| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Given area find angle/side |
| Difficulty | Moderate -0.8 This is a straightforward two-part question requiring direct application of the area formula (½ab sin C) to find sin θ, then using the cosine rule with the obtuse angle. Both parts are standard textbook exercises with no problem-solving insight required, making it easier than average but not trivial since it requires careful handling of the obtuse angle condition. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempts \(A=\frac{1}{2}ab\sin C \Rightarrow 100=\frac{1}{2}\times 25\times 15\sin BAC\) | M1 | Attempt area formula with given information. Angle used should be \(\theta\), \(BAC\), \(CAB\) or \(A\). |
| \(\sin\theta° = \frac{8}{15}\) | A1 | Or exact equivalent e.g. \(\frac{200}{375}\), \(\frac{100}{187.5}\). Note: finishing with \(\sin C = \frac{8}{15}\) is A0 but allow all marks in (b). |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(BC^2 = 15^2+25^2-2\times15\times25\times\cos\theta°\) where \(\theta°=\arcsin\frac{8}{15}\) | M1 | Attempt cosine rule using acute angle from (a). Implied by \(BC \approx 14.7\) cm. |
| \(BC^2=15^2+25^2-2\times15\times25\times\cos(180-\text{their }32.2°)\) | dM1 | Full use of cosine rule with obtuse angle correct to nearest degree: \((180-\text{their }32.2°)\). |
| \(BC^2=1484.4\ldots \Rightarrow BC = \text{awrt } 38.5\text{ cm}\) | A1 | CSO. Condone missing units. Special case: radians give \(BC=38.5\) scores M1 dM1 A0. |
## Question 2(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempts $A=\frac{1}{2}ab\sin C \Rightarrow 100=\frac{1}{2}\times 25\times 15\sin BAC$ | M1 | Attempt area formula with given information. Angle used should be $\theta$, $BAC$, $CAB$ or $A$. |
| $\sin\theta° = \frac{8}{15}$ | A1 | Or exact equivalent e.g. $\frac{200}{375}$, $\frac{100}{187.5}$. Note: finishing with $\sin C = \frac{8}{15}$ is A0 but allow all marks in (b). |
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## Question 2(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $BC^2 = 15^2+25^2-2\times15\times25\times\cos\theta°$ where $\theta°=\arcsin\frac{8}{15}$ | M1 | Attempt cosine rule using acute angle from (a). Implied by $BC \approx 14.7$ cm. |
| $BC^2=15^2+25^2-2\times15\times25\times\cos(180-\text{their }32.2°)$ | dM1 | Full use of cosine rule with obtuse angle correct to nearest degree: $(180-\text{their }32.2°)$. |
| $BC^2=1484.4\ldots \Rightarrow BC = \text{awrt } 38.5\text{ cm}$ | A1 | CSO. Condone missing units. **Special case:** radians give $BC=38.5$ scores M1 dM1 A0. |\begin{enumerate}
\item The triangle $A B C$ is such that
\end{enumerate}
\begin{itemize}
\item $A B = 15 \mathrm {~cm}$
\item $A C = 25 \mathrm {~cm}$
\item angle $B A C = \theta ^ { \circ }$
\item area triangle $A B C = 100 \mathrm {~cm} ^ { 2 }$\\
(a) Find the value of $\sin \theta ^ { \circ }$
\end{itemize}
Given that $\theta > 90$\\
(b) find the length of $B C$, in cm , to 3 significant figures.
\hfill \mbox{\textit{Edexcel P1 2024 Q2 [5]}}