| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | January |
| Marks | 11 |
| 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 multi-part question requiring cosine rule to find an angle, then basic circular sector calculations. All techniques are standard C2 material with no novel problem-solving required—slightly easier than average due to clear structure and routine application of formulas. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(16^2 = 10^2 + 12^2 - 2\times10\times12\cos\angle YXZ\) | M1 | Uses cosine rule; must be correct statement |
| \(\cos\angle YXZ = \frac{10^2+12^2-16^2}{2\times10\times12}\) or \(\frac{-12}{240}\) or \(-0.05\) | A1 | Correct value or correct numerical expression |
| \(\angle BOC = 1.62(08...)\) (N.B. 92.87 degrees is A0) | A1 | Accept awrt 1.62; must be seen in part (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(s = 5\theta\) with their \(\theta\) from part (a) | M1 | Uses \(s=r\theta\) with \(\theta\) in radians, or correct formula for degrees |
| awrt 8.1 | A1 | May be implied by their perimeter |
| Perimeter \(= r\theta + 28 = 28 +\) their arc length | M1 | Adds arc length to 28 or \((16+7+5)\) |
| awrt 36.1 | A1 | Do not need units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area of sector \(= \frac{1}{2}(5)^2\theta\) | B1ft | Formula used with \(\theta\) in radians or correct formula for degrees |
| Area of triangle \(= \frac{1}{2}\times10\times12\sin\theta\) \((= 59.92\) or \(59.93)\) | B1ft | Correct formula for area used; may use half base times height |
| Area of shaded region \(= \frac{1}{2}\times10\times12\sin\theta - \frac{1}{2}(5)^2\theta = 59.9... - 20.2... = 39.7\ (\text{cm}^2)\) | M1 A1 | M1: subtracts sector area from triangle area this way round; A1: awrt 39.7; do not need units |
## Question 11:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $16^2 = 10^2 + 12^2 - 2\times10\times12\cos\angle YXZ$ | M1 | Uses cosine rule; must be correct statement |
| $\cos\angle YXZ = \frac{10^2+12^2-16^2}{2\times10\times12}$ or $\frac{-12}{240}$ or $-0.05$ | A1 | Correct value or correct numerical expression |
| $\angle BOC = 1.62(08...)$ (N.B. 92.87 degrees is A0) | A1 | Accept awrt 1.62; must be seen in part (a) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $s = 5\theta$ with their $\theta$ from part (a) | M1 | Uses $s=r\theta$ with $\theta$ in radians, or correct formula for degrees |
| awrt 8.1 | A1 | May be implied by their perimeter |
| Perimeter $= r\theta + 28 = 28 +$ their arc length | M1 | Adds arc length to 28 or $(16+7+5)$ |
| awrt 36.1 | A1 | Do not need units |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area of sector $= \frac{1}{2}(5)^2\theta$ | B1ft | Formula used with $\theta$ in radians or correct formula for degrees |
| Area of triangle $= \frac{1}{2}\times10\times12\sin\theta$ $(= 59.92$ or $59.93)$ | B1ft | Correct formula for area used; may use half base times height |
| Area of shaded region $= \frac{1}{2}\times10\times12\sin\theta - \frac{1}{2}(5)^2\theta = 59.9... - 20.2... = 39.7\ (\text{cm}^2)$ | M1 A1 | M1: subtracts sector area from triangle area this way round; A1: awrt 39.7; do not need units |11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{88ed9a17-97a5-4548-80bb-70b4b901a39d-13_625_1155_285_456}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a triangle $X Y Z$ with $X Y = 10 \mathrm {~cm} , Y Z = 16 \mathrm {~cm}$ and $Z X = 12 \mathrm {~cm}$.
\begin{enumerate}[label=(\alph*)]
\item Find the size of the angle $Y X Z$, giving your answer in radians to 3 significant figures.
The point $A$ lies on the line $X Y$ and the point $B$ lies on the line $X Z$ and $A X = B X = 5 \mathrm {~cm} . A B$ is the arc of a circle with centre $X$.
The shaded region $S$, shown in Figure 1, is bounded by the lines $B Z , Z Y , Y A$ and the arc $A B$.
Find
\item the perimeter of the shaded region to 3 significant figures,
\item the area of the shaded region to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q11 [11]}}