June 2025 Paper 2 Q4

Notes

It is acceptable in this question to work in degrees and convert if necessary.
NB Angle ABC in degrees is 22.591…°

M1: Attempts to use the cosine rule with values correctly placed. May be implied by, e.g., \(\cos\theta = \dfrac{8^2 + 13^2 - 6.4^2}{2 \times 8 \times 13}\) or \(\cos^{-1}\left(\dfrac{6.4^2 - 8^2 - 13^2}{-2 \times 8 \times 13}\right)\)
Condone slips in substitution if a correct cosine formula is seen. The placement of the values should be correct. So, condone e.g. a missing 2 once a correct cosine formula is seen.
Alt 1: attempts the cosine rule and finds angle \(ADB\) (= awrt 0.5 or \(29^\circ\)) or angle \(BAD\) (= awrt 2.2 or 2.3 or \(129^\circ\)) and uses the sine rule correctly with angle \(ABC\) involved.
Alt 2: attempts the cosine rule and finds angle \(ADB\) (= awrt 0.5 or \(29^\circ\)) and angle \(BAD\) (= awrt 2.2 or 2.3 or \(129^\circ\)) and sums angles \(ADB\), \(BAD\), and \(ABC\) to \(\pi\) (or \(180^\circ\))
Alt 3: using Pythagoras and simultaneous equations to find the length of \(AN\) or \(BN\) (see diagram below) and uses e.g. \(\cos ABC = \dfrac{BN}{8}\) or \(\sin ABC = \dfrac{AN}{8}\)

A1*: cso Correct proof.
If starting with \(6.4^2 = 8^2 + 13^2 - 2(8)(13)\cos ABC\) then there must be an intermediate line such as

• \(\cos ABC = \dfrac{8^2 + 13^2 - 6.4^2}{2 \times 8 \times 13}\) or angle \(ABC = \cos^{-1}\left(\dfrac{8^2 + 13^2 - 6.4^2}{2 \times 8 \times 13}\right)\)
• \(\cos ABC =\) awrt 0.92 or \(\dfrac{4801}{5200}\)
• angle \(ABC =\) awrt 0.3943

The minimum required is e.g. \(\cos^{-1}\left(\dfrac{8^2 + 13^2 - 6.4^2}{2 \times 8 \times 13}\right)\) or \(\cos\theta = \dfrac{8^2 + 13^2 - 6.4^2}{2 \times 8 \times 13} \rightarrow \theta = 0.394\) both of which score M1A1*.
The final value must be 0.394 and not e.g. 0.3943. There should be no obvious incorrect statements in the proof e.g. \(\cos(0.92) = 0.394\) and no obvious incorrect work.
Allow the use of, e.g., \(\theta\), \(x\), or even e.g. \(A\) throughout. Mention of radians is not required.
Those working in degrees and achieve 22.6 can convert to 0.394 without calculation for M1A1
Attempts via verification e.g. \(AD^2 = 8^2 + 13^2 - 2(8)(13)\cos 0.394 \rightarrow AD =\) awrt 6.4 score maximum M1A0*.

Helpful Diagram:

Notes

It is acceptable in this question to work in degrees and convert if necessary.
NB Angle ABC in degrees is 22.591…°

Note: you may need to check the diagram for working.

M1: Attempts the area of the sector or the area of the triangle \(ABD\) via a correct method.
May be implied by a full attempt at the area of \(R\).
The sector area may be implied by \(\dfrac{1576}{125}\) or by \(\dfrac{0.394}{2\pi} \times \pi \times 8^2\) or \(\dfrac{22.6}{360} \times \pi \times 8^2\)
The angle \(ABC\) is given in the question as 0.394 and should be used (or a more accurate value). Do not allow use of a different value for angle \(ABC\) (other than the angle in degrees).

There are many acceptable alternative approaches to find the area of triangle \(ABD\), e.g. using their angle \(ADB\) or their angle \(BAD\) or using right-angled triangles
e.g. \(\dfrac{1}{2}(13)(\text{``}3.07\text{''})\) or \(\dfrac{1}{2}(\text{``}7.39\text{''})(\text{``}3.07\text{''}) + \dfrac{1}{2}(\text{``}5.614\text{''})(\text{``}3.07\text{''})\) and could be implied by e.g. 11.3 + 8.6

Alternatively, attempts the area of the triangle \(ACD\) \(\left(\text{e.g. } \dfrac{1}{2}(5)(6.4)\sin\text{``}0.501\text{''} = \text{awrt } 7.68\right)\)
or the area of the segment (in sector \(ABC\)) \(\left(\dfrac{1}{2}\left(8^2\right)(0.394) - \dfrac{1}{2}\left(8^2\right)\sin 0.394 = \text{awrt } 0.324\right)\)
via a correct method. The area of the triangle \(ABC\) alone is insufficient for this mark.
Note that \(\sin 0.394\) might be seen as \(\sqrt{1 - \left(\dfrac{4801}{5200}\right)^2}\)
For values, see the diagram below.

dM1: Complete and correct method for the area of \(R\). Requires correct attempts at both the area of the sector and the area of the triangle \(ABD\) and subtraction of the two values (or expressions). Allow sector – triangle \(ABD\) if this is then recovered by making the area positive.

Alternatively, correct attempts at both the area of the triangle \(ACD\) and the area of the segment (in sector \(ABC\)) and subtraction of the two values (or expressions).

Condone slips in substitution provided a correct formula is seen.
May be implied by a correct answer.

A1: awrt 7.34 to 7.37 no units required but penalise incorrect units.
ISW after an acceptable answer is seen.

Helpful Diagram: