1993 - WAEC Mathematics Past Questions and Answers - page 3
In the diagram above, |QR| = 12cm and |QS| = 10cm. If ∠PQR = 90°, ∠RSQ = 90° and PSQ is a straight line, find |PS|
\(\frac{PQ}{RQ} = \frac{RQ}{SQ}\) (Similar triangles)
\(\therefore \frac{PS + 10}{12} = \frac{12}{10}\)
\(10 (PS + 10) = 12 \times 12\)
\(10 PS + 100 = 144 \implies 10 PS = 44\)
\(PS = 4.4 cm\)
QRP = QPR
= 180 - 80 = 100/20 = 50o
SRP = SPR
= 180 - 30 = 150/2 = 75o
∴ SPQ = SPR - QPR
= 75 - 50 = 25o
QRK + RKQ + KQR = 180
31 + 42 + KQR = 180o
KQR = 180 - 73 = 107o
In the diagram above, QRS is a straight line, QP//RT, PRQ = 56°, ∠QPR =84° and ∠TRS = x°. Find x
< PQR = 180° - (84° + 56°)
= 40°
\(\therefore\) x = 40° (corresponding angles QP// RT)
Simplify \(\frac{2.25}{0.015}\) leaving your answer in standard form
\(\frac{2.25}{0.015}\)
= \(\frac{2250}{15}\)
= \(150\)
= \(1.5 \times 10^{2}\)
State the fifth and seventh terms of the sequence \(-2, -3, -4\frac{1}{2}, ...\)
\(-2, -3, -4\frac{1}{2}, ...\)
This is a G.P with r = 1\(\frac{1}{2}\).
\(T_{n} = ar^{n - 1}\) (terms of a G.P)
\(T_{5} = (-2)(\frac{3}{2})^{5 - 1}\)
= \(-2 \times \frac{81}{16}\)
= \(-\frac{81}{8}\)
\(T_{7} = (-2)(\frac{3}{2})^{7 - 1}\)
= \(-2 \times \frac{729}{64}\)
= \(-\frac{729}{32}\)
x = 90 x sin30 = 45o
∴Dist = 150 + 45 = 195km
Given that tan x = 5/12, what is the value of sin x + cos x ?
\(\tan x = \frac{opp}{adj} = \frac{5}{12}\)
\(Hyp^2 = opp^2 + adj^2\)
\(Hyp^2 = 5^2 + 12^2\)
= \(25 + 144 = 169\)
\(Hyp = \sqrt{169} = 13\)
\(\sin x = \frac{5}{13}; \cos x = \frac{12}{13}\)
\(\sin x + \cos x = \frac{5}{13} + \frac{12}{13}\)
= \(\frac{17}{13}\)