Consider ∆XOB and using Pythagoras theorem 13\(^2\) = 12\(^2\) + h\(^2\) ⇒ 169 = 144 + h\(^2\) ⇒ 169 - 144 =h\(^2\) ⇒ 25 = h\(^2\) ⇒ h = \(\sqrt25\) = 5cm tan \theta = \(\frac {opp}{adj}\) ⇒ tan \theta = \(\frac{12}{5}\) = 2.4 ⇒ \theta = tan\(^{-1}\)(2.4) ⇒ \theta = 67.38\(^0\) ∠AOB = 2\theta = 2 x 67.38\(^o\) = 134.76\(^o\) L = \(\frac{\theta}{360^o} \times 2\pi r\) ⇒ L = \(\frac {134.76}{360} \times 2 \times \frac {22}{7} \times 13 = \frac {77082.72}{2520}\) ∴ L = 30.6cm (to 3 s.f)