JAMB Mathematics 2023 Paper

Consider ∆XOB and using Pythagoras theorem

13${}^2$ = 12${}^2$ + h${}^2$

⇒ 169 = 144 + h${}^2$

⇒ 169 - 144 =h${}^2$

⇒ 25 = h${}^2$

⇒ h = $\sqrt{25}$ = 5cm

tan θ = $\frac{opp}{adj}$

⇒ tan θ = $\frac{12}{5}$ = 2.4

⇒ θ = tan${} ^{-1}$(2.4)

⇒ θ = 67.38${} ^0$

∠AOB = 2θ = 2 x 67.38${} ^{\circ}$ = 134.76${} ^{\circ}$

L = $\frac{\theta}{360^{\circ}} \times 2\pi r$

⇒ L = $\frac{134.76}{360} \times 2 \times \frac{22}{7} \times 13 = \frac{77082.72}{2520}$

∴ L = 30.6cm (to 3 s.f)