Given that sin \(P = \frac{5}{13}\), where p is acute, find the value of cos p - tan p
\(\frac{79}{156}\)
\(\frac{85}{156}\)
\(\frac{7}{13}\)
\(\frac{8}{1}\)
Correct answer is A
If \(sin P = \frac{5}{13}\) from right angled triangle from pythagoras theorem
\(BC^2 = 13^2 - 5^2\\
=169-25\\
BC = \sqrt{144} = 12\\
∴ cos P - tan P = \frac{12}{13} - \frac{5}{12}\\
=\frac{79}{156}\)
The bearing S40°E is the same as
040o
050o
130o
140o
Correct answer is D
= 90° + 50°
= 140°
If 2x : (x +1) = 3:2, what is the value of x?
\(\frac{1}{2}\)
1
\(1\frac{1}{2}\)
3
Correct answer is D
\(2x : (x + 1) = 3:2\\
\frac{2x}{x+1}=\frac{3}{2}\\
∴ 4x = 3x + 3 x =3\)
If \(y \propto \frac{1}{\sqrt{x}}\) and x = 16 when y = 2, find x when y = 24
\(\frac{1}{9}\)
\(\frac{1}{6}\)
\(\frac{1}{3}\)
\(\frac{2}{3}\)
Correct answer is A
\(y \propto \frac{1}{\sqrt{x}}\)
\(y = \frac{k}{\sqrt{x}}\)
When x = 16, y = 2.
\(2 = \frac{k}{\sqrt{16}} \implies 2 = \frac{k}{4}\)
\(k = 8\)
\(y = \frac{8}{\sqrt{x}}\)
When y = 24,
\(24 = \frac{8}{\sqrt{x}}\)
\(\sqrt{x} = \frac{8}{24} = \frac{1}{3}\)
\(\therefore x = (\frac{1}{3})^2\)
\(x = \frac{1}{9}\)
Subtract (-y + 3x + 5z) from (4y - x - 2z)
5y - 4x - 7z
3y + 2x + 3z
-5y + 4x + 7z
2x - 5y + 3z
Correct answer is A
(4y - x - 2z) - (-y + 3x + 5z)
= 4y + y - x - 3x - 2z - 5z
= 5y - 4x - 7z
WAEC Subjects
Aptitude Tests