| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Tangent and sector - two tangents from external point |
| Difficulty | Standard +0.8 This question requires understanding of tangent properties, sector areas, and arc length formulas. Part (i) involves setting up an equation by equating areas and using geometric relationships to derive a non-trivial result involving tan θ = 2θ. Part (ii) requires finding the angle from arc length, then calculating areas involving both circular sectors and triangles. The multi-step reasoning and geometric insight needed place this above average difficulty. |
| Spec | 1.03f Circle properties: angles, chords, tangents1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(AT\) or \(BT = r\tan\theta\) or \(OT = \frac{r}{\cos\theta}\) | B1 | May be seen on diagram |
| \(\frac{1}{2}r^2 2\theta\), & \(\frac{1}{2} \times r \times (r\tan\theta\) or \(AT)\) or \(\frac{1}{2} \times r \times \left(\frac{r}{\cos\theta}\ \text{or}\ OT\right)\sin\theta\) | M1 | Both formulae, (\(\frac{1}{2}r^2\theta\), \(\frac{1}{2}bh\) or \(\frac{1}{2}ab\sin\theta\)), seen with \(2\theta\) used when needed |
| \(\frac{1}{2}r^2 2\theta = 2 \times \frac{1}{2} \times r \times r\tan\theta - \frac{1}{2}r^2 2\theta\) oe \(\rightarrow 2\theta = \tan\theta\) | A1 | Fully correct working from a correct statement. Note: \(\frac{1}{2}r^2 2\theta = \frac{1}{2}r^2\tan\theta\) is a valid statement |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta = 1.2\) or sector area \(= 76.8\) | B1 | |
| Area of kite \(= 165\) awrt | B1 | |
| \(164.6 - 76.8 = 87.8\) awrt | B1 | awrt 87.8 with little or no working can be awarded 3/3. SC Final answers that round to 88 with little or no working can be awarded 2/3 |
## Question 6(i):
$AT$ or $BT = r\tan\theta$ or $OT = \frac{r}{\cos\theta}$ | B1 | May be seen on diagram
$\frac{1}{2}r^2 2\theta$, & $\frac{1}{2} \times r \times (r\tan\theta$ or $AT)$ or $\frac{1}{2} \times r \times \left(\frac{r}{\cos\theta}\ \text{or}\ OT\right)\sin\theta$ | M1 | Both formulae, ($\frac{1}{2}r^2\theta$, $\frac{1}{2}bh$ or $\frac{1}{2}ab\sin\theta$), seen with $2\theta$ used when needed
$\frac{1}{2}r^2 2\theta = 2 \times \frac{1}{2} \times r \times r\tan\theta - \frac{1}{2}r^2 2\theta$ oe $\rightarrow 2\theta = \tan\theta$ | A1 | Fully correct working from a correct statement. Note: $\frac{1}{2}r^2 2\theta = \frac{1}{2}r^2\tan\theta$ is a valid statement
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## Question 6(ii):
$\theta = 1.2$ or sector area $= 76.8$ | B1 |
Area of kite $= 165$ awrt | B1 |
$164.6 - 76.8 = 87.8$ awrt | B1 | awrt 87.8 with little or no working can be awarded 3/3. SC Final answers that round to 88 with little or no working can be awarded 2/3
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\includegraphics[max width=\textwidth, alt={}, center]{58d65166-2b1a-4b58-9859-afe919c0a3a9-08_454_684_255_726}
The diagram shows points $A$ and $B$ on a circle with centre $O$ and radius $r$. The tangents to the circle at $A$ and $B$ meet at $T$. The shaded region is bounded by the minor $\operatorname { arc } A B$ and the lines $A T$ and $B T$. Angle $A O B$ is $2 \theta$ radians.\\
(i) In the case where the area of the sector $A O B$ is the same as the area of the shaded region, show that $\tan \theta = 2 \theta$.\\
(ii) In the case where $r = 8 \mathrm {~cm}$ and the length of the minor $\operatorname { arc } A B$ is 19.2 cm , find the area of the shaded region.\\
\hfill \mbox{\textit{CAIE P1 2018 Q6 [6]}}