| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Proving angle or length value |
| Difficulty | Challenging +1.2 This is a multi-part trigonometry problem requiring systematic application of sine/cosine rules across multiple triangles. While it involves several steps and careful angle tracking, the path is clearly signposted (triangle BED → triangle ABC → triangle BCD), and each step uses standard techniques without requiring novel geometric insight. The 'show that' format provides target values to work towards, reducing problem-solving demand. Slightly above average difficulty due to length and coordination required, but well within reach of a competent A-level student. |
| 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 angles1.05l Double angle formulae: and compound angle formulae |
8 In this question you must show detailed reasoning.
The diagram shows triangle $A B C$.\\
\includegraphics[max width=\textwidth, alt={}, center]{ec83c2c5-f8f8-4357-abfa-d40bc1d026b4-06_737_1383_456_342}
The angles $C A B$ and $A B C$ are each $45 ^ { \circ }$, and angle $A C B = 90 ^ { \circ }$.\\
The points $D$ and $E$ lie on $A C$ and $A B$ respectively. $A E = D E = 1 , D B = 2$. Angle $B E D = 90 ^ { \circ }$, angle $E B D = 30 ^ { \circ }$ and angle $D B C = 15 ^ { \circ }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $B C = \frac { \sqrt { 2 } + \sqrt { 6 } } { 2 }$.
\item By considering triangle $B C D$, show that $\sin 15 ^ { \circ } = \frac { \sqrt { 6 } - \sqrt { 2 } } { 4 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 Q8 [6]}}