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}<br /> \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°