| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Proving angle or length value |
| Difficulty | Standard +0.8 This question requires multiple steps combining trigonometry (sine/cosine rules, right-angled triangles) with exact value manipulation involving surds. Part (i) needs finding BX using right-angled triangle properties with 150° angle, then proving a specific inverse tan expression. Part (ii) requires showing AC equals a nested surd expression, likely using cosine rule or Pythagoras with the previous results. The 'show that' format and exact surd arithmetic elevate this above routine sine/cosine rule applications, but it follows a clear geometric setup without requiring novel insight. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05g Exact trigonometric values: for standard angles1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(BX = 6\cos30 = 3\sqrt{3}\) | B1 | co |
| \(CX = 6\sin30 = 3\) | B1 | co |
| \(\tan CAB = \frac{\text{opp}}{\text{adj}} = \frac{3}{4 + 3\sqrt{3}}\) | M1 | Must be tan in correct 90° triangle |
| \(CAB = \tan^{-1}\left(\frac{3}{4+3\sqrt{3}}\right)\) | A1 | Answer given – beware fortuitous answers. |
| (ii) Pythagoras with his AX and CX or cosine rule used correctly → \(AC = \sqrt{52 + 24\sqrt{3}}\) | M1 A1 | For any correct method. Answer given – beware fortuitous answers. |
**(i)** $BX = 6\cos30 = 3\sqrt{3}$ | B1 | co
$CX = 6\sin30 = 3$ | B1 | co
$\tan CAB = \frac{\text{opp}}{\text{adj}} = \frac{3}{4 + 3\sqrt{3}}$ | M1 | Must be tan in correct 90° triangle
$CAB = \tan^{-1}\left(\frac{3}{4+3\sqrt{3}}\right)$ | A1 | Answer given – beware fortuitous answers.
**(ii)** Pythagoras with his AX and CX or cosine rule used correctly → $AC = \sqrt{52 + 24\sqrt{3}}$ | M1 A1 | For any correct method. Answer given – beware fortuitous answers.
**Total: [4] [2]**
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In the diagram, $A B C$ is a triangle in which $A B = 4 \mathrm {~cm} , B C = 6 \mathrm {~cm}$ and angle $A B C = 150 ^ { \circ }$. The line $C X$ is perpendicular to the line $A B X$.\\
(i) Find the exact length of $B X$ and show that angle $C A B = \tan ^ { - 1 } \left( \frac { 3 } { 4 + 3 \sqrt { } 3 } \right)$.\\
(ii) Show that the exact length of $A C$ is $\sqrt { } ( 52 + 24 \sqrt { } 3 ) \mathrm { cm }$.
\hfill \mbox{\textit{CAIE P1 2006 Q6 [6]}}