| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Triangle with circular sector |
| Difficulty | Standard +0.3 This is a straightforward application of the cosine rule to find an angle, followed by standard sector area and kite area calculations. All formulas are directly applicable with no problem-solving insight required. The multi-part structure and need for radians adds slight complexity, but it remains easier than average as it's purely procedural with clear geometric setup. |
| Spec | 1.03f Circle properties: angles, chords, tangents1.05b Sine and cosine rules: including ambiguous case1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(5^2 = 10^2 + 12^2 - 2\times10\times12\cos\angle XAB\), or \(\cos\angle XAB = \frac{10^2+12^2-5^2}{2\times10\times12}\) or \(\frac{219}{240}\) or \(0.9125\) or \(\frac{73}{80}\) | M1 | Must be correct cosine rule statement; answers in degrees gain M1 A0 |
| \(\angle XAB = 0.421\) or \(0.134\pi\) | A1 | Accept awrt 0.421; also 0.42 is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of sector is \(\frac{1}{2}r^2\theta = \frac{1}{2}\times10^2\times\theta\) | M1 | Uses area formula with \(r=10\) and angle in radians |
| Area of major sector is \(\frac{1}{2}\times r^2(2\pi - 2\times\text{"0.421"})\) or \(\pi\times r^2 - \frac{1}{2}\times r^2\times2\times\text{"0.421"}\) | M1 | Finds angle in major sector and uses sector formula or subtracts minor area from circle |
| \(= 272\) | A1 | Accept awrt 272 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Way 1: Area of triangle \(AXB = \frac{1}{2}\times10\times12\times\sin XAB\) | M1 | Finds area of triangle \(AXB\) using 10, 12 and angle \(XAB\) |
| Area of kite \(= 2\times\) triangle \(AXB\) | dM1 | Doubles area of triangle \(AXB\) |
| \(=\) awrt 49 | A1 | Do not need units |
| Way 2: Find angle \(XBA\) and hence area \(XBY\): Area of kite \(=\) area of \(XBY +\) Area \(XAY = 37.298 + 11.76 = 49\) | M1, dM1, A1 | Finds angle \(XBA\) (0.958) by cosine rule (NOT \(90-XAB\)); adds areas of triangles \(XBY\) and \(XAY\) |
| Way 3: Finds length \(XY\) by cosine rule or elementary trigonometry (8.173); Uses area of kite \(= \frac{1}{2}\times\text{"8.173"}\times12\) | M1, dM1, A1 |
## Question 9:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $5^2 = 10^2 + 12^2 - 2\times10\times12\cos\angle XAB$, or $\cos\angle XAB = \frac{10^2+12^2-5^2}{2\times10\times12}$ or $\frac{219}{240}$ or $0.9125$ or $\frac{73}{80}$ | M1 | Must be correct cosine rule statement; answers in degrees gain M1 A0 |
| $\angle XAB = 0.421$ or $0.134\pi$ | A1 | Accept awrt 0.421; also 0.42 is A0 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of sector is $\frac{1}{2}r^2\theta = \frac{1}{2}\times10^2\times\theta$ | M1 | Uses area formula with $r=10$ and angle in radians |
| Area of major sector is $\frac{1}{2}\times r^2(2\pi - 2\times\text{"0.421"})$ or $\pi\times r^2 - \frac{1}{2}\times r^2\times2\times\text{"0.421"}$ | M1 | Finds angle in major sector and uses sector formula or subtracts minor area from circle |
| $= 272$ | A1 | Accept awrt 272 |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| **Way 1:** Area of triangle $AXB = \frac{1}{2}\times10\times12\times\sin XAB$ | M1 | Finds area of triangle $AXB$ using 10, 12 and angle $XAB$ |
| Area of kite $= 2\times$ triangle $AXB$ | dM1 | Doubles area of triangle $AXB$ |
| $=$ awrt 49 | A1 | Do not need units |
| **Way 2:** Find angle $XBA$ and hence area $XBY$: Area of kite $=$ area of $XBY +$ Area $XAY = 37.298 + 11.76 = 49$ | M1, dM1, A1 | Finds angle $XBA$ (0.958) by cosine rule (NOT $90-XAB$); adds areas of triangles $XBY$ and $XAY$ |
| **Way 3:** Finds length $XY$ by cosine rule or elementary trigonometry (8.173); Uses area of kite $= \frac{1}{2}\times\text{"8.173"}\times12$ | M1, dM1, A1 | |
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b99072a-cd16-4c1d-9e44-085926a3ba24-13_460_698_269_625}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
In Figure 3, the points $A$ and $B$ are the centres of the circles $C _ { 1 }$ and $C _ { 2 }$ respectively. The circle $C _ { 1 }$ has radius 10 cm and the circle $C _ { 2 }$ has radius 5 cm . The circles intersect at the points $X$ and $Y$, as shown in the figure.
Given that the distance between the centres of the circles is 12 cm ,
\begin{enumerate}[label=(\alph*)]
\item calculate the size of the acute angle $X A B$, giving your answer in radians to 3 significant figures,
\item find the area of the major sector of circle $C _ { 1 }$, shown shaded in Figure 3,
\item find the area of the kite $A Y B X$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2015 Q9 [8]}}