(i)     Show that ${p^2}(41 - 40\cos 0.7) = 36$ .

(ii)    Find p .

Write down the length of BD.

Find ${\rm{A}}\widehat {\rm{D}}{\rm{B}}$ .

(i)     Show that ${\rm{C}}\widehat {\rm{B}}{\rm{D}} = 1.29$ radians, correct to 2 decimal places.

(ii)    Hence, find the area of the shaded region.

(i) evidence of valid approach     (M1)

e.g. choosing cosine rule

correct substitution     (A1)

e.g. ${6^2} = {(5p)^2} + {(4p)^2} - 2 \times (4p) \times (5p)\cos 0.7$

simplification      A1

e.g. $36 = 25{p^2} + 16{p^2} - 40{p^2}\cos 0.7$

${p^2}(41 - 40\cos 0.7) = 36$     AG     N0

(ii) $1.85995 \ldots$

$p = 1.86$     A1     N1

Note: Award A0 for $p = \pm 1.86$ , i.e. not rejecting the negative value.

[4 marks]

${\text{BD}} = 6$     A1     N1

[1 mark]

evidence of valid approach     (M1)

e.g. choosing sine rule

correct substitution     A1

e.g. $\frac{{\sin {\rm{A}}\widehat {\rm{D}}{\rm{B}}}}{{4p}} = \frac{{\sin 0.7}}{6}$

${\text{acute }}{\rm{A}}\widehat {\rm{D}}{\rm{B = 0}}{\rm{.9253166}} \ldots$     (A1)

$\pi  - 0.9253166 \ldots = 2.216275 \ldots$

${\rm{A}}\widehat {\rm{D}}{\rm{B}} = 2.22$     A1     N3

[4 marks]

(i) evidence of valid approach     (M1)

e.g. recognize isosceles triangle, base angles equal

$\pi  - 2(0.9253 \ldots )$     A1

${\rm{C}}\widehat {\rm{B}}{\rm{D}} = 1.29$     AG     N0

(ii) area of sector BCD     (A1)

e.g. $0.5 \times (1.29) \times {(6)^2}$

area of triangle BCD     (A1)

e.g. $0.5 \times {(6)^2}\sin 1.29$

evidence of subtraction     M1

$5.92496 \ldots$

$5.937459 \ldots$

${\text{area}} = 5.94$     A1     N3

[6 marks]

There were mixed results with this question. Most candidates could access part (a) and made the correct choice with the cosine rule but sloppy notation often led to candidates not being able to show the desired result.

There were mixed results with this question. Most candidates could access part (a) and made the correct choice with the cosine rule but sloppy notation often led to candidates not being able to show the desired result.

In part (c), candidates again correctly identified an appropriate method but failed to recognize that their result of 0.925 was acute and not obtuse as required.

In (d) (i), many attempted to use the sine rule under the incorrect assumption that DC was equal to 5p, rather than rely on some basic isosceles triangle geometry. Consequently, the result of 1.29 for ${\rm{C}}\widehat {\rm{B}}{\rm{D}}$ was not easy to show. There was a great deal of success with (d) (ii) with candidates using appropriate techniques to find the area of the shaded region although some stopped after finding the area of the sector.