Find AB.

Find the area of the shaded region.

METHOD 1

choosing cosine rule     (M1)

substituting correctly     A1

e.g. \({\rm{AB}} = \sqrt {{{3.9}^2} + {{3.9}^2} - 2(3.9)(3.9)\cos 1.8} \)

\({\rm{AB}} = 6.11\) (cm)     A1     N2

METHOD 2

evidence of approach involving right-angled triangles     (M1)

substituting correctly     A1

e.g. \(\sin 0.9 = \frac{x}{{3.9}}\) , \(\frac{1}{2}{\rm{AB}} = 3.9\sin 0.9\)

\({\rm{AB}} = 6.11\) (cm)     A1     N2

METHOD 3

choosing the sine rule     (M1)

substituting correctly     A1

e.g. \(\frac{{\sin 0.670 \ldots }}{{3.9}} = \frac{{\sin 1.8}}{{{\rm{AB}}}}\)

\({\rm{AB}} = 6.11\) (cm)     A1     N2

[3 marks]

METHOD 1

reflex \({\rm{A}}\widehat {\rm{O}}{\rm{B}} = 2\pi  - 1.8\) \(( = 4.4832)\)     (A2)

correct substitution \(A = \frac{1}{2}{(3.9)^2}(4.4832 \ldots )\)     A1

area =34.1 (cm²)     A1     N2

METHOD 2

finding area of circle \(A = \pi {(3.9)^2}\) \(( = 47.78 \ldots )\)     (A1)

finding area of (minor) sector \(A = \frac{1}{2}{(3.9)^2}(1.8)\) \(( = 13.68 \ldots )\)    (A1)

subtracting     M1

e.g. \(\pi {(3.9)^2} - 0.5{(3.9)^2}(1.8)\) , \(47.8 - 13.7\)

area = 34.1 (cm²)    A1     N2

METHOD 3

finding reflex \({\rm{A}}\widehat {\rm{O}}{\rm{B}} = 2\pi  - 1.8\) \(( = 4.4832)\)     (A2)

finding proportion of total area of circle     A1

e.g. \(\frac{{2\pi  - 1.8}}{{2\pi }} \times \pi {(3.9)^2}\) , \(\frac{\theta }{{2\pi }} \times \pi {r^2}\)

area = 34.1 (cm²)     A1     N2

[4 marks]

This question was well answered by the majority of candidates. Full solutions were common in both parts, and a variety of successful approaches were used. Radians were well handled with few candidates working with the angle in degrees. Some candidates incorrectly found the length of the arc subtended by the central angle rather than the length of segment [AB].

In part (b), some candidates incorrectly subtracted the area of the triangle or even a length. Many candidates failed to give answers to 3 significant figures and therefore lost an accuracy mark.