 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/5/a/solution

The two labelled regions in the original diagram are complementary circular sectors. Their areas add to

π r^{2}

and satisfy

A = 2 B

, so

B = π r^{2} /3

. The sector with central angle

θ

has area

r^{2} θ /2

. Hence

θ = \frac{2 π}{3} = 12 0^{\circ} .

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