| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | October |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Parallelogram problems |
| Difficulty | Standard +0.3 This is a straightforward application of the parallelogram area formula (Area = ab sin θ) to find an obtuse angle, followed by using the cosine rule for the diagonal. Both are standard techniques with clear pathways, making it slightly easier than average but requiring multiple steps and careful angle consideration. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Area \(ABCD\) is 40 cm\(^2 \Rightarrow 40 = 6\times10\times\sin\theta\) oe | M1 | Score for \(40=6\times10\times\sin\theta\) or \(20=\frac{1}{2}\times6\times10\times\sin\theta\) where \(\theta\) is one of the corner angles |
| \(\sin\theta = \frac{2}{3} \Rightarrow \theta = 180° - 41.8°\) | M1 | Score for \(\sin\theta = k \Rightarrow \theta = 180° - \arcsin k\) |
| \(\angle DAB =\) awrt \(138.19°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Attempts \(DB^2 = 10^2 + 6^2 - 2\times10\times6\cos"138.19°"\) | M1 | Allow if the angle used is acute as long as it is clearly their attempt at angle \(DAB\). Allow use of \(41.8°\) unless they correctly found angle \(DAB\) and chose wrong one |
| \(DB =\) awrt \(15.0\) (cm) | A1 | Accept 15 in place of 15.0. Allow from attempts using awrt \(138°\) |
# Question 4:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| Area $ABCD$ is 40 cm$^2 \Rightarrow 40 = 6\times10\times\sin\theta$ oe | M1 | Score for $40=6\times10\times\sin\theta$ or $20=\frac{1}{2}\times6\times10\times\sin\theta$ where $\theta$ is one of the corner angles |
| $\sin\theta = \frac{2}{3} \Rightarrow \theta = 180° - 41.8°$ | M1 | Score for $\sin\theta = k \Rightarrow \theta = 180° - \arcsin k$ |
| $\angle DAB =$ awrt $138.19°$ | A1 | |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts $DB^2 = 10^2 + 6^2 - 2\times10\times6\cos"138.19°"$ | M1 | Allow if the angle used is acute as long as it is clearly their attempt at angle $DAB$. Allow use of $41.8°$ unless they correctly found angle $DAB$ and chose wrong one |
| $DB =$ awrt $15.0$ (cm) | A1 | Accept 15 in place of 15.0. Allow from attempts using awrt $138°$ |
---4. A parallelogram $A B C D$ has area $40 \mathrm {~cm} ^ { 2 }$
Given that $A B$ has length $10 \mathrm {~cm} , B C$ has length 6 cm and angle $D A B$ is obtuse, find
\begin{enumerate}[label=(\alph*)]
\item the size of angle $D A B$, in degrees, to 2 decimal places,
\item the length of diagonal $B D$, in cm , to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2019 Q4 [5]}}