2020 - JAMB Mathematics Past Questions and Answers - page 4

\(\frac{ \frac{r}{2}}{r}\) Sin \(\theta\) = \(\frac{1}{2}\) 

\(\theta\) = sin\(^{-1}\) (\(\frac{1}{2}\)) = 30\(^o\) = 60\(^o\) 

Length of the minor arc:

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

\(\frac{60}{360} \times 2 \pi \times r = \frac{\pi}{3}\) 

\(\frac{Q}{360}\) = \(\frac{r}{L}\)

\(\frac{300}{360}\) = \(\frac{r}{7.2}\)

r = \(\frac{300 \times 7.2}{360}\) 

= 8cm 

***

V = 2080cm\(^3\), h = ? 

r = 7cm

Volume:

v= \(\pi r^2h\)

h = \(\frac{V}{\pi r^2} = \frac{3080}{\frac{22}{7} \times 49}\) 

\(\frac{3080}{54}\) = 20cm

h = 20cm

Total number of acres = 2 + 5 + 3 + 11 + 9 = 30

The angle of acres = 2 + 5 + 3 + 11 + 9 = 30

The angle of the cassava sector = \(\frac{3}{30} \times 360^o = 36^o\) 

\(\frac{h}{n - 2} = \frac{n + 3}{n}\)

n\(^2\) = (n + 3) (n - 2) 

n\(^2\) = n\(^2\) + n - 6

n\(^2\) + n - 6 - n\(^2\) = 0

n - 6 = 0

n = 6

The common ratio => \(\frac{n}{n - 2} = \frac{6}{6 - 2} = \frac{6}{4}\) = \(\frac{3}{2}\) 

32 - x + x + 24 + 4 = 40

60 - x = 40

x = 60 - 40 

x = 20

Sum of the interior angles = (2n - 4) 90\(^o\) 

A pentagon has 5 sides(n), hence, n = 5 

Therefore, the sum of interior angles = 6 x 90\(^o\) = 540\(^o\) 

6x + 6y = 540\(^o\) 

6(x + y) = 540\(^o\) 

x + y = \(\frac{540^o}{6}\) = 90\(^o\) 

y = 90\(^o\) 

y = 90 - x

x \(\frac{\sum x}{N}\)

15 = \(\frac{\sum x }{N}\)

\(\sum x\) = 15N........(i)

y = \(\frac{\sum y}{Ny} = \frac{\sum x + 45}{N + 1}\)

\(\frac{18}{1} = \frac{15N + 45}{N + 1}\) 

18(N + 1) = 15N + 45

18N + 18 = 15N + 45

18N - 15N = 45 - 18 

3N = 27 

N = \(\frac{27}{3}\) 

= 9

\(\frac{h}{500}\) = sin 30\(^o\) 

= 500 sin 30\(^o\) 

= 500 x \(\frac{1}{2}\) 

= 250m 

h = 250m 

The non-zero positive value of x is \(\sqrt{2}\)