Show that the shaded area can be expressed as \(50\theta  - 50\sin \theta \).

Find the value of \(\theta \) for which the shaded area is equal to half that of the unshaded area, giving your answer correct to four significant figures.

\(A = \frac{1}{2} \times {10^2} \times \theta  - \frac{1}{2} \times {10^2} \times \sin \theta \)     M1A1

Note:     Award M1 for use of area of segment = area of sector – area of triangle.

\( = 50\theta  - 50\sin \theta \)     AG

[2 marks]

METHOD 1

unshaded area \( = \frac{{\pi  \times {{10}^2}}}{2} - 50(\theta  - \sin \theta )\)

(or equivalent eg \(50\pi  - 50\theta  + 50\sin \theta )\)     (M1)

\(50\theta  - 50\sin \theta  = \frac{1}{2}(50\pi  - 50\theta  + 50\sin \theta )\)     (A1)

\(3\theta  - 3\sin \theta  - \pi  = 0\)

\( \Rightarrow \theta  = 1.969{\text{ (rad)}}\)     A1

METHOD 2

\(50\theta  - 50\sin \theta  = \frac{1}{3}\left( {\frac{{\pi  \times {{10}^2}}}{2}} \right)\)     (M1)(A1)

\(3\theta  - 3\sin \theta  - \pi  = 0\)

\( \Rightarrow \theta  = 1.969{\text{ (rad)}}\)     A1

[3 marks]

Part (a) was very well done. Most candidates knew how to calculate the area of a segment. A few candidates used \(r = 20\).

Part (b) challenged a large proportion of candidates. A common error was to equate the unshaded area and the shaded area. Some candidates expressed their final answer correct to three significant figures rather than to the four significant figures specified.