22M.2.SL.TZ2.1

EITHER

uses the cosine rule           (M1)

${\text{AB}}^2=5^2+5^2-2\times 5\times 5\times \cos 1.9$           (A1)

OR

uses right-angled trigonometry           (M1)

$\frac{{\displaystyle \frac{\text{AB}}{2}}}{5}\sin 0.95$           (A1)

OR

uses the sine rule           (M1)

$\alpha =\frac{1}{2}\left(\mathrm{\pi }-1.9\right)\left(=0.6207\dots \right)$

$\frac{\text{AB}}{\sin 1.9}=\frac{5}{\sin 0.6207\dots }$           (A1)

THEN

$\text{AB}=8.13415\dots$

$\text{AB}=8.13 \left(\text{m}\right)$           A1

 

[3 marks]

let the shaded area be $A$

METHOD 1

attempt at finding reflex angle           (M1)

$\text{A}\hat{\text{O}}\text{B}=2\mathrm{\pi }-1.9 \left(=4.3831\dots \right)$

substitution into area formula           (A1)

$A=\frac{1}{2}\times 5^2\times 4.3831\dots$  OR  $\left(\frac{2\mathrm{\pi }-1.9}{2\mathrm{\pi }}\right)\times \mathrm{\pi }\left(5^2\right)$

$=54.7898\dots$

$=54.8 \left({\text{m}}^2\right)$           A1

 

METHOD 2

let the area of the circle be $A_C$ and the area of the unshaded sector be $A_U$

$A=A_C-A_U$           (M1)

$A=\mathrm{\pi }\times 5^2-\frac{1}{2}\times 5^2\times 1.9 \left(=78.5398\dots -23.75\right)$           (A1)

$=54.7898\dots$

$=54.8 \left({\text{m}}^2\right)$           A1

 

[3 marks]

Most students used the cosine rule to correctly find AB in part (a), although many found the arc length instead of the chord.

Part (b) was generally correctly solved. Some candidates found the area of the unshaded region rather than the shaded one.

[N/A]