| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Exact trigonometric values |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring standard techniques: (a) uses Pythagorean identity to find cos from sin, (b) applies cosine rule with exact values, and (c) uses sine rule. All steps are routine C2 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\cos^2 P = 1 - \left(\frac{2}{3}\right)^2 = \frac{5}{9}\) | M1 | |
| acute \(\therefore \cos \angle QPR = \sqrt{\frac{5}{9}} = \frac{1}{3}\sqrt{5}\) | A1 | |
| (b) \(QR^2 = 7^2 + (3\sqrt{5})^2 - (2 \times 7 \times 3\sqrt{5} \times \frac{1}{3}\sqrt{5})\) | M1 A1 | |
| \(QR^2 = 49 + 45 - 70 = 24\) | ||
| \(QR = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}\) | M1 A1 | |
| (c) \(\frac{\sin Q}{3\sqrt{5}} = \frac{\frac{1}{3}\sqrt{5}}{2\sqrt{6}}\) | M1 | |
| \(\sin Q = \frac{\sqrt{5}}{\sqrt{6}}\) | ||
| \(\angle PQR = 65.9°\) (1dp) | M1 A1 | (9) |
**(a)** $\cos^2 P = 1 - \left(\frac{2}{3}\right)^2 = \frac{5}{9}$ | M1 |
acute $\therefore \cos \angle QPR = \sqrt{\frac{5}{9}} = \frac{1}{3}\sqrt{5}$ | A1 |
**(b)** $QR^2 = 7^2 + (3\sqrt{5})^2 - (2 \times 7 \times 3\sqrt{5} \times \frac{1}{3}\sqrt{5})$ | M1 A1 |
$QR^2 = 49 + 45 - 70 = 24$ | |
$QR = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$ | M1 A1 |
**(c)** $\frac{\sin Q}{3\sqrt{5}} = \frac{\frac{1}{3}\sqrt{5}}{2\sqrt{6}}$ | M1 |
$\sin Q = \frac{\sqrt{5}}{\sqrt{6}}$ | |
$\angle PQR = 65.9°$ (1dp) | M1 A1 | **(9)**5.
Figure 2
Figure 2 shows triangle $P Q R$ in which $P Q = 7$ and $P R = 3 \sqrt { 5 }$.\\
Given that $\sin ( \angle Q P R ) = \frac { 2 } { 3 }$ and that $\angle Q P R$ is acute,
\begin{enumerate}[label=(\alph*)]
\item find the exact value of $\cos ( \angle Q P R )$ in its simplest form,
\item show that $Q R = 2 \sqrt { 6 }$,
\item find $\angle P Q R$ in degrees to 1 decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q5 [9]}}