# Question 3(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2} \times 3^2 \times \alpha = 7.2 \Rightarrow \alpha = \ldots$ or $\frac{1}{2} \times 3^2 \times 1.6 = 7.2 \Rightarrow \alpha = 1.6$ | M1 | Uses correct sector area formula and 7.2 to find $\alpha$. Must show values embedded in equation |
| $\alpha = 1.6^*$ | A1* | Correct proof starting with $\frac{1}{2} \times 3^2 \times \alpha = 7.2$ with at least one intermediate line and no errors. Must conclude $\alpha = 1.6$. If different variable used (e.g. $\theta$) must state $\alpha = 1.6$ |

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# Question 3(b)(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| Angle $COA = \frac{1}{2}(2\pi - 1.6) (= 2.34\ldots) \approx 134°$ | M1 | |
| Area $COA = \frac{1}{2} \times 5 \times 3 \sin("2.34") (= 5.38\ldots)$ | M1 | |
| Total Area $= 2 \times \frac{1}{2} \times 5 \times 3\sin("2.34") + 7.2$ | dM1 | Dependent on previous M marks |
| $= 18 \text{ cm}^2$ Awrt $18 \text{ cm}^2$ (Ans $= 17.96$) | A1 | |

**Alternative (b)(i):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $AB = 2 \times 3\sin 0.8$ | M1 | |
| $ON = 3\cos 0.8$ | M1 | |
| Total Area $= \frac{1}{2}(5 + ON) \times AB + 7.2 - \frac{1}{2} \times 3\cos 0.8 \times 2 \times 3\sin 0.8$ | dM1 | |
| $= 18 \text{ cm}^2$ Awrt $18 \text{ cm}^2$ | A1 | |

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# Question 3(b)(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| Arc $AB = 3 \times 1.6 (= 4.8)$ | B1 | |
| $AC^2 = 5^2 + 3^2 - 2 \times 5 \times 3\cos("2.34")$ | M1 | |
| Total perimeter $= 2 \times \sqrt{5^2 + 3^2 - 2 \times 5 \times 3\cos("2.34")} + 3 \times 1.6$ | dM1 | |
| $=$ Awrt $19.6$ cm | A1 | |

# Question b(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}(2\pi - 1.6)$ | M1 | Correct method for angle $COA$; sight of awrt 2.34 is sufficient; may be implied by $134°$ |
| $\frac{1}{2} \times 5 \times 3 \sin(\text{"2.34"})$ (= awrt 5.38/5.39) | M1 | Uses correct method for area of triangle $COA$ or $COB$; angle may be in degrees; if state $\frac{1}{2}ab\sin C$ but embed as $\frac{1}{2}abC$ condone as slip |
| $2 \times$ area of triangle $COA + 7.2$ | dM1 | Fully correct strategy for area; dependent on previous M mark; embedded values sufficient |
| awrt $18 \text{ cm}^2$ | A1 | **Must come from a correct method** |

**Alt b(i):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB = 2 \times 3\sin 0.8$ (awrt 4.30) | M1 | Find length $AB$ |
| Find length $ON$ where $N$ is midpoint of $AB$ (awrt 2.09) | M1 | — |
| $7.2 +$ area of triangle $ABC -$ area of triangle $AOB$ | dM1 | Fully correct strategy |
| awrt $18 \text{ cm}^2$ | A1 | **Must come from a correct method** |

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# Question b(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Arc length $= 4.8$ | B1 | Correct expression or value |
| Uses correct method for $AC^2$, $AC$, $CB^2$ or $CB$; awrt 54.9 or awrt 7.41 | M1 | Embedded values sufficient; condone confusion of $AC^2/CB^2$ and $AC/CB$; angle in degrees $\approx 134°$ |
| $2\times\sqrt{5^2+3^2-2\times5\times3\cos(\text{"2.34"})}+3\times1.6$ | dM1 | Dependent on B and M marks; must have square rooted $AC^2$ or $CB^2$ |
| awrt $19.6$ cm (Ans $= 19.619$) | A1 | **Must come from a correct method** |

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