Question 3 - A-Level Maths

Edexcel P1 2021 June — Question 3 9 marks

Exam Board	Edexcel
Module	P1 (Pure Mathematics 1)
Year	2021
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Sine and Cosine Rules
Type	Algebraic side lengths
Difficulty	Standard +0.3 This is a straightforward application of the cosine rule with algebraic side lengths, followed by solving a quadratic equation. The steps are standard: apply cosine rule, substitute q = p + 2, solve the resulting quadratic, then use the triangle area formula. While it requires careful algebraic manipulation and working with surds, it follows a predictable pattern with no novel insights required, making it slightly easier than average.
Spec	1.02c Simultaneous equations: two variables by elimination and substitution 1.02f Solve quadratic equations: including in a function of unknown 1.05b Sine and cosine rules: including ambiguous case 1.05c Area of triangle: using 1/2 ab sin(C)

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

Figure 1 shows the plan view of a flower bed.
The flowerbed is in the shape of a triangle $A B C$ with

$A B = p$ metres
$A C = q$ metres
$B C = 2 \sqrt { 2 }$ metres
angle $B A C = 60 ^ { \circ }$
1. Show that

$$p ^ { 2 } + q ^ { 2 } - p q = 8$$ Given that side $A C$ is 2 metres longer than side $A B$, use algebra to find

the exact value of $p$,
the exact value of $q$. Using the answers to part (b),

calculate the exact area of the flower bed.

Show mark scheme Show mark scheme source

Question 3:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$(2\sqrt{2})^2 = p^2 + q^2 - 2pq\cos60°$	M1	Correct application of cosine rule
$p^2 + q^2 - pq = 8$ *	A1*	Completion to given result (proof)

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$q = p+2 \Rightarrow 8 = p^2 + (p+2)^2 - p(p+2)$	M1	Substitutes $q = p+2$ into their equation
$p^2 + 2p - 4 = 0$ or $q^2 - 2q - 4 = 0$	A1
$p = \frac{-2 \pm \sqrt{2^2 - 4(1)(-4)}}{2}$ or $q = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-4)}}{2}$	M1	Correct use of quadratic formula (or completing the square)
$p = -1 + \sqrt{5}$ or $q = 1 + \sqrt{5}$	B1 (A1 on EPEN)
$p = -1+\sqrt{5}$ and $q = 1+\sqrt{5}$ only	A1cso	Both values, only positive solutions accepted

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\text{Area} = \frac{1}{2} \times (-1+\sqrt{5})(1+\sqrt{5}) \times \sin60°$	M1	Correct formula with their $p$ and $q$ values
$\text{Area} = \sqrt{3}\ (\text{m}^2)$	A1

Question (Cosine Rule/Triangle - Part a):

Answer	Marks	Guidance
$(2\sqrt{2})^2 = p^2 + q^2 - 2pq\cos 60°$ or equivalent	M1	Award for correct application of cosine rule; condone missing brackets
$p^2 + q^2 - pq = 8$	A1*	No errors including omission of brackets; one of the M1 lines must have been seen; if = 8 stated without working, A1 cannot be scored

Question (Triangle - Part b):

Answer	Marks	Guidance
Substitutes $q = p + 2$ into the given equation	M1
$p^2 + 2p - 4 = 0$ or equivalent in $q$: $(q^2 - 2q - 4 = 0)$	A1
Attempts to solve using formula or completing the square	M1	Usual rules for solving quadratics apply; if quadratic formula used values must be embedded
$p = -1 + \sqrt{5}$ or $q = 1 + \sqrt{5}$	B1	Ignore any other solutions; must be exact; independent of previous method mark
$p = -1 + \sqrt{5}$ and $q = 1 + \sqrt{5}$	A1	Only cso; all other marks must have been scored

Question (Triangle - Part c):

Answer	Marks	Guidance
Attempts area using $\frac{1}{2} \times (-1+\sqrt{5})(1+\sqrt{5}) \times \sin 60°$	M1	Must see at least one stage of working using their $p$ and $q$
$\sqrt{3}$ (m²)	A1	Condone lack of units; do not accept rounded answers

# Question 3:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(2\sqrt{2})^2 = p^2 + q^2 - 2pq\cos60°$ | M1 | Correct application of cosine rule |
| $p^2 + q^2 - pq = 8$ * | A1* | Completion to given result (proof) |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $q = p+2 \Rightarrow 8 = p^2 + (p+2)^2 - p(p+2)$ | M1 | Substitutes $q = p+2$ into their equation |
| $p^2 + 2p - 4 = 0$ or $q^2 - 2q - 4 = 0$ | A1 | |
| $p = \frac{-2 \pm \sqrt{2^2 - 4(1)(-4)}}{2}$ or $q = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-4)}}{2}$ | M1 | Correct use of quadratic formula (or completing the square) |
| $p = -1 + \sqrt{5}$ or $q = 1 + \sqrt{5}$ | B1 (A1 on EPEN) | |
| $p = -1+\sqrt{5}$ and $q = 1+\sqrt{5}$ only | A1cso | Both values, only positive solutions accepted |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Area} = \frac{1}{2} \times (-1+\sqrt{5})(1+\sqrt{5}) \times \sin60°$ | M1 | Correct formula with their $p$ and $q$ values |
| $\text{Area} = \sqrt{3}\ (\text{m}^2)$ | A1 | |

# Question (Cosine Rule/Triangle - Part a):

$(2\sqrt{2})^2 = p^2 + q^2 - 2pq\cos 60°$ or equivalent | M1 | Award for correct application of cosine rule; condone missing brackets

$p^2 + q^2 - pq = 8$ | A1* | No errors including omission of brackets; one of the M1 lines must have been seen; if = 8 stated without working, A1 cannot be scored

---

# Question (Triangle - Part b):

Substitutes $q = p + 2$ into the given equation | M1 |

$p^2 + 2p - 4 = 0$ or equivalent in $q$: $(q^2 - 2q - 4 = 0)$ | A1 |

Attempts to solve using formula or completing the square | M1 | Usual rules for solving quadratics apply; if quadratic formula used values must be embedded

$p = -1 + \sqrt{5}$ **or** $q = 1 + \sqrt{5}$ | B1 | Ignore any other solutions; must be exact; independent of previous method mark

$p = -1 + \sqrt{5}$ **and** $q = 1 + \sqrt{5}$ | A1 | Only cso; all other marks must have been scored

---

# Question (Triangle - Part c):

Attempts area using $\frac{1}{2} \times (-1+\sqrt{5})(1+\sqrt{5}) \times \sin 60°$ | M1 | Must see at least one stage of working using their $p$ and $q$

$\sqrt{3}$ (m²) | A1 | Condone lack of units; do not accept rounded answers

---

Show LaTeX source

\begin{enumerate}
  \item In this question you must show all stages of your working.
\end{enumerate}

Solutions relying on calculator technology are not acceptable.

\begin{tikzpicture}[scale=0.9]

  % Coordinates
  \coordinate[label=below left:{$A$}]  (A) at (0,0);
  \coordinate[label=above:{$B$}]       (B) at (1.5,4);
  \coordinate[label=below right:{$C$}] (C) at (8,0);

  % Triangle
  \draw (A) -- (B) -- (C) -- cycle;

  % Angle arc at A (60°)
  \pic[draw, angle radius=10mm, angle eccentricity=1.4, "$60^\circ$" font=\small]
      {angle = C--A--B};
      

  % Side labels
  \node[left]  at ($(A)!0.5!(B)$) {$p\,\mathrm{m}$};
  \node[below] at ($(A)!0.5!(C)$) {$q\,\mathrm{m}$};
  \node[above right] at ($(B)!0.5!(C)$) {$2\sqrt{2}\,\mathrm{m}$};

\end{tikzpicture}

Figure 1 shows the plan view of a flower bed.\\
The flowerbed is in the shape of a triangle $A B C$ with

\begin{itemize}
  \item $A B = p$ metres
  \item $A C = q$ metres
  \item $B C = 2 \sqrt { 2 }$ metres
  \item angle $B A C = 60 ^ { \circ }$\\
(a) Show that
\end{itemize}

$$p ^ { 2 } + q ^ { 2 } - p q = 8$$

Given that side $A C$ is 2 metres longer than side $A B$, use algebra to find\\
(b) (i) the exact value of $p$,\\
(ii) the exact value of $q$.

Using the answers to part (b),\\
(c) calculate the exact area of the flower bed.\\

\hfill \mbox{\textit{Edexcel P1 2021 Q3 [9]}}

This paper (9 questions)

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Q1 8 Q2 10 Q3 9 Q4 6 Q5 8 Q6 8 Q7 10 Q8 9 Q9 7

Answer	Marks	Guidance
\((2\sqrt{2})^2 = p^2 + q^2 - 2pq\cos 60°\) or equivalent	M1	Award for correct application of cosine rule; condone missing brackets
\(p^2 + q^2 - pq = 8\)	A1*	No errors including omission of brackets; one of the M1 lines must have been seen; if = 8 stated without working, A1 cannot be scored

Answer	Marks	Guidance
Substitutes \(q = p + 2\) into the given equation	M1
\(p^2 + 2p - 4 = 0\) or equivalent in \(q\): \((q^2 - 2q - 4 = 0)\)	A1
Attempts to solve using formula or completing the square	M1	Usual rules for solving quadratics apply; if quadratic formula used values must be embedded
\(p = -1 + \sqrt{5}\) or \(q = 1 + \sqrt{5}\)	B1	Ignore any other solutions; must be exact; independent of previous method mark
\(p = -1 + \sqrt{5}\) and \(q = 1 + \sqrt{5}\)	A1	Only cso; all other marks must have been scored

Answer	Marks	Guidance
Attempts area using \(\frac{1}{2} \times (-1+\sqrt{5})(1+\sqrt{5}) \times \sin 60°\)	M1	Must see at least one stage of working using their \(p\) and \(q\)
\(\sqrt{3}\) (m²)	A1	Condone lack of units; do not accept rounded answers