| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Real-world application problems |
| Difficulty | Standard +0.3 This is a straightforward application of the sine rule in part (a) requiring a single calculation, followed by part (b) which needs the cosine rule to find the remaining sides and simple rounding up. The context is accessible, the methods are standard AS-level techniques with no novel problem-solving required, making it slightly easier than average. |
| 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 |
| States \(\frac{\sin\theta}{12} = \frac{\sin 27}{7}\) | M1 | Sine rule with sides and angles in correct positions |
| Finds \(\theta =\) awrt \(51°\) or awrt \(129°\) | A1 | Either value acceptable |
| \(= \) awrt \(128.9°\) | A1 | Must be \(128.9°\) only, must be seen in part (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to find part or all of \(AD\), e.g. \(AD^2 = 7^2 + 12^2 - 2\times12\times7\cos101.9\) or \((AC)^2 = 7^2+12^2-2\times12\times7\cos(180-"128.9"-27)\) or \(12\cos27\) or \(7\cos"51"\) | M1 | Correct method for finding \(AD\) or part of \(AD\) (e.g. \(AC\) or \(CD\)); condone incorrect labelling |
| Full method for total length \(= 12+7+7+"15.09"\) | dM1 | Complete method for total length; must use correct combination of angles and sides |
| \(= 42\) m | A1 | Rounds correct \(41.09\) m up to \(42\) m |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States $\frac{\sin\theta}{12} = \frac{\sin 27}{7}$ | M1 | Sine rule with sides and angles in correct positions |
| Finds $\theta =$ awrt $51°$ or awrt $129°$ | A1 | Either value acceptable |
| $= $ awrt $128.9°$ | A1 | Must be $128.9°$ only, must be seen in part (a) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to find part or all of $AD$, e.g. $AD^2 = 7^2 + 12^2 - 2\times12\times7\cos101.9$ or $(AC)^2 = 7^2+12^2-2\times12\times7\cos(180-"128.9"-27)$ or $12\cos27$ or $7\cos"51"$ | M1 | Correct method for finding $AD$ or part of $AD$ (e.g. $AC$ or $CD$); condone incorrect labelling |
| Full method for total length $= 12+7+7+"15.09"$ | dM1 | Complete method for total length; must use correct combination of angles and sides |
| $= 42$ m | A1 | Rounds correct $41.09$ m up to $42$ m |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bcbd842f-b2e2-4587-ab4c-15a57a449e5d-10_360_1164_260_607}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the design for a structure used to support a roof.\\
The structure consists of four steel beams, $A B , B D , B C$ and $A D$.\\
Given $A B = 12 \mathrm {~m} , B C = B D = 7 \mathrm {~m}$ and angle $B A C = 27 ^ { \circ }$
\begin{enumerate}[label=(\alph*)]
\item find, to one decimal place, the size of angle $A C B$.
The steel beams can only be bought in whole metre lengths.
\item Find the minimum length of steel that needs to be bought to make the complete structure.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2020 Q5 [6]}}