| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector with attached triangle |
| Difficulty | Standard +0.3 This is a straightforward application of sine rule, arc length, and area formulas with clear given information. Part (a) uses sine rule in triangle OAD, part (b) applies arc length formula s=rθ, and part (c) combines sector and triangle area formulas. While multi-step, each step follows standard procedures with no conceptual challenges beyond routine P1 trigonometry and radian geometry. |
| 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 |
|---|---|---|
| \(\frac{\sin\alpha}{14}=\frac{\sin 0.43}{6}\) | M1 | Sight of values embedded or awrt 1.33/1.34 implies mark |
| \(\alpha=1.337\) radians, accept awrt 1.33/1.34 or awrt 76.6/76.7° | A1 | |
| Angle \(AOD = \pi - 1.337 =\) awrt \(1.805\) radians | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Attempts \(s=r\theta\) with \(r=6\) and an allowable \(\theta\) | M1 | Accept allowable angle: \((a)\), \(\pi-{(a)}\), or \(2\pi-{(a)}\) |
| Arc length \(ABC =\) awrt \(26.9\) m | A1 | (2) Note: using acute angle gives 29.7m → M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Attempts \(\frac{1}{2}r^2\theta\) with \(r=6\) and allowable \(\theta\) in radians (\(=80.6\)) | M1 | Accept \(\frac{1}{2}r\times\) arc length. Condone use of \(\frac{1}{2}r^2(\theta-\sin\theta)\) with allowable \(\theta\) |
| Attempts area \(AOD = \frac{1}{2}\times6\times14\times\sin(0.91)\) oe (\(=33.1\)) | M1 | Angle \(OAD\) found by correct method: \(\pi-0.43-{(a)}\); or two right-angled triangles method |
| Attempts sector \(+\) triangle with correct attempt at angles | dM1 | Dependent on both previous M marks |
| \(=113.7\ \text{m}^2\) | A1 | (4) (9 marks) Note: using acute angle gives \(130\ \text{m}^2\) → M1M1dM1A0 |
# Question 5(a):
$\frac{\sin\alpha}{14}=\frac{\sin 0.43}{6}$ | M1 | Sight of values embedded or awrt 1.33/1.34 implies mark
$\alpha=1.337$ radians, accept awrt 1.33/1.34 or awrt 76.6/76.7° | A1 |
Angle $AOD = \pi - 1.337 =$ awrt $1.805$ radians | A1 | **(3)**
---
# Question 5(b):
Attempts $s=r\theta$ with $r=6$ and an allowable $\theta$ | M1 | Accept allowable angle: $(a)$, $\pi-{(a)}$, or $2\pi-{(a)}$
Arc length $ABC =$ awrt $26.9$ m | A1 | **(2)** Note: using acute angle gives 29.7m → M1A0
---
# Question 5(c):
Attempts $\frac{1}{2}r^2\theta$ with $r=6$ and allowable $\theta$ in radians ($=80.6$) | M1 | Accept $\frac{1}{2}r\times$ arc length. Condone use of $\frac{1}{2}r^2(\theta-\sin\theta)$ with allowable $\theta$
Attempts area $AOD = \frac{1}{2}\times6\times14\times\sin(0.91)$ oe ($=33.1$) | M1 | Angle $OAD$ found by correct method: $\pi-0.43-{(a)}$; or two right-angled triangles method
Attempts sector $+$ triangle with correct attempt at angles | dM1 | Dependent on both previous M marks
$=113.7\ \text{m}^2$ | A1 | **(4) (9 marks)** Note: using acute angle gives $130\ \text{m}^2$ → M1M1dM1A0
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6a5d0ffc-a725-404b-842a-f3b6000e6fed-14_470_940_246_500}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows the plan view of a viewing platform at a tourist site.
The shape of the viewing platform consists of a sector $A B C O A$ of a circle, centre $O$, joined to a triangle $A O D$.
Given that
\begin{itemize}
\item $O A = O C = 6 \mathrm {~m}$
\item $A D = 14 \mathrm {~m}$
\item angle $A D C = 0.43$ radians
\item angle $A O D$ is an obtuse angle
\item $O C D$ is a straight line\\
find
\begin{enumerate}[label=(\alph*)]
\item the size of angle $A O D$, in radians, to 3 decimal places,
\item the length of arc $A B C$, in metres, to one decimal place,
\item the total area of the viewing platform, in $\mathrm { m } ^ { 2 }$, to one decimal place.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2021 Q5 [9]}}