JAMB Mathematics Past Questions & Answers - Page 240

The required locus is angle bisector of the two lines

$\frac{h_1}{h_2}$ = $\frac{2}{3}$

h₂ = $\frac{2h_1}{3}$

$\frac{r_1}{r_2}$ = $\frac{9}{8}$

r₂ = $\frac{9r_1}{8}$

v₁ = $\pi$($\frac{9r_1}{8}$)₂($\frac{2h_1}{3}$)

= $\pi$r₁ 2h₁ x $\frac{27}{32}$

v = $\frac{\pi r_1 2h_1 \times \frac{27}{32}}{\pi r_1 2h_1}$ = $\frac{27}{32}$

v₂ : v₁ = 27 : 32

Volume of a cylinder = πr$^2$h

First convert 3m to cm by multiplying by 100

Volume of External cylinder = π \times 13$^2$ \times 300

Volume of Internal cylinder = π \times 10$^2$ \times 300

Hence; Volume of External cylinder - Volume of Internal cylinder

Total volume (v) = π (169 - 100) \times 300

V = π \times 69 \times 300

V = 20700πcm$^3$

Cos $\theta$ = $\frac{12}{13}$

x² + 12² = 13²

x² = 169- 144 = 25

x = 25

= 5

Hence, tan$\theta$ = $\frac{5}{12}$ and cos$\theta$ = $\frac{12}{13}$

If cos²$\theta$ = 1 + $\frac{1}{tan^2\theta}$

= 1 + $\frac{1}{\frac{(5)^2}{12}}$

= 1 + $\frac{1}{\frac{25}{144}}$

= 1 + $\frac{144}{25}$

= $\frac{25 + 144}{25}$

= $\frac{169}{25}$

For a regular polygon of n sides

n = $\frac{360}{\text{Exterior angle}}$

Exterior < = 180° - 140°

= 40°

n = $\frac{360}{40}$

= 9 sides