| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Proving angle or length value |
| Difficulty | Standard +0.8 This is a multi-step trigonometry problem requiring coordinate geometry or multiple applications of trigonometric rules in a right-angled triangle context. Part (i) is straightforward using Pythagoras, but part (ii) requires finding an angle expression involving inverse tan, which demands careful manipulation of trigonometric identities and angle relationships—more challenging than routine sine/cosine rule applications but not requiring exceptional insight. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan\!\left(\frac{\pi}{3}\right) = \frac{AC}{2x}\) or \(\cos\!\left(\frac{\pi}{3}\right)\!\left(= \sin\frac{\pi}{6}\right) = \frac{2x}{AB}\) → \(AC = 2\sqrt{3}x\) or \(AB = 4x\) | B1 | Either trig ratio |
| \(AM = \sqrt{13x^2},\ \sqrt{13}x,\ 3.61x\) | M1A1 | Complete method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan(M\hat{A}C) = \frac{x}{\text{Their } AC}\) | M1 | "Their \(AC\)" must be \(f(x)\), \(M\hat{A}C \neq \theta\) |
| \(\theta = \frac{1}{6}\pi - \tan^{-1}\frac{1}{2\sqrt{3}}\) AG | A1 | Justifies \(\frac{\pi}{6}\) and links MAC & \(\theta\) |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\!\left(\frac{\pi}{3}\right) = \frac{AC}{2x}$ or $\cos\!\left(\frac{\pi}{3}\right)\!\left(= \sin\frac{\pi}{6}\right) = \frac{2x}{AB}$ → $AC = 2\sqrt{3}x$ or $AB = 4x$ | B1 | Either trig ratio |
| $AM = \sqrt{13x^2},\ \sqrt{13}x,\ 3.61x$ | M1A1 | Complete method |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan(M\hat{A}C) = \frac{x}{\text{Their } AC}$ | M1 | "Their $AC$" must be $f(x)$, $M\hat{A}C \neq \theta$ |
| $\theta = \frac{1}{6}\pi - \tan^{-1}\frac{1}{2\sqrt{3}}$ **AG** | A1 | Justifies $\frac{\pi}{6}$ and links MAC & $\theta$ |5\\
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In the diagram, triangle $A B C$ is right-angled at $C$ and $M$ is the mid-point of $B C$. It is given that angle $A B C = \frac { 1 } { 3 } \pi$ radians and angle $B A M = \theta$ radians. Denoting the lengths of $B M$ and $M C$ by $x$,\\
(i) find $A M$ in terms of $x$,\\
(ii) show that $\theta = \frac { 1 } { 6 } \pi - \tan ^ { - 1 } \left( \frac { 1 } { 2 \sqrt { 3 } } \right)$.
\hfill \mbox{\textit{CAIE P1 2016 Q5 [5]}}