Edexcel C2 — Question 2 4 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Marks4
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Mark schemeDownload PDF ↗
TopicSine and Cosine Rules
TypeAlgebraic side lengths
DifficultyModerate -0.3 This is a straightforward application of the cosine rule with a standard angle (60°). Students substitute given expressions into the formula, expand a quadratic, and solve. It's slightly easier than average because it's a direct single-method question with no conceptual challenges, though the algebra requires care.
Spec1.05b Sine and cosine rules: including ambiguous case
\includegraphics{figure_1} Figure 1 shows triangle \(PQR\) in which \(PQ = x\), \(PR = 7 - x\), \(QR = x + 1\) and \(\angle PQR = 60°\). Using the cosine rule, find the value of \(x\). [4]
AnswerMarks Guidance
\((7-x)^2 = x^2 + (x+1)^2 - [2 \times x \times (x+1) \times \cos 60°]\)M1 A1
\(49 - 14x + x^2 = x^2 + x^2 + 2x + 1 - x^2 - x\)M1 A1
\(15x = 48\), \(x = \frac{16}{5}\)M1 A1 (4) 
$(7-x)^2 = x^2 + (x+1)^2 - [2 \times x \times (x+1) \times \cos 60°]$ | M1 A1 |
$49 - 14x + x^2 = x^2 + x^2 + 2x + 1 - x^2 - x$ | M1 A1 |
$15x = 48$, $x = \frac{16}{5}$ | M1 A1 | (4)
\includegraphics{figure_1}

Figure 1 shows triangle $PQR$ in which $PQ = x$, $PR = 7 - x$, $QR = x + 1$ and $\angle PQR = 60°$.

Using the cosine rule, find the value of $x$. [4]

\hfill \mbox{\textit{Edexcel C2  Q2 [4]}}
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