1999 - WAEC Mathematics Past Questions and Answers - page 2

x = 180° - 123° = 57°

z = 180° - 105° = 75°

y = 180° - 57° - 75° = 48°

< QOR = 50° x 2 = 100°

Reflex < QOR = 360° - 100° = 260°

\(\therefore\) 30° + 50° + 260° + m = (4 - 2) x 180°

340° + m = 360°

m = 360° - 340° = 20°

Given that m = -3, n = 2, the value of
\(\frac{3n^2 - 2n^3}{m}\
\frac{3(2)^2 -2(-3)^2}{-3}= \frac{12+54}{-3}=-22\)

\(\frac{2-18m^2}{1+3m}=\frac{2(1-(3m)^2)}{1+3m}\

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

\(y = \sqrt{ax-b}\)

\(y^2 = ax-b\)

\(y^2 +b = ax\)

\(x = \frac{y^2 + b}{a}\)

\(K = 81 \times 2^{2n - 2}\)

\(\sqrt{K} = \sqrt{81 \times 2^{2n - 2}}\)

= \(9 \times 2^{n - 1}\)

\(\frac{4}{x+1}-\frac{3}{x-1} \

=\frac{4x - 4 - 3x - 3}{(x+1)(x-1)}=\frac{x-7}{x^2 - 1}\)

\(y \propto \frac{1}{x^2}\)

\(y = \frac{k}{x^2}\)

\(x^2 = \frac{k}{y}\)

\(x = \frac{\sqrt{k}}{\sqrt{y}}\)

Since k is a constant, then \(\sqrt{k}\) is also a constant.

\(\therefore x \propto \frac{1}{\sqrt{y}}\)

\(\cos x = \frac{\sqrt{3}}{2}\)

\(\sin x = \frac{1}{2}\)

\(\cos x - \sin x = \frac{\sqrt{3} - 1}{2}\)

\(\tan x = \frac{20}{75} = 0.267\)

\(x = \tan^{-1} 0.267 = 14.93°\)

\(\approxeq\) 14.9°