| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Proving angle relationships |
| Difficulty | Standard +0.3 This is a straightforward application of basic trigonometry and sector area formulas. Part (a) requires using inverse sine with given sides (sin⁻¹(9/15)), and part (b) involves finding a sector area and subtracting a triangle area using standard formulas. The geometry is clearly presented and the methods are routine for P1 level, 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)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EITHER By trigonometry: \(B\hat{A}C=0.6435\ldots\) and \(A\hat{B}C=\frac{\pi-0.6435}{2}\) OR By Pythagoras: \(AP=12\Rightarrow BP=3\) so \(\tan A\hat{B}C=\frac{9}{3}\) OR Using \(\triangle PBC\) and either the sine or cosine rule: \(\sin A\hat{B}C=\frac{3}{\sqrt{10}}\) or \(\cos A\hat{B}C=\frac{\sqrt{10}}{10}\) | M1 | \(\frac{3}{\sqrt{10}}=0.9486\ldots\quad\frac{\sqrt{10}}{10}=0.3162\ldots\) |
| \(A\hat{B}C=\frac{\pi-0.6435}{2}\) or \(\tan^{-1}\frac{9}{3}\) or \(\sin^{-1}\frac{3}{\sqrt{10}}\) or \(\cos^{-1}\frac{\sqrt{10}}{10}\) or \(1.249(04\ldots)\) or \(71.56°\) \(= 1.25\) radians (3 sf) | A1 | AG. Final answer must be 1.25, more accurate value 1.24904… with no rounding to 3sf seen as the final answer gets M1A0. If decimals are used all values must be given to at least 4sf for A1 |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(BC=\sqrt{(\textit{their}\,3)^2+9^2}\) or \(\frac{9}{\sin 1.25}\) \([=\sqrt{90},\;3\sqrt{10}\) or \(9.48697\ldots]\) | M1 | Using correct method(s) to find \(BC\) |
| Area of sector \(=\frac{1}{2}\times(\textit{their}\,BC)^2\times\tan^{-1}3\) \([=56.207\text{ or }56.25]\) | M1 | Using \(\tan^{-1}3\) or \(1.25\) and *their* \(BC\), but not 9 or 15, in correct area of sector formula |
| Area of triangle \(PBC=13.4\) to \(13.6\) or \(\frac{1}{2}\times9\times3\) | B1 | |
| \([\text{Area}=(56.207\text{ or }56.25)-\textit{their}\,13.5=]\;42.7\) or \(42.8\) | A1 | AWRT |
| Total | 4 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **EITHER** By trigonometry: $B\hat{A}C=0.6435\ldots$ and $A\hat{B}C=\frac{\pi-0.6435}{2}$ **OR** By Pythagoras: $AP=12\Rightarrow BP=3$ so $\tan A\hat{B}C=\frac{9}{3}$ **OR** Using $\triangle PBC$ and either the sine or cosine rule: $\sin A\hat{B}C=\frac{3}{\sqrt{10}}$ or $\cos A\hat{B}C=\frac{\sqrt{10}}{10}$ | M1 | $\frac{3}{\sqrt{10}}=0.9486\ldots\quad\frac{\sqrt{10}}{10}=0.3162\ldots$ |
| $A\hat{B}C=\frac{\pi-0.6435}{2}$ or $\tan^{-1}\frac{9}{3}$ or $\sin^{-1}\frac{3}{\sqrt{10}}$ or $\cos^{-1}\frac{\sqrt{10}}{10}$ or $1.249(04\ldots)$ or $71.56°$ $= 1.25$ radians (3 sf) | A1 | AG. Final answer must be 1.25, more accurate value 1.24904… with no rounding to 3sf seen as the final answer gets M1A0. If decimals are used all values must be given to at least 4sf for A1 |
| **Total** | **2** | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $BC=\sqrt{(\textit{their}\,3)^2+9^2}$ or $\frac{9}{\sin 1.25}$ $[=\sqrt{90},\;3\sqrt{10}$ or $9.48697\ldots]$ | M1 | Using correct method(s) to find $BC$ |
| Area of sector $=\frac{1}{2}\times(\textit{their}\,BC)^2\times\tan^{-1}3$ $[=56.207\text{ or }56.25]$ | M1 | Using $\tan^{-1}3$ or $1.25$ and *their* $BC$, but not 9 or 15, in correct area of sector formula |
| Area of triangle $PBC=13.4$ to $13.6$ or $\frac{1}{2}\times9\times3$ | B1 | |
| $[\text{Area}=(56.207\text{ or }56.25)-\textit{their}\,13.5=]\;42.7$ or $42.8$ | A1 | AWRT |
| **Total** | **4** | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{10b2ec29-adca-4313-ae24-bab8b2d9f8a4-08_556_751_255_696}
In the diagram the lengths of $A B$ and $A C$ are both 15 cm . The point $P$ is the foot of the perpendicular from $C$ to $A B$. The length $C P = 9 \mathrm {~cm}$. An arc of a circle with centre $B$ passes through $C$ and meets $A B$ at $Q$.
\begin{enumerate}[label=(\alph*)]
\item Show that angle $A B C = 1.25$ radians, correct to 3 significant figures.
\item Calculate the area of the shaded region which is bounded by the $\operatorname { arc } C Q$ and the lines $C P$ and $P Q$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q7 [6]}}