Show that \({\rm{OC}} = r\cos 1.4\) .

The area of the shaded region is \(25\) cm² . Find the value of \(r\) .

use right triangle trigonometry     A1

eg   \(\cos 1.4 = \frac{{{\rm{OC}}}}{r}\)

\({\rm{OC}} = r\cos 1.4\)     AG     N0

[1 mark]

correct value for BC

eg   \({\rm{BC}} = r\sin 1.4\) , \(\sqrt {{r^2} - {{(r\cos 1.4)}^2}} \)     (A1)

area of \(\Delta {\rm{OBC}} = \frac{1}{2}r\sin 1.4 \times r\cos 1.4\) \(\left( { = \frac{1}{2}{r^2}\sin 1.4 \times \cos 1.4} \right)\)     A1

area of sector \({{\rm{OAB}} = \frac{1}{2}{r^2} \times 1.4}\)     A1

attempt to subtract in any order     (M1)

eg sector – triangle, \({\frac{1}{2}{r^2}\sin 1.4 \times \cos 1.4 - 0.7{r^2}}\)

correct equation     A1

eg   \({0.7{r^2} - \frac{1}{2}r\sin 1.4 \times r\cos 1.4 = 25}\)

attempt to solve their equation     (M1)

eg sketch, writing as quadratic, \(\frac{{25}}{{0.616 \ldots }}\)

\(r = 6.37\)     A1     N4

[7 marks]

Note: Exception to FT rule. Award A1FT for a correct FT answer from a quadratic equation involving two trigonometric functions.

As to be expected, candidates found this problem challenging. In part (a), many were able to use right angle trigonometry to find the length of OC.

As to be expected, candidates found this problem challenging. Those who used a systematic approach in part (b) were more successful than those whose work was scattered about the page. While a pleasing number of candidates successfully found the area of sector AOB, far fewer were able to find the area of triangle BOC. Candidates who took an analytic approach to solving the resulting equation were generally less successful than those who used their GDC. Candidates who converted the angle to degrees generally were not very successful.