2017 - WAEC Mathematics Past Questions and Answers - page 3

Tan 60 = 3; Tan 30 = 1

\(\frac{\tan 60^o - 1}{1 - tan 30^o}\)  = \(\frac{\sqrt{3 - 1}}{1 - \frac{1}{\sqrt{3}}}\) = \(\frac{\sqrt{3 - 1}}{\frac{3 - 1}{\sqrt{3}}}\)

= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3 - 1}}{\sqrt{3}}\)

= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= \(\sqrt{3}\)

 

\(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\)

\(\sqrt{12}\) + 2 x 2 - 2\(\sqrt{3}\)

2 \(\sqrt{3}\) - 2 \(\sqrt{3}\) + 4

= 4

The sum of the exterior angles of any polygon is always 360 degrees. Let's calculate the sum of the given exterior angles and determine the number of sides. 1. **Sum of known exterior angles:** 30\(^o\) + 40\(^o\) + 60\(^o\) = 130\(^o\) 2. **Let *n* be the number of remaining angles, each being 46\(^o\):** 46\(^o\) * n 3. **Total sum of exterior angles:** 130\(^o\) + 46\(^o\) * n = 360\(^o\) 4. **Solve for *n*:** * 46n = 360 - 130 * 46n = 230 * n = 230 / 46 * n = 5 5. **Total number of exterior angles (and sides):** 3 + 5 = 8 Since the polygon has 8 sides, it is an octagon.

a = \(\frac{2}{3}\), d = \(\frac{7}{15}\) - \(\frac{2}{3}\)

= 7 - 10

= \(\frac{-3}{15}\)

d = - \(\frac{-1}{5}\)

T6 = a + 5d

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

= \(\frac{2}{3}\) - 1

= \(\frac{2 - 3}{3}\)

= \(\frac{-1}{3}\)

 

 

(x + \(\frac{1}{2}\)) (n - \(\frac{2}{3}\))

\(x^2 - \frac{2}{3^x} + \frac{x}{2} - \frac{1}{3}\)

\(6x^2 - 4n + 3n - 2 = 0\)

\(6x^2 - x - 2 = 0\)

 

d = \(\sqrt{\frac{6}{x} - \frac{y}{2}}\)

\(d^2 = \frac{6}{x} - \frac{y}{2}\)

\(2xd^2 = 12 - xy\)

\(2xd^2 + xy = 12\)

x = \(\frac{6 + 12}{d^2 + y}\)

t = \(2^{-x} = \frac{1}{2^{x}}\)

\(\implies 2^{x} =\frac{1}{t}\)

\(2^{x+1} = 2^{x} \times 2^{1}\)

= \(\frac{1}{t} \times 2 = \frac{2}{t}\)

Total = 8 + 12 + 4

= 24

\(\frac{8}{24} \times (\frac{12}{24} + \frac{4}{24}\))

= \(\frac{1}{3} \times (\frac{1}{2} + \frac{1}{6}\))

= \(\frac{1}{3} \times \frac{3 + 1}{6}\)

\(\frac{1}{3} \times \frac{4}{6} = \frac{2}{9}\)