**(a)**

| Attempts $r\theta = 5 \times 1.2$ | M1 | Attempts $r\theta = 5 \times 1.2$ |
|---|---|---|
| Perimeter $= 5 + 5 + 6 = 16$ (km) | A1 (2) | Achieves 16 (km) |

**(b)**

| Attempts $\frac{1}{2}r^2\theta = \frac{1}{2} \times 5^2 \times 1.2$ | M1 | Attempts $\frac{1}{2}r^2\theta = \frac{1}{2} \times 5^2 \times 1.2$ (or $\frac{1}{2}r l = \frac{1}{2} \times 5 \times 6$) May be implied by Area of sector = 15 seen. |
| Area $AOP = \frac{1}{4} \times \left(\frac{1}{2} \times 5^2 \times 1.2\right) = 3.75$ km² * | M1, A1* (3) | Attempts $\frac{1}{4} \times \left(\frac{1}{2} \times 5^2 \times 1.2\right)$ or $\frac{3}{4} \times \left(\frac{1}{2} \times 5^2 \times 1.2\right)$ or other full method to find the area for one of the two regions. Achieves given answer, including units, showing all steps with no errors and full accuracy kept throughout. Score A0 if rounded values were used during working. |

**(c)**

| Sets $\frac{1}{2} \times 5 \times OP \times \sin 1.2 = 3.75 \Rightarrow OP = ...$ | M1 | Sets $\frac{1}{2} \times 5 \times OP \times \sin 1.2 = 3.75 \Rightarrow OP = ...$ There are some longer methods to find OP, so this may be scored for any complete method. E.g finds area of triangle $APB$ from difference of area triangle $OAB = \left(\frac{1}{2}5^2 \sin 1.2\right)$ and 3.75, the uses ratio $OP:PB$ is same as ratio of these areas to form and solve an equation for OP. Other method may be possible. Or $\frac{1}{2} OP \cdot h = 3.75, h = 5 \sin 1.2 \Rightarrow OP = ...$ |
| $OP = 1.6...$ | A1 | Achieves distance $OP =$ awrt 1.6 (allow if called $AP$). May be implied by subsequent work. |
| $AP^2 = 5^2 + "1.6..."^2 - 2 \times 5 \times "1.6..." \times \cos 1.2$ | M1 | Attempts the cosine rule, or other full method, to find at least $AP^2$ using angle 1.2, $OA = 5$ and their $OP$. Condone if $AP^2$ is missing or labelled $AP$. |
| $AP = 4.7\text{km}$ or $4700\text{m}$ | A1cso (4) | $AP = 4.7\text{km}$ or $4700\text{m}$ (isw once seen if further rounding occurs). Must include the correct units and is given to nearest 100m. |

**Total: 9 marks**

**Guidance for (a):**
- M1: Attempts $r\theta = 5 \times 1.2$
- A1: Achieves 16 (km)

**Guidance for (b):**
- M1: Attempts $\frac{1}{2}r^2\theta = \frac{1}{2} \times 5^2 \times 1.2$ (or $\frac{1}{2}r l = \frac{1}{2} \times 5 \times 6$) May be implied by Area of sector = 15 seen.
- M1: Attempts $\frac{1}{4} \times \left(\frac{1}{2} \times 5^2 \times 1.2\right)$ or $\frac{3}{4} \times \left(\frac{1}{2} \times 5^2 \times 1.2\right)$ or other full method to find the area for one of the two regions.
- A1*: Achieves given answer, including units, showing all steps with no errors and full accuracy kept throughout. Score A0 if rounded values were used during working.

Approaches by verification are acceptable. The first M1 will apply the same way, a clear attempt at the sector area formula will be needed to determine the area.
The second M may then be gained for attempting Area Sector $AOB$ = Area $R_1$ + Area $R_2$ = $3 \times 3.75 + 3.75 = ...$

Then for the A mark the calculations need to both be correct (giving 15), and a conclusion made: E.g. (so) Area $R_2 = 3.75$ km².

**Guidance for (c):**
- M1: Sets $\frac{1}{2} \times 5 \times OP \times \sin 1.2 = 3.75 \Rightarrow OP = ...$ There are some longer methods to find OP, so this may be scored for any complete method. E.g finds area of triangle $APB$ from difference of area triangle $OAB = \left(\frac{1}{2}5^2 \sin 1.2\right)$ and 3.75, the uses ratio $OP:PB$ is same as ratio of these areas to form and solve an equation for OP. Other method may be possible. Or $\frac{1}{2} OP \cdot h = 3.75, h = 5 \sin 1.2 \Rightarrow OP = ...$

- A1: Achieves distance $OP =$ awrt 1.6 (allow if called $AP$). May be implied by subsequent work.

- M1: Attempts the cosine rule, or other full method, to find at least $AP^2$ using angle 1.2, $OA = 5$ and their $OP$. Condone if $AP^2$ is missing or labelled $AP$.

- A1cso: $AP = 4.7\text{km}$ or $4700\text{m}$ (isw once seen if further rounding occurs). Must include the correct units and is given to nearest 100m.

Watch out for using of $\frac{1}{2} \times 5 \times OP \times 1.2 = 3.75 \Rightarrow OP = 1.25$ which also leads to 4.7km. This scores at most M0A0M1A0 despite giving the same answer.

SC If $R_1$ and $R_2$ are mixed up in part (c) then score as a misread if the work is otherwise correct – so M1 for attempting $\frac{1}{2} \times 5 \times OP \times \sin 1.2 = 11.25 \Rightarrow OP = ...$ , and dM1 for a correct use of the cosine rule to find $AP$.

Some methods do not find $OP$, but a different intermediate length for a triangle containing $AP$. Such methods score the M1 for the attempt to find two other side lengths for a triangle with side $AP$ and the A1 for a correct length, then second M for full method to find at least $AP^2$. For example:

| Let $X$ be base of perpendicular to $OA$ though $P$, then $\frac{1}{2} \times 5 \times XP = 3.75 \Rightarrow XP = 1.5 \Rightarrow OX = \frac{\text{"1.5"}}{\tan 1.2}$ so | M1 |
|---|---|
| $AX = 5 - ... = ...$ | A1 |
| $AP^2 = "4.416..."^2 + "1.5"^2 = ...$ | M1 |
| $AP = 4.7\text{km}$ or $4700\text{m}$ | A1 |
| (4) (9 marks) | |

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## Question 5(c) Alt:

| Let $X$ be base of perpendicular to $OA$ though $P$, then $\frac{1}{2} \times 5 \times XP = 3.75 \Rightarrow XP = 1.5 \Rightarrow OX = \frac{\text{"1.5"}}{\tan 1.2}$ so | M1 |
|---|