2002 - WAEC Mathematics Past Questions and Answers - page 4

\(\frac{2^{\frac{1}{2}}\times 8^{\frac{1}{2}}}{4}\=\frac{2^{\frac{1}{2}}\times 2^{\frac{3}{2}}}{2^2}\=2^{\frac{1}{2}}+2^{\frac{3}{2}}-2=2^0 = 1\)

\(\tan 40° = \frac{30}{|PQ|}\)

\(|PQ| = \frac{30}{\tan 40°} = \frac{30}{0.839}\)

= 35.75m

\(V = \pi r^2 h\
210 = \frac{22}{7} \times 14^2 \times h \
h = \frac{210}{22 \times 28}\)
Curved surface area \(= 2r\pi h\
= 2 \times \frac{22}{7} \times 14 \times \frac{210}{22 \times 26} = 30cm^2\)

\(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\)

= \(3(\sqrt{4 \times 3}) + 10\sqrt{3} - (\frac{6}{\sqrt{3}})(\frac{\sqrt{3}}{\sqrt{3}})\)

= \(6\sqrt{3} + 10\sqrt{3} - 2\sqrt{3}\)

= \(14\sqrt{3}\)

m(2a - b) - 2n(b - 2a)

= m(2a - b) - (-2n)(2a - b)

= m(2a - b) + 2n(2a - b)

= (m + 2n)(2a - b)

q oranges = t naira

1 naira = \(\frac{q}{t}\)

p naira = \(p(\frac{q}{t})\)

= \(\frac{pq}{t}\) oranges

In \(\Delta\) QPT,

\(\frac{PT}{6\sqrt{3}} = \sin 30°\)

PT = \(6\sqrt{3} \times \frac{1}{2} = 3\sqrt{3} cm\)

In \(\Delta\) RPT,

\(\frac{PT}{RT} = \tan 60°\)

\(\frac{3\sqrt{3}}{RT} = \tan 60°\)

\(RT = \frac{3\sqrt{3}}{\sqrt{3}} = 3 cm\)

x - 35° = 65° (corresponding angles)

x = 65° + 35° = 100°