| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2017 |
| Session | Specimen |
| Marks | 6 |
| Topic | Sine and Cosine Rules |
| Type | Exact trigonometric values |
| Difficulty | Standard +0.8 This is a multi-step geometry problem requiring careful application of trigonometry across multiple triangles with non-standard angles (15°, 30°, 45°). Part (a) requires working through triangle BED using sine rule and Pythagoras, then relating to the 45-45-90 triangle ABC. Part (b) requires deriving an exact value for sin 15° through geometric reasoning rather than standard formulas. The problem demands systematic reasoning across interconnected triangles and algebraic manipulation of surds, making it notably harder than routine mechanics questions but still within standard A-level scope. |
| 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 angles |
In this question you must show detailed reasoning.
The diagram shows triangle $ABC$.
\includegraphics{figure_8}
The angles $CAB$ and $ABC$ are each $45°$, and angle $ACB = 90°$.
The points $D$ and $E$ lie on $AC$ and $AB$ respectively. $AE = DE = 1$, $DB = 2$.
Angle $BED = 90°$, angle $EBD = 30°$ and angle $DBC = 15°$.
\begin{enumerate}[label=(\alph*)]
\item Show that $BC = \frac{\sqrt{2} + \sqrt{6}}{2}$. [3]
\item By considering triangle $BCD$, show that $\sin 15° = \frac{\sqrt{6} - \sqrt{2}}{4}$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2017 Q8 [6]}}