| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Algebraic side lengths |
| Difficulty | Standard +0.8 This question requires multiple sophisticated steps: deriving an algebraic constraint from the area formula (part a), then combining this with the circle geometry theorem (angle in semicircle = 90°) and cosine rule to solve a system of equations (part b). The algebraic manipulation is non-trivial, requiring students to connect geometric properties across different contexts rather than apply a single standard technique. |
| Spec | 1.03f Circle properties: angles, chords, tangents1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}xy\sin60 = \sqrt{3}(x+y)\) Must have \(xy\), not \(ab\) | M1 | Verification method: eg substitute \(4x+4y\) for \(xy\) in \(\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)xy=\sqrt{3}(x+y)\): M1; Condone incorrect use of brackets eg \(x+y(\sqrt{3})\) if next line shows intention is correct |
| \(\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)xy = \sqrt{3}(x+y)\) oe \(\Rightarrow 4x+4y=xy\) (AG) | A1 [2] | obtain \(\sqrt{3}(x+y)=\sqrt{3}(x+y)\): A1; Must show sufficient working to justify the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Angle \(ABC\) is \(90°\) | B1* | May be implied by correct trig statement; Possibly on diagram |
| \(\cos60=\frac{x}{y}\) or \(\frac{x}{y}=\frac{1}{2}\) or \(y=2x\) | B1 | or any correct \(x,y\) equation eg \(\frac{x}{\sin30}=\frac{y}{\sin90}\); Dep B1* |
| \(4x+4(2x)=x(2x)\) oe or \(4\left(\frac{y}{2}\right)+4y=y\times\frac{y}{2}\) oe | M1 | or \(\sqrt{3}(x+2x)=\frac{1}{2}\times x\times 2x\times\frac{\sqrt{3}}{2}\); Obtain a correct equation in \(x\) or \(y\) only |
| Answer | Marks | Guidance |
|---|---|---|
| \(x=6,\ y=12\) | A1 [4] | No need to consider \(x=0\) or \(y=0\) |
## Question 6:
### Part (a):
$\frac{1}{2}xy\sin60 = \sqrt{3}(x+y)$ Must have $xy$, not $ab$ | **M1** | Verification method: eg substitute $4x+4y$ for $xy$ in $\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)xy=\sqrt{3}(x+y)$: M1; Condone incorrect use of brackets eg $x+y(\sqrt{3})$ if next line shows intention is correct
$\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)xy = \sqrt{3}(x+y)$ oe $\Rightarrow 4x+4y=xy$ **(AG)** | **A1** [2] | obtain $\sqrt{3}(x+y)=\sqrt{3}(x+y)$: A1; Must show sufficient working to justify the given answer
### Part (b):
Angle $ABC$ is $90°$ | **B1*** | May be implied by correct trig statement; Possibly on diagram
$\cos60=\frac{x}{y}$ or $\frac{x}{y}=\frac{1}{2}$ or $y=2x$ | **B1** | or any correct $x,y$ equation eg $\frac{x}{\sin30}=\frac{y}{\sin90}$; Dep B1*
$4x+4(2x)=x(2x)$ oe or $4\left(\frac{y}{2}\right)+4y=y\times\frac{y}{2}$ oe | **M1** | or $\sqrt{3}(x+2x)=\frac{1}{2}\times x\times 2x\times\frac{\sqrt{3}}{2}$; Obtain a correct equation in $x$ or $y$ only
$(\Rightarrow x(x-6)=0)$ or $\Rightarrow y(y-12)=0$
$x=6,\ y=12$ | **A1** [4] | No need to consider $x=0$ or $y=0$
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{68f1107f-f188-4698-934e-8fd593b25418-4_442_661_840_260}
The diagram shows triangle $A B C$, with $A B = x \mathrm {~cm} , A C = y \mathrm {~cm}$ and angle $B A C = 60 ^ { \circ }$. It is given that the area of the triangle is $( x + y ) \sqrt { 3 } \mathrm {~cm} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $4 x + 4 y = x y$.
When the vertices of the triangle are placed on the circumference of a circle, $A C$ is a diameter of the circle.
\item Determine the value of $x$ and the value of $y$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q6 [6]}}