| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Basic sine rule application |
| Difficulty | Easy -1.2 This is a straightforward sine rule application requiring direct substitution into the formula and calculator work. Part (a) is pure recall and computation, while part (b) adds minimal complexity by requiring recognition that the longest side constraint determines which solution to choose from the ambiguous case. No problem-solving insight needed beyond standard textbook procedure. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\sin x°}{21} = \frac{\sin 25°}{13}\) | M1 | Correct statement of sine rule with sides and angles in correct position. Implied by \(\sin x° =\) awrt 0.68 |
| \(\sin x° = 0.6827\) | A1 | awrt 0.6827. ISW if they go on to find value of \(x\) in (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin^{-1}(0.6827) = \ldots (43.05°)\) | M1 | Applies inverse sine to value from part (a) to find angle in degrees. For correct (a), awrt \(43°\) is sufficient. Implied by correct answer for acute or obtuse angle |
| \(AC < AB\) so \(\angle ABC < \angle ACB\), so required angle is \(180° - \sin^{-1}(0.6827) = \ldots\) | M1 | Attempts to find correct angle \(180° - \arcsin(0.6827)\). Award even if no reasoning given. No diagram required |
| \(x =\) awrt \(136.95\) | A1 | awrt 136.95. A0 if two angles are given |
# Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\sin x°}{21} = \frac{\sin 25°}{13}$ | M1 | Correct statement of sine rule with sides and angles in correct position. Implied by $\sin x° =$ awrt 0.68 |
| $\sin x° = 0.6827$ | A1 | awrt 0.6827. ISW if they go on to find value of $x$ in (a) |
# Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin^{-1}(0.6827) = \ldots (43.05°)$ | M1 | Applies inverse sine to value from part (a) to find angle in degrees. For correct (a), awrt $43°$ is sufficient. Implied by correct answer for acute or obtuse angle |
| $AC < AB$ so $\angle ABC < \angle ACB$, so required angle is $180° - \sin^{-1}(0.6827) = \ldots$ | M1 | Attempts to find correct angle $180° - \arcsin(0.6827)$. Award even if no reasoning given. No diagram required |
| $x =$ awrt $136.95$ | A1 | awrt 136.95. A0 if two angles are given |
---2. In the triangle $A B C$,
\begin{itemize}
\item $A B = 21 \mathrm {~cm}$
\item $B C = 13 \mathrm {~cm}$
\item angle $B A C = 25 ^ { \circ }$
\item angle $A C B = x ^ { \circ }$
\begin{enumerate}[label=(\alph*)]
\item Use the sine rule to find the value of $\sin x ^ { \circ }$, giving your answer to 4 decimal places.
\end{itemize}
Given also that $A B$ is the longest side of the triangle,
\item find the value of $x$, giving your answer to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2022 Q2 [5]}}