1997 - WAEC Mathematics Past Questions and Answers - page 3

From the diagram, the angles in the triangle are:

(180 - a)°, (180 - b)° and x°.

$\therefore$ (180 - a) + (180 - b) + x = 180°

360 + x - a - b = 180°

360 - 180 + x = a + b

180 + x = a + b.

Sum of the angle of a pentagon
= (2 x 5 - 4) x 90
= 540^o
126 + 114 + y + 92 + 83 = 540
y = 540 - 415 = 125^o

3x - 8 ≥ 5x
3x - 5x ≥ 8
-2x ≥ 8
x ≤ -4

$\frac{3.6721}{4}=0.918025$

$\approxeq$ 0.9180

a + 10d = 25
-3 + 10d = 25
d = 28/10 = 2⁴/₅

r = log9/log3 = 2log3/log3 = 2

$\frac{64}{27}=(\frac{3}{4}{)}^{t-1}$

$(\frac{3}{4}{)}^t=\frac{64}{27}\times \frac{3}{4}=\frac{16}{9}$

$(\frac{3}{4}{)}^t=(\frac{9}{16}{)}^{-1}$

$(\frac{3}{4}{)}^t=(\frac{3}{4}{)}^{-2}$

t = -2

TQR = 90 - 42 = 48^o
QRT = 180 - (64 - 48)
    = 180 - 112
    = 68^o