JAMB Mathematics Past Questions & Answers - Page 9

\(\frac {\sin A}{a} = \frac {\sin B}{b} = \frac {\sin C}{c}\)

\(\implies \frac {\sin 82^0}{43.2} = \frac {\sin 56^0}{AB}\)

\(\implies AB \times \sin 82^0 = 43.2 \times \sin 56^0\)

\(\therefore AB = \frac {43.2 \times \sin 56^0}{\sin 82^0}\) = 36.2cm (to 3 s.f)

The height of the second building H = h + 24

tan θ = \(\frac {opp}{adj}\)

tan 30\(^o = \frac {h}{x}\)

\(\implies\frac{\sqrt 3}{3} = \frac {h}{x}\)

\(\implies x = \sqrt 3 = 3h\)

\(\implies x = \frac {3h}{\sqrt 3}\) ....(i)

tan 60\(^o = \frac {24}{x}\)

\(\implies\sqrt 3 = \frac {24}{x}\)

\(\implies x\sqrt 3 = 24\)

\(\implies x = \frac {24}{\sqrt 3}\) ....(ii)

Equate equation (i) and (ii)

\(\implies \frac {3h}{\sqrt 3} = \frac {24}{\sqrt 3}\)

\(\implies\) 3h = 24

\(\implies h = \frac {24}{3}\) = 8m

∴The height of the second building = 8 + 24 = 32m

Let the third number = \(x\)

Then the first number = 100% \(x + 35%x = 135\)%\(x = \frac {135x}{100} = 1.35x\) (Note: 100% \(x = x\))

The second number = 180% \(x = \frac {180x}{100} = 1.80x\)

∴ The ratio of the first number to the second number = \(1.35x : 1.80x = 3 : 4\)

The locus of a point equidistant from two intersecting lines is pair of bisectors of the angles between the two lines.

Given the sequence 3, 8, 13, 18, ... which is an arithmetic sequence

a = 3

d = T\(_2\) - T\(_1\) = 8 - 3 = 5

The general term of an A.P is:

T\(_n\) = a + (n - 1)d

⇒ T\(_n\) = 3 + (n - 1)5

= T\(_n\) = 3 + 5n - 5

∴ T\(_n\) = 5n - 2