JAMB Mathematics Past Questions & Answers - Page 44

\(\frac{2\sin 30 + 5\tan 60}{\sin 60}\)

\(\sin 30 = \frac{1}{2}; \tan 60 = \sqrt{3}; \sin 60 = \frac{\sqrt{3}}{2}\)

\(\therefore \frac{2\sin 30 + 5\tan 60}{\sin 60} = \frac{2(\frac{1}{2}) + 5(\sqrt{3})}{\frac{\sqrt{3}}{2}}\)\)

= \(\frac{1 + 5\sqrt{3}}{\frac{\sqrt{3}}{2}}\)

= \(\frac{2(1 + 5\sqrt{3})}{\sqrt{3}}\)

= \(\frac{2 + 10\sqrt{3}}{\sqrt{3}}\)

Rationalizing, we get

= \(\frac{2\sqrt{3} + 30}{3}\)

= \(\frac{2}{3} \sqrt{3} + 10\)

\(\sin x = \frac{opp}{Hyp} = \frac{4}{5}\)

5\(^2\) = 4\(^2\) + adj\(^2\)

adj\(^2\) = 25 - 16 = 9

adj = \(\sqrt{9}\) = 3

\(\tan x = \frac{4}{3}\)

\(\cot x = \frac{1}{\frac{4}{3}} = \frac{3}{4}\)

\(\cot^2 x = (\frac{3}{4})^2 = \frac{9}{16}\)

\(\csc x = \frac{1}{\sin x}\)

= \(\frac{1}{\frac{4}{5}} = \frac{5}{4}\)

\(\csc^2 x = (\frac{5}{4})^2 = \frac{25}{16}\)

\(\therefore \frac{1 + \cot^2 x}{\csc^2 x - 1} = \frac{1 + \frac{9}{16}}{\frac{25}{16} - 1}\)

= \(\frac{25}{16} \div \frac{9}{16}\)

= \(\frac{25}{9}\)

|AC| = |DF| = 13 cm

Using Pythagoras theorem,

|AC|\(^2\) = |AB|\(^2\) + |BC|\(^2\)

13\(^2\) = 12\(^2\) + |BC|\(^2\)

|BC|\(^2\) = 169 - 144 = 25

|BC| = \(\sqrt{25}\)

= 5 cm

Volume of triangular prism = \(\frac{1}{2} \times base \times length \times height\)

= \(\frac{1}{2} \times 5 \times 12 \times 18\)

= 540 cm\(^3\)

Length of arc = \(\frac{\theta}{360°} \times 2 \pi r\)

= \(\frac{57}{360} \times 2 \times \frac{22}{7} \times 7\)

= 6.97 cm

Number of students that scored above 40 = 55 + 45 + 30 + 15 + 5 = 150 students.