| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Quadrilateral with diagonal |
| Difficulty | Standard +0.3 Part (a) is a straightforward application of the cosine rule to express cos θ in terms of x, requiring algebraic manipulation but no insight. Part (b) requires substituting x = 2√3, using the sine rule in triangle ABD to find AB, then calculating the area of ABC using ½ab sin C. While multi-step, each step follows standard procedures for P1 sine/cosine rule questions with no novel problem-solving required. Slightly easier than average due to clear structure and routine techniques. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Scheme | Marks | Guidance |
| e.g. \((\cos\theta) = \frac{5^2+(1+x)^2-x^2}{2 \times 5(1+x)} = \frac{25+1+2x+x^2-x^2}{2 \times 5(1+x)}\) or e.g. \(x^2 = 5^2+(1+x)^2-2 \times 5(1+x)\cos\theta\) | M1 | Uses the cosine rule to form an equation for \(\cos\theta\). Condone just an expression for \(\cos\theta\). (You may not see \(\cos\theta =\)). Condone invisible brackets for this mark. Be aware of longer versions which may involve splitting triangle \(BCD\) into two right angled triangles. They would still need to proceed to an equation involving \(\cos\theta\) to score this mark. |
| \(\cos\theta = \frac{13+x}{5+5x}\) * | A1* | Achieves the given answer with no errors seen including invisible brackets. There must be at least one stage of intermediate working between their starting equation and achieving the given answer. Condone \(\cos\theta\) not appearing in the answer line, provided it is seen correctly as the subject on an earlier line. |
| (a) Total: 2 marks | ||
| \(\theta = \text{awrt } 42°\) (\(42.470747...\)) | B1 | \(\theta = \text{awrt } 42°\) seen or implied. Accept 42.5. May work in radians (awrt 0.73/0.74 radians). |
| Attempts to find \(AB\), \(AD\) or \(AC\): e.g. \(\frac{AB}{\sin"42"} = \frac{5}{\sin 30} \Rightarrow AB = ...\), or e.g. \(\angle ABC = 180-30-"42.5" = "107.6"\), \(\angle DBC = \frac{\sin"42.5"}{\sin\frac{2\sqrt{3}}{2\sqrt{3}}} \Rightarrow \angle DBC = "60.5"\), \(\angle ABD = "107.6"-"60.5" = "47.1"\), or e.g. \(\frac{AD}{\sin("47.1")} = \frac{2\sqrt{3}}{\sin 30} \Rightarrow AD = ...\), or e.g. \(\frac{AC}{\sin"108"} = \frac{5}{\sin 30}\) | M1 | Attempts to find \(AB\), \(AD\) or \(AC\) using a correct method. Angles and lengths must be in the correct positions in the relevant formula or formulae. Condone slips in any rearrangement, calculations and substituting in \(x = 2\sqrt{3}\) provided the method is correct. Condone working in radians provided the angles are consistently in degrees or radians within the expression or formula. |
| \(AB = 6.75...\) or \(AD = 5.07...\) or \(AC = 9.53...\) | A1 | awrt 6.75/6.76 or awrt 9.53/9.54 or awrt 5.07/5.08 (may be implied by further work). |
| Area \(= \frac{1}{2} \times 5 \times "6.75" \times \sin(180-30-"42.5")\) or \(= \frac{1}{2} \times 5 \times "9.54" \times \sin("42.5")\) | dM1 | Correct full method to find the total area (the expression is sufficient). It is dependent on the previous method mark. Condone use of incorrectly rounded angles/slips and may work in radians. e.g. Area \(= \frac{1}{2} \times 5 \times "6.75" \times \sin(\pi - \frac{\pi}{6} - "0.74")\). May find the areas of the two separate triangles \(ABD\) and \(BDC\) and add them together. e.g. Area \(= \frac{1}{2} \times "5.07" \times "6.75" \times \sin 30 + \frac{1}{2} \times "4.46" \times 5 \times \sin"42.5"\). Use the diagrams above to help with the various methods. Invisible brackets may be implied by further work or their answer. Note that if, as part of their method to find the total area, they find angle \(ADB\) but incorrectly deduce this as an acute angle then this is dM0. |
| \(= \text{awrt } 16.1 \text{ (m}^2\text{)}\) | A1 | awrt 16.1 m\(^2\) (condone lack of units). |
| (b) Total: 5 marks | ||
| Total for Question 5: 7 marks |
| Answer/Scheme | Marks | Guidance |
|---|---|---|
| e.g. $(\cos\theta) = \frac{5^2+(1+x)^2-x^2}{2 \times 5(1+x)} = \frac{25+1+2x+x^2-x^2}{2 \times 5(1+x)}$ or e.g. $x^2 = 5^2+(1+x)^2-2 \times 5(1+x)\cos\theta$ | M1 | Uses the cosine rule to form an equation for $\cos\theta$. Condone just an expression for $\cos\theta$. (You may not see $\cos\theta =$). Condone invisible brackets for this mark. Be aware of longer versions which may involve splitting triangle $BCD$ into two right angled triangles. They would still need to proceed to an equation involving $\cos\theta$ to score this mark. |
| $\cos\theta = \frac{13+x}{5+5x}$ * | A1* | Achieves the given answer with no errors seen including invisible brackets. There must be at least one stage of intermediate working between their starting equation and achieving the given answer. Condone $\cos\theta$ not appearing in the answer line, provided it is seen correctly as the subject on an earlier line. |
| **(a) Total: 2 marks** | | |
| $\theta = \text{awrt } 42°$ ($42.470747...$) | B1 | $\theta = \text{awrt } 42°$ seen or implied. Accept 42.5. May work in radians (awrt 0.73/0.74 radians). |
| Attempts to find $AB$, $AD$ or $AC$: e.g. $\frac{AB}{\sin"42"} = \frac{5}{\sin 30} \Rightarrow AB = ...$, or e.g. $\angle ABC = 180-30-"42.5" = "107.6"$, $\angle DBC = \frac{\sin"42.5"}{\sin\frac{2\sqrt{3}}{2\sqrt{3}}} \Rightarrow \angle DBC = "60.5"$, $\angle ABD = "107.6"-"60.5" = "47.1"$, or e.g. $\frac{AD}{\sin("47.1")} = \frac{2\sqrt{3}}{\sin 30} \Rightarrow AD = ...$, or e.g. $\frac{AC}{\sin"108"} = \frac{5}{\sin 30}$ | M1 | Attempts to find $AB$, $AD$ or $AC$ using a correct method. Angles and lengths must be in the correct positions in the relevant formula or formulae. Condone slips in any rearrangement, calculations and substituting in $x = 2\sqrt{3}$ provided the method is correct. Condone working in radians provided the angles are consistently in degrees or radians within the expression or formula. |
| $AB = 6.75...$ or $AD = 5.07...$ or $AC = 9.53...$ | A1 | awrt 6.75/6.76 or awrt 9.53/9.54 or awrt 5.07/5.08 (may be implied by further work). |
| Area $= \frac{1}{2} \times 5 \times "6.75" \times \sin(180-30-"42.5")$ or $= \frac{1}{2} \times 5 \times "9.54" \times \sin("42.5")$ | dM1 | Correct full method to find the total area (the expression is sufficient). It is dependent on the previous method mark. Condone use of incorrectly rounded angles/slips and may work in radians. e.g. Area $= \frac{1}{2} \times 5 \times "6.75" \times \sin(\pi - \frac{\pi}{6} - "0.74")$. May find the areas of the two separate triangles $ABD$ and $BDC$ and add them together. e.g. Area $= \frac{1}{2} \times "5.07" \times "6.75" \times \sin 30 + \frac{1}{2} \times "4.46" \times 5 \times \sin"42.5"$. Use the diagrams above to help with the various methods. Invisible brackets may be implied by further work or their answer. Note that if, as part of their method to find the total area, they find angle $ADB$ but incorrectly deduce this as an acute angle then this is dM0. |
| $= \text{awrt } 16.1 \text{ (m}^2\text{)}$ | A1 | awrt 16.1 m$^2$ (condone lack of units). |
| **(b) Total: 5 marks** | | |
| **Total for Question 5: 7 marks** | | |
---5.
Figure 2
Diagram NOT accurately drawn
Figure 2 shows the plan view of a frame for a flat roof.\\
The shape of the frame consists of triangle $A B D$ joined to triangle $B C D$.\\
Given that
\begin{itemize}
\item $B D = x \mathrm {~m}$
\item $C D = ( 1 + x ) \mathrm { m }$
\item $B C = 5 \mathrm {~m}$
\item angle $B C D = \theta ^ { \circ }$
\begin{enumerate}[label=(\alph*)]
\item show that $\cos \theta ^ { \circ } = \frac { 13 + x } { 5 + 5 x }$
\end{itemize}
Given also that
\begin{itemize}
\item $x = 2 \sqrt { 3 }$
\item angle $B A C = 30 ^ { \circ }$
\item $A D C$ is a straight line
\item find the area of triangle $A B C$, giving your answer, in $\mathrm { m } ^ { 2 }$, to one decimal place.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2023 Q5 [7]}}