2026-08-31T11:35:07.751532 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Figure 6 shows the sector \(OAB\) of a circle wit...

(a) Length of the arc \(AB\) [2]

The formulae \(s = r\theta\) and \(A = \tfrac{1}{2}r^{2}\theta\) hold only when \(\theta\) is measured in radians, which is why the question states the angle as \(1.2\) radians rather than in degrees. No conversion is needed, and none should be attempted.

\[\text{arc } AB = r\theta = 8 \times 1.2 \quad \textbf{[M1]}\] \[= 9.6 \ \mathrm{cm} \quad \textbf{[A1]}\]

The M mark is for using \(r\theta\) with the given values and survives an arithmetic slip; the A mark needs the value with its unit.

(b) Area of the sector \(OAB\) [2]

\[\text{Area of sector} = \frac{1}{2}r^{2}\theta = \frac{1}{2} \times 8^{2} \times 1.2 = \frac{1}{2} \times 64 \times 1.2 \quad \textbf{[M1]}\] \[= 38.4 \ \mathrm{cm^{2}} \quad \textbf{[A1]}\]

Squaring the radius before halving is where slips occur: \(\tfrac{1}{2}(8)^{2} = 32\), not \(16\).

(c) Area of the segment, to 3 significant figures [3]

The chord \(AB\) cuts the sector into the triangle \(OAB\) and the segment between the chord and the arc. So the segment is what is left when the triangle is removed from the sector:

\[\text{segment} = \text{sector} - \text{triangle}.\]

The triangle has two sides equal to the radius with the known angle between them, so use \(\tfrac{1}{2}ab\sin C\) with \(a = b = r\):

\[\text{Area of } OAB = \frac{1}{2}r^{2}\sin\theta = \frac{1}{2} \times 64 \times \sin 1.2 = 32 \times 0.93203\ldots = 29.825\ldots \quad \textbf{[M1][A1]}\]

The M mark is for the subtraction structure together with a correct triangle formula, and it survives an evaluation slip; the A mark is for the triangle's area.

\[\text{Area of segment} = 38.4 - 29.825\ldots = 8.5747\ldots\] \[= 8.57 \ \mathrm{cm^{2}} \ (3\ \mathrm{s.f.}) \quad \textbf{[A1]}\]

The wrong turn this question is built to catch. Evaluating \(\sin 1.2\) with the calculator in degree mode. That returns \(0.02094\), giving a triangle area of \(0.670\) and a segment of \(37.7\) cm\(^{2}\), which is almost the whole sector and is plainly impossible. Set the calculator to radians for the whole question. A second common loss is premature rounding: using \(29.8\) rather than \(29.825\ldots\) gives \(8.6\), which fails the 3 significant figure requirement, so keep full accuracy until the final line.

Checks. The segment must be a modest fraction of the sector, since the triangle occupies most of it when the angle is well under \(\pi\); here \(8.57\) out of \(38.4\), roughly \(22\%\), is entirely reasonable for an angle of \(1.2\) radians. A further check: \(1.2\) radians is about \(68.75^{\circ}\), so the sector is a little under a fifth of the full circle of area \(64\pi \approx 201\) cm\(^{2}\), and one fifth of that is about \(40\) cm\(^{2}\), agreeing with part (b). Finally the perimeter of the sector, \(2(8) + 9.6 = 25.6\) cm, is a sensible magnitude beside the chord \(2 \times 8 \times \sin 0.6 = 9.04\) cm, which is slightly shorter than the arc as it must be.