Mathematics Questions and Answers

Circumference of Latitude 0oS where R is the radius of the earth is = 2\(\pi\) R cos \(\theta\)

ABCD is the floor, by pythagoras

AC² = 144 + 81 = \(\sqrt{225}\)

AC = 15cm

Height of room is 8m, diagonal of floor is 15m

∴ The cosine of the angle which a diagonal of the room makes with the floor is EC² = 15²

cosine = \(\frac{\text{adj}}{\text{hyp}}\) = \(\frac{15}{17}\)

From the diagram, WYZ = 60^o, XYW = 180^o - 60^o

= 120^o

LX = 30^o

XWY = 180^o - 120^o + 30^o

XWY = 30^o

WXY = XYW = 30^o

Side XY = YW

YW = 8m, sin 60^o = \(\frac{3}{2}\)

∴ sin 60^o = \(\frac{h}{YW}\), sin 60^o = \(\frac{h}{8}\)

h = 8 x \(\frac{3}{2}\)

= 4\(\sqrt{3}\)

Tan \(\theta\) = \(\frac{m^2 - n^2}{2mn}\)

\(\frac{\text{Opp}}{\text{Adj}}\) by pathagoras theorem

= Hyp² = Opp² + Adj²

Hyp² = (m² - n²) + (2mn)²

Hyp² = m⁴ - 2m²n⁴ - 4m² - n²

Hyp² = m⁴ + 2m² + n²n

Hyp² = (m² - n²)²

Hyp² = \(\frac{m^2 + n^2}{2mn}\)

210o = 180o - 210o = - 30o

From ratio of sides, sin -30o = -\(\frac{1}{2}\)

Cos 210o = 180o - 210o = -30o

= cos -30o = \(\frac{-3}{2}\)

But tan 30o = \(\frac{1}{\sqrt{3}}\), rationalizing this

= \(\frac{1}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{3}}{3}\)

∴ = \(\frac{-1}{2}\), \(\frac{\sqrt{-3}}{2}\), \(\frac{\sqrt{3}}{3}\)