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Length of chord = \(2r \sin (\frac{\theta}{2})\) = \(2 \times 8 \times \sin (\frac{90}{2})\) = \(16 \times \frac{\sqrt{2}}{2}\) = \(8\sqrt{2} cm\)

Angle of major sector = 360^\circ - 120^\circ = 240^\circ Area of major sector : \(\frac{\theta}{360} \times \pi r^{2}\) r = \(\frac{4\sqrt{3}}{2} = 2\sqrt{3} cm\) Area : \(\frac{240}{360} \times \pi \times (2\sqrt{3})^{2}\) = \(8\pi cm^{2}\)

Using the cosine rule, we have \(p^{2} = q^{2} + r^{2} - 2qr \cos P\) \(p^{2} = 8^{2} + 6^{2} - 2(8)(6)(\frac{1}{12})\) = \(64 + 36 - 8\) \(p^{2} = 92 \therefore p = \sqrt{92} cm\)

\(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{3} \times 3\sqrt{2}}\) = \(\frac{\sqrt{5} \times 2\sqrt{6}}{\sqrt{2} \times \sqrt{3}}\) = \(\frac{\sqrt{5} \times 2 \sqrt{6}}{\sqrt{6}}\) = 2\(\sqrt{5}\)

\(27^{x + 2} \div 9^{x + 1} = 3^{2x}\) \(3^{3(x + 2)} \div 3^{2(x + 1)} = 3^{2x}\) \(3^{3x + 6} \div 3^{2x + 2} = 3^{2x}\) \(\therefore (3x + 6) - (2x + 2) = 2x\) \(x + 4 = 2x \implies x = 4\)

log3x2 = -8 x2 = 3-8 x = \(3^{-8 \times \frac{1}{2}}\) x = 3-4 x = \(\frac{1}{3^4}\) x = \(\frac{1}{81}\)

\((\sqrt{6} + 2)^2 - (\sqrt{6} - 2)^2\) = \([(\sqrt{6} + 2) + (\sqrt{6} - 2)][(\sqrt{6} + 2) - (\sqrt{6} - 2)]\) = \((\sqrt{6} + 2 + \sqrt{6} - 2)(\sqrt{6} + 2 - \sqrt{6} + 2)]\) = \((2\sqrt{6})(4)\) = \(8\sqrt{6}\)

y \(\propto \sqrt{n}\) y = k\(\sqrt{n}\) when y = 4, n = 4 4 = k\(\sqrt{4}\) 4 = 2k k = 2 Therefore, y = 2\(\sqrt{n}\) y = 2\(\sqrt{\frac{16}{9}}\) y = 2\((\frac{4}{3})\) y = \(\frac{8}{3}\)

P1 (4, 3), P2 (x, y) y = 2x + 4 .....(1) y = 7 - x .....(2) Substitute (2) in (1) 7 - x = 2x + 4 7 - 4 = 2x + x 3 = 3x x = 1 Substitute in eqn (2) y = 7 - x y = 7 - 1 y = 6 P2 (1, 6) Distance between 2 points is given as D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) D = \(\sqrt{(1 - 4)^2 + (6 - 3)^2}\) D = \(\sqrt{(-3)^2 + (3)^2}\) D = \(\sqrt{9 + 9}\) D = \(\sqrt{18}\) D = \(\sqrt{9 \times 2}\) D = \(3\sqrt{2}\)

\(\theta\) = 135o Cos 135o = Cos(90 + 45)o = cos90ocos45o - sin90osin45o = 0cos45o - (1 x \(\frac{\sqrt{2}}{2}\)) = \(-\frac{\sqrt{2}}{2}\)

y = \(\sqrt{8m}\); log y = log \(\sqrt{8m}\) log y = log \(\sqrt{8}\) + log \(\sqrt{m}\) log y = \(\frac{1}{2}\)log 2\(^3\) + log m\(\frac{1}{2}\) log y = \(\frac{3}{2}\) log 2 + \(\frac{1}{2}\) log m

(\(\frac{27}{125}\))-\(\frac{1}{3}\) x (\(\frac{4}{9}\))\(\frac{1}{2}\) = (\(\frac{3^3}{5^3}\))-\(\frac{1}{3}\)-\(\frac{1}{3}\) x (\(\frac{3^2}{3^2}\))\(\frac{1}{2}\) -\(\frac{1}{2}\) = \(\frac{3^{-1}}{3^{-1}} \times \frac{2}{3}\) = \(\frac{\frac{1}{3}}{\frac{1}{5}} \times \frac{2}{3}\) \(\frac{1}{3} \times \frac{5}{1} \times {2}{3} = \frac{10}{9}\)

Area of sector = \(\frac{\theta}{360^\circ} \times \pi r^{2}\) = \(\frac{150}{360} \times \pi \times 12^{2}\) = 60\(\pi\) cm\(^{2}\)

\((\frac{16}{81})^{\frac{1}{4}} \div (\frac{9}{16})^{-\frac{1}{2}}\) \((\frac{16}{81})^{\frac{1}{4}} \div (\frac{16}{9})^{\frac{1}{2}}\) \((\frac{2^4}{3^4})^{\frac{1}{4}} \div (\frac{4^2}{3^2})^{\frac{1}{2}}\) \(\frac{2^{4 \times \frac{1}{4}}}{3^{4 \times \frac{1}{4}}} \div \frac{4^{2 \times \frac{1}{2}}}{3^{2 \times \frac{1}{2}}}\) \(\frac{2}{3} \div \frac{4}{3}\) \(\frac{2}{3} \times \frac{3}{4}\) \(\frac{2}{4}\) \(\frac{1}{2}\)

log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = 3 log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = 3log33 log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = log333 log3(\(\frac{18 \times 3}{X}\)) = log333 \(\frac{18 \times 3}{X}\) = 33 18 x 3 = 27 x X x = \(\frac{18 \times 3}{27}\) = 2