| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Real-world application problems |
| Difficulty | Moderate -0.3 This is a straightforward application of sine and cosine rules in two triangles with clearly labeled sides and angles. Part (a) uses sine rule directly, part (b) requires finding one more side length then summing. The context is simple, steps are routine, and it's slightly easier than average due to the clear setup and standard technique application. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(\frac{\sin \angle ACB}{6.5} = \frac{\sin 35}{4.7}\) | M1 | Uses the sine rule with angles and sides in correct positions. Alt: cosine rule on ACB then solve quadratic to find AC, then use cosine rule again |
| \(\angle ACB = \text{awrt}(52 \text{ or } 53)°\) or \(\text{awrt}(127 \text{ or } 128)°\) | A1 | |
| \(\angle ACB = 127.5°\) only | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Eg. \(\frac{(AC)}{\sin\text{"}{17.5°\text{"} }} = \frac{6.5}{\sin\text{"}{127.5°\text{"} }}\) or \(= \frac{4.7}{\sin 35°}\) | M1 | Uses a formula that finds part or all of length \(AD\) (eg \(AC\), \(CD\), \(AX\), \(XD\), \(AD\)). Sight of awrt 8.2, awrt 2.46 or awrt 5.72 implies this mark |
| \(\left[\frac{(CD)}{\sin\text{"}{75°\text{"} }} = \frac{4.7}{\sin\text{"}{127.5°\text{"} }} \Rightarrow (CD) = \ldots \Rightarrow (AC)+(CD)\right] = \text{awrt } 8.2\) | A1 | Sight of awrt 8.2 implies length \(AD\) found. Accept \(\text{"..."}=8.2\). May be implied by sum totalling awrt 8.2 |
| Total length of wood \(= 8.1 + 6.5 + 4.7 + 4.7 = \text{awrt } 24.1\) | A1 | Do not accept 24 or 25 without seeing awrt 24.1 or calculation totalling awrt 24.1 |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $\frac{\sin \angle ACB}{6.5} = \frac{\sin 35}{4.7}$ | M1 | Uses the sine rule with angles and sides in correct positions. Alt: cosine rule on ACB then solve quadratic to find AC, then use cosine rule again |
| $\angle ACB = \text{awrt}(52 \text{ or } 53)°$ or $\text{awrt}(127 \text{ or } 128)°$ | A1 | |
| $\angle ACB = 127.5°$ only | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Eg. $\frac{(AC)}{\sin\text{"}{17.5°\text{"} }} = \frac{6.5}{\sin\text{"}{127.5°\text{"} }}$ or $= \frac{4.7}{\sin 35°}$ | M1 | Uses a formula that finds part or all of length $AD$ (eg $AC$, $CD$, $AX$, $XD$, $AD$). Sight of awrt 8.2, awrt 2.46 or awrt 5.72 implies this mark |
| $\left[\frac{(CD)}{\sin\text{"}{75°\text{"} }} = \frac{4.7}{\sin\text{"}{127.5°\text{"} }} \Rightarrow (CD) = \ldots \Rightarrow (AC)+(CD)\right] = \text{awrt } 8.2$ | A1 | Sight of awrt 8.2 implies length $AD$ found. Accept $\text{"..."}=8.2$. May be implied by sum totalling awrt 8.2 |
| Total length of wood $= 8.1 + 6.5 + 4.7 + 4.7 = \text{awrt } 24.1$ | A1 | Do not accept 24 or 25 without seeing awrt 24.1 or calculation totalling awrt 24.1 |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c8f8d35d-c2dd-4a1f-a4bb-a4fa06413d12-14_327_595_251_676}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Not to scale
Figure 3 shows the design for a structure used to support a roof.
The structure consists of four wooden beams, $A B , B D , B C$ and $A D$.
Given $A B = 6.5 \mathrm {~m} , B C = B D = 4.7 \mathrm {~m}$ and angle $B A C = 35 ^ { \circ }$
\begin{enumerate}[label=(\alph*)]
\item find, to one decimal place, the size of angle $A C B$,
\item find, to the nearest metre, the total length of wood required to make this structure.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2019 Q7 [6]}}