1dep | |
| 16.55(…) or 16.6 or \(\sqrt{274}\) | A1 | accept 17 with M2 awarded |
| Alternative method 2 | ||
| \(\tan^{-1} \dfrac{7}{15}\) or 25.0… | M1 | |
| \(\dfrac{15}{\cos(\text{their } 25\ldots)}\) or \(\dfrac{7}{\sin(\text{their } 25\ldots)}\) | M1dep | |
| 16.55(…) or 16.6 | A1 | accept 17 with M2 awarded |
| Alternative method 3 | ||
| \(\tan^{-1} \dfrac{15}{7}\) or 64.98… or 65 | M1 | |
| \(\dfrac{15}{\sin(\text{their } 64.98\ldots)}\) or \(\dfrac{7}{\cos(\text{their } 64.98\ldots)}\) | M1dep | |
| 16.55(…) or 16.6 | A1 | accept 17 with M2 awarded |
Additional guidance
| Allow rounding or truncation after correct answer seen eg1 16.55, Answer 16 eg2 \(\sqrt{274}\), Answer 16.5 | M2A1 M2A1 |
| Misconception of square root eg \(\sqrt{274} = 137\) | M2A0 |
| \(15^2 - 7^2\) | M1M0A0 |
| \(\sqrt{176}\) without seeing \(15^2\) or 225 and \(7^2\) or 49 | M0M0A0 |
| Answer | Mark | Comments |
|---|---|---|
| It is more than \(90^\circ\) | B1 |