WAEC Mathematics Past Questions & Answers - Page 111

cos(x + 40)^o = 0.0872

x + 40 = cos^-10.0872

x + 40^o = 84.99^o

x = 84.99^o - 40^o

x = 44.99

x = 45^o

No explanation has been provided for this answer.

Let 0 = angle of the minor sector

angle of the major sector = 360 - \(\theta\)(angle at a point)

2 \(\pi r\) = 54 + 126(i.e circumference of minor and major arc)

2\(\pi r = 180^o\)

r = \(\frac{180}{2\pi}\) = \(\frac{90}{\pi}\)

Lenght of ninor arc

= \(\frac{\theta}{360} \times 2 \pi r\)

54 = \(\frac{\theta}{360} \times 3 \pi r\)

\(\theta = \frac{360 \times 54}{2 \pi r}\)

but r = \(\frac{90}{\pi}\) substituting \(\frac{90}{\pi}\) for r

\(\theta = \frac{360 \times 54 \times \pi}{2 \times \pi \times 90}\)

\(\theta = 2 \times 54 = 108^o\)

angle of the major sector = 360 - 108^o

= 252^o

Curved surface area = 2\(\pi h\)

704 = 2 x \(\frac{22}{7} \times 8 \times h\)

704 = \(\frac{352}{7}\)h

352h = 704 x 7

h = \(\frac{704 \times 7}{352}\)

= \(\frac{4928}{352}\)

h = 14cm

\(\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6}\)

= \(\frac{-2m - 5}{4} \geq \frac{5m - 14}{12}\)

12(-2m - 5) \(\geq\) 4(5m - 14)

-24m - 60 \(\geq\) 20m - 56

-24m - 20m \(\geq\) -56 + 60

44m \(\geq\) 4

m \(\leq \frac{4}{-44}\)

m \(\leq \frac{-1}{11}\)