1999 - WAEC Mathematics Past Questions and Answers - page 2
Calculate the value of Y in the diagram
x = 180° - 123° = 57°
z = 180° - 105° = 75°
y = 180° - 57° - 75° = 48°
In the diagram, POR is a circle with center O. ∠QPR = 50°, ∠PQO = 30° and ∠ORP = m. Find m.
< 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°
\(\frac{3n^2 - 2n^3}{m}\
\frac{3(2)^2 -2(-3)^2}{-3}= \frac{12+54}{-3}=-22\)
Simplify \(\frac{2-18m^2}{1+3m}\)
\(\frac{2-18m^2}{1+3m}=\frac{2(1-(3m)^2)}{1+3m}\
\frac{2(1-(3m)(1+3m)}{1+3m}=2(1-3m)\)
If \(y = \sqrt{ax-b}\) express x in terms of y, a and b
\(y = \sqrt{ax-b}\)
\(y^2 = ax-b\)
\(y^2 +b = ax\)
\(x = \frac{y^2 + b}{a}\)
Given that \(81\times 2^{2n-2} = K, find \sqrt{K}\)
\(K = 81 \times 2^{2n - 2}\)
\(\sqrt{K} = \sqrt{81 \times 2^{2n - 2}}\)
= \(9 \times 2^{n - 1}\)
Simplify \(\frac{4}{x+1}-\frac{3}{x-1}\)
\(\frac{4}{x+1}-\frac{3}{x-1} \
=\frac{4x - 4 - 3x - 3}{(x+1)(x-1)}=\frac{x-7}{x^2 - 1}\)
If y varies inversely as x\(^2\), how does x vary with y?
\(y \propto \frac{1}{x^2}\)
\(y = \frac{k}{x^2}\)
\(x^2 = \frac{k}{y}\)
\(x = \frac{\sqrt{k}}{\sqrt{y}}\)
Since k is a constant, then \(\sqrt{k}\) is also a constant.
\(\therefore x \propto \frac{1}{\sqrt{y}}\)
If \(tan x = \frac{1}{\sqrt{3}}\), find cos x - sin x such that \(0^o \leq x \leq 90^o\)
\(\cos x = \frac{\sqrt{3}}{2}\)
\(\sin x = \frac{1}{2}\)
\(\cos x - \sin x = \frac{\sqrt{3} - 1}{2}\)
From the top of a cliff 20m high, a boat can be sighted at sea 75m from the foot of the cliff. Calculate the angle of depression of the boat from the top of the cliff
\(\tan x = \frac{20}{75} = 0.267\)
\(x = \tan^{-1} 0.267 = 14.93°\)
\(\approxeq\) 14.9°