1993 - WAEC Mathematics Past Questions and Answers - page 3

\(\frac{PQ}{RQ} = \frac{RQ}{SQ}\) (Similar triangles)

\(\therefore \frac{PS + 10}{12} = \frac{12}{10}\)

\(10 (PS + 10) = 12 \times 12\)

\(10 PS + 100 = 144 \implies 10 PS = 44\)

\(PS = 4.4 cm\)

< PQR = 180° - (84° + 56°)

= 40°

\(\therefore\) x = 40° (corresponding angles QP// RT)

\(\frac{2.25}{0.015}\)

= \(\frac{2250}{15}\)

= \(150\)

= \(1.5 \times 10^{2}\)

\(-2, -3, -4\frac{1}{2}, ...\)

This is a G.P with r = 1\(\frac{1}{2}\).

\(T_{n} = ar^{n - 1}\) (terms of a G.P)

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

= \(-2 \times \frac{81}{16}\)

= \(-\frac{81}{8}\)

\(T_{7} = (-2)(\frac{3}{2})^{7 - 1}\)

= \(-2 \times \frac{729}{64}\)

= \(-\frac{729}{32}\)

\(\tan x = \frac{opp}{adj} = \frac{5}{12}\)

\(Hyp^2 = opp^2 + adj^2\)

\(Hyp^2 = 5^2 + 12^2\)

= \(25 + 144 = 169\)

\(Hyp = \sqrt{169} = 13\)

\(\sin x = \frac{5}{13}; \cos x = \frac{12}{13}\)

\(\sin x + \cos x = \frac{5}{13} + \frac{12}{13}\)

= \(\frac{17}{13}\)