| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2005 |
| Session | November |
| Marks | 5 |
| 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 coordinate geometry problem requiring systematic application of trigonometry and the tangent formula. Part (i) involves straightforward right-angle trigonometry to find CD. Part (ii) requires finding coordinates or using angle relationships to verify a given result. While multi-step, it follows standard techniques without requiring novel insight—slightly easier than average due to the 'show that' format providing the target answer. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(2d\sin30 + 2d\sqrt{3}\sin60 = 2d.\frac{1}{2} + 2d\sqrt{3}.\frac{\sqrt{3}}{2} = 4d\) | M1 A1 | For one - allow even if decimals used Co – exact answer only. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = \frac{2}{\sqrt{3}}\) | M1 DM1 A1 | Use of \(\tan = \text{opp}/\text{adj}\) in correct triangle For horizontal step attempted Co. Lost if decimals used. |
(i) $2d\sin30 + 2d\sqrt{3}\sin60 = 2d.\frac{1}{2} + 2d\sqrt{3}.\frac{\sqrt{3}}{2} = 4d$ | M1 A1 | For one - allow even if decimals used Co – exact answer only. | [2]
(ii) $\tan\theta = \frac{\text{ans to (i)}}{2d\cos30 + 2\sqrt{3}d\cos60}$
$\tan\theta = \frac{2}{\sqrt{3}}$ | M1 DM1 A1 | Use of $\tan = \text{opp}/\text{adj}$ in correct triangle For horizontal step attempted Co. Lost if decimals used. | [3]3\\
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In the diagram, $A B E D$ is a trapezium with right angles at $E$ and $D$, and $C E D$ is a straight line. The lengths of $A B$ and $B C$ are $2 d$ and $( 2 \sqrt { 3 } ) d$ respectively, and angles $B A D$ and $C B E$ are $30 ^ { \circ }$ and $60 ^ { \circ }$ respectively.\\
(i) Find the length of $C D$ in terms of $d$.\\
(ii) Show that angle $C A D = \tan ^ { - 1 } \left( \frac { 2 } { \sqrt { 3 } } \right)$.
\hfill \mbox{\textit{CAIE P1 2005 Q3 [5]}}