If \(\sin x = \frac{4}{5}\), find \(\frac{1 + \cot^2 x}{\csc^2 x - 1}\).
\(\frac{13}{2}\)
\(\frac{25}{9}\)
\(\frac{3}{13}\)
\(\frac{4}{11}\)
Correct answer is B
\(\sin x = \frac{opp}{Hyp} = \frac{4}{5}\)
5\(^2\) = 4\(^2\) + adj\(^2\)
adj\(^2\) = 25 - 16 = 9
adj = \(\sqrt{9}\) = 3
\(\tan x = \frac{4}{3}\)
\(\cot x = \frac{1}{\frac{4}{3}} = \frac{3}{4}\)
\(\cot^2 x = (\frac{3}{4})^2 = \frac{9}{16}\)
\(\csc x = \frac{1}{\sin x}\)
= \(\frac{1}{\frac{4}{5}} = \frac{5}{4}\)
\(\csc^2 x = (\frac{5}{4})^2 = \frac{25}{16}\)
\(\therefore \frac{1 + \cot^2 x}{\csc^2 x - 1} = \frac{1 + \frac{9}{16}}{\frac{25}{16} - 1}\)
= \(\frac{25}{16} \div \frac{9}{16}\)
= \(\frac{25}{9}\)
394 cm\(^3\)
425 cm\(^3\)
268 cm\(^3\)
540 cm\(^3\)
Correct answer is D
|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\(^3\)
In the circle above, with centre O and radius 7 cm. Find the length of the arc AB, when < AOB = 57°
5.32 cm
4.39 cm
7.33 cm
6.97 cm
Correct answer is D
Length of arc = \(\frac{\theta}{360°} \times 2 \pi r\)
= \(\frac{57}{360} \times 2 \times \frac{22}{7} \times 7\)
= 6.97 cm
100
79
150
90
Correct answer is C
Number of students that scored above 40 = 55 + 45 + 30 + 15 + 5 = 150 students.
| Marks | 1 | 2 | 3 | 4 | 5 |
| Frequency | 2y - 2 | y - 1 | 3y - 4 | 3 - y | 6 - 2y |
The table above is the distribution of data with mean equals to 3. Find the value of y.
5
2
3
6
Correct answer is B
| Marks (x) | 1 | 2 | 3 | 4 | 5 | |
| Frequency (f) | 2y - 2 | y - 1 | 3y - 4 | 3 - y | 6 - 2y | 3y + 2 |
| fx | 2y - 2 | 2y - 2 | 9y - 12 | 12 - 4y | 30 - 10y | 26 - y |
Mean = \(\frac{\sum fx}{\sum f}\)
\(3 = \frac{26 - y}{3y + 2}\)
\(3(3y + 2) = 26 - y\)
\(9y + 6 = 26 - y\)
\(9y + y = 26 - 6\)
\(10y = 20 \implies y = 2\)
JAMB Subjects
Aptitude Tests