Mathematics Questions and Answers

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

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

In PNO, ONP

= 180 - (35 + 86)

= 180 - 121

= 59°

PRQ = QNP = 59°(angles in the same segment of a circle are equal)

Sum of interior angles of any polygon is (2n - 4) right angle; n angles of the Nonagon = 9

Where 3 are equal and 6 other angles = 1110o

(2 x 9 - 4)90o = (18 - 4)90o

14 x 90o = 1260o

9 angles = 1260°; 6 angles = 110o

Remaining 3 angles = 1260o - 1110o = 150o

Size of one of the 3 angles \(\frac{150}{3}\) = 50o

a = 4, r = \(\frac{4}{2}\)

\(\frac{8}{4}\) = 2

n = 11

T_n = ar^{n - 1}

T₁₁ = 4(2)^{11 - 1}

4(2)¹⁰ = 2¹²

since 4 = 2²

= 2¹²