| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2002 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Verify shape type from coordinates |
| Difficulty | Standard +0.3 This is a straightforward coordinate-free geometry problem requiring basic trigonometry in right triangles and the application of Pythagoras' theorem. Part (i) uses simple trig ratios (AC = l/cos30°, then finding BC and AB), while part (ii) applies tan to angle BAD. The 'show that' format provides the answer, requiring only verification through standard A-level techniques with no novel insight needed. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(AC = \cos 30 = \sqrt{3}/2\). \(BC = 2\sin 30 = 1\). \(AB = \sqrt{(1^2 + 3/4)} = \frac{\sqrt{7}}{2}\) | B1, B1, M1 A1 | Correct only – not decimal; Correct only; Use of Pythagoras. Correct only. Answer given. Could be cosine rule |
| (ii) \(\tan(x + 30) = BC ÷ AC = 1 + (\sqrt{3}/2)\). \(x = \tan^{-1}(2/\sqrt{3}) - 30\) | M1, A1 | Use of tangent in 90° triangle – \(\tan = \text{opp/adj}\), x the subject – beware fortuitous answers |
**(i)** $AC = \cos 30 = \sqrt{3}/2$. $BC = 2\sin 30 = 1$. $AB = \sqrt{(1^2 + 3/4)} = \frac{\sqrt{7}}{2}$ | B1, B1, M1 A1 | Correct only – not decimal; Correct only; Use of Pythagoras. Correct only. Answer given. Could be cosine rule
**(ii)** $\tan(x + 30) = BC ÷ AC = 1 + (\sqrt{3}/2)$. $x = \tan^{-1}(2/\sqrt{3}) - 30$ | M1, A1 | Use of tangent in 90° triangle – $\tan = \text{opp/adj}$, x the subject – beware fortuitous answers6\\
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In the diagram, triangle $A B C$ is right-angled and $D$ is the mid-point of $B C$. Angle $D A C = 30 ^ { \circ }$ and angle $B A D = x ^ { \circ }$. Denoting the length of $A D$ by $l$,\\
(i) express each of $A C$ and $B C$ exactly in terms of $l$, and show that $A B = \frac { 1 } { 2 } l \sqrt { } 7$,\\
(ii) show that $x = \tan ^ { - 1 } \left( \frac { 2 } { \sqrt { } 3 } \right) - 30$.
\hfill \mbox{\textit{CAIE P1 2002 Q6 [6]}}