C2 January 2006 Q5
5.
In Figure 2 \(OAB\) is a sector of a circle radius 5 m. The chord \(AB\) is 6 m long.
(a) Show that \(\cos A\hat{O}B = \dfrac{7}{25}\). (2)
(b) Hence find the angle \(A\hat{O}B\) in radians, giving your answer to 3 decimal places. (1)
(c) Calculate the area of the sector \(OAB\). (2)
(d) Hence calculate the shaded area. (3)
| Scheme | Marks |
|---|---|
| \(\cos A\hat{O}B = \dfrac{5^2 + 5^2 - 6^2}{2 \times 5 \times 5}\) or \(\sin\theta = \tfrac{3}{5}\) with use of \(\cos 2\theta = 1 - 2\sin^2\theta\) attempted | M1 |
| \(= \underline{\dfrac{7}{25}} \quad *\) | A1cso |
| (2) |
Notes
M1 for a full method leading to \(\cos A\hat{O}B\) [N.B. Use of calculator is M0] (usual rules about quoting formulae)
| Scheme | Marks |
|---|---|
| \(A\hat{O}B = 1.2870022\ldots\) radians 1.287 or better | B1 |
| (1) |
Notes
Use of (b) in degrees is M0
| Scheme | Marks |
|---|---|
| Sector \(= \tfrac{1}{2} \times 5^2 \times (b), \ = 16.087\ldots\) (AWRT) \(\underline{16.1}\) | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| Triangle \(= \tfrac{1}{2} \times 5^2 \times \sin(b)\) or \(\tfrac{1}{2} \times 6 \times \sqrt{5^2 - 3^2}\) | M1 |
| Segment \(=\) (their sector) \(-\) their triangle | dM1 |
| \(=\) (sector from c) \(- 12 =\) (AWRT) \(\underline{4.1}\) (ft their part(c)) | A1ft |
| (3) | |
| (8 marks) |
Notes
1st M1 for full method for the area of triangle \(AOB\)
2nd M1 for their sector \(-\) their triangle. Dependent on 1st M1 in part (d).
A1ft for their sector from part (c) \(- 12\) [or 4.1 following a correct restart].