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Diameter = 8cm ∴ Radius = 4cm Length of arc = \(\frac{\theta}{360}\) x 2 \(\pi\)r but Q = 22\(\frac{1}{2}\) ∴ Length \(\frac{22\frac{1}{2}}{360}\) x 2 x \(\pi\) x 4 = \(\frac{22\frac{1}{2} \times 8\pi}{360}\) = \(\frac{180}{360}\) = \(\frac{\pi}{2}\)

Area of sector \(\frac{\theta}{360}\) x \(\pi\)r2, \(\theta\) = 60o, r = 3m = \(\frac{60}{360}\) x \(\frac{12}{7}\) x 3 x 3 \(\frac{1}{6}\) x \(\frac{22}{7}\) x 9 = \(\frac{33}{7}\) = 4.7m2

The angle between 2 latitudes one in northern hemisphere and the other in southern hemisphere and the other in southern hemisphere is the sum of their latitudes. ∴ Total angle difference = (30 + 13) = 43o

sin \(\theta\) = cos \(\theta\) 0 \(\leq\) \(\theta\) \(\leq\) 360o The acute angle where sin \(\theta\) = cos \(\theta\) = 45o But at the third Quadrant Cos \(\theta\) = -ve; sin \(\theta\) = -ve. at the 3rd quadrant, value with respect to Q is (180 + Q) where Q = acute angle (180 + 45) = 225^\circ The two solution are 45^\circ, 225^\circ

Let each of the unknown side be x. \(10^{2} = x^{2} + x^{2} - 2(x)(x) \cos 120\) \(100 = 2x^{2} - 2x^{2} \cos 120\) \(100 = 2x^{2} + x^{2} = 3x^{2}\) x = \(\sqrt{\frac{100}{3}}\) = \(\frac{10}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) x = \(\frac{10\sqrt{3}}{3}\)cm

Angle corresponding to 7 in a pie chart will be \(\frac{7 \times 360}{\text{sum of items}}\) = \(\frac{7 }{18}\) x 360 = 140o

\(\begin{array}{c|c} x & x - x & (x - x)^2\\ \hline 7 & -3 & 9\\8 & -2 & 4 \\9 & -1 & 1\\10 & 0 & 1\\11 & 1 & 1\\ 12 & 2 & 4\\13 & 3 & 9\\ \hline & & 28\end{array}\) S.D = \(\sqrt{\frac{\sum(x - x)^2}{N}}\) = \(\sqrt{\frac{\sum d^2}{N}}\) = \(\sqrt{\frac{28}{7}}\) = \(\sqrt{4}\) = 2

The total angle in a regular pentagon = \((2(5) - 4) \times 90\) = \(6 \times 90 = 540^\circ\) Each interior angle = \(\frac{540}{5} = 108^\circ\) While the interior of a quadrilateral = \(\frac{360}{4} = 90^\circ\) \(PTU + TPU + TUP = 180^\circ\) \(108^\circ + (180^\circ - 90^\circ) + TUP = 180^\circ\) \(TUP = 180^\circ - (108^\circ + 18^\circ) = 54^\circ\)

< PSO = \(\frac{1}{2}\) < SOQ = \(\frac{1}{2}\)(180) = 90^\circ < RTS = < PQS = 32^\circ (Alternative angle) < PSQ = 90 - < PSQ = 90^\circ - 32^\circ = 58^\circ

OQ = OR = radii < ROQ = 180 - 96 = 84º \(\bigtriangleup\)OQR = Isosceles R = Q R + Q + 84 = 180(angle in a \(\bigtriangleup\)) 2R = 96 since R = Q R = 48º ORQ = 48º

Construction joins R to P such that SRP = straight line R = 180o - 107o < p = 180o - (107o - 34o) 108 - 141o = 39o Angle < S = 39o (corr. Ang.) But in \(\bigtriangleup\)SRT < S = < T = 39o SRT = 180 - (39o + 39o) = 180o - 78o = 102o

Since QRU= Xo RSU = Xo, But RSU = 180o - (50o + 35o) = 180o - 85o = 95o x = 95o

Using angle bisector theorem: line PS bisects angle QPR QS/QP = SR/PR 3/4 = SR/9 4SR = 27 SR = \(\frac{27}{4}\) = 6\(\frac{3}{4}\)cm

SQ2 + OS2 = OQ2 + 52 = 142 SQ2 = 142 - 52 196 - 25 = 171 ST2 + TQ2 = SQ2 ST2 + 22 = 171 ST2 = 171 - 4 = 167 ST = \(\sqrt{167}\) = 12.92 = 12.9cm

(a) \(\log_{10} (\frac{30}{16}) - 2 \log_{10} (\frac{5}{9}) + \log_{10} (\frac{400}{243})\) = \(\log_{10} (\frac{30}{16}) - \log_{10} (\frac{5}{9})^{2} + \log_{10} (\frac{400}{243})\) = \(\log_{10} (\frac{30}{16} \times \frac{400}{243} \div \frac{25}{81})\) = \(\log_{10} (\frac{30}{16} \times \frac{400}{243} \times \frac{81}{25})\) = \(\log_{10} (10) = 1\) (b) \(\sqrt{\frac{P}{Q}} = \sqrt{\frac{3.6 \times 10^{-3}}{2.25 \times 10^{6}}\) = \(\sqrt{\frac{36 \times 10^{-4}}{2.25 \times 10^{6}}\) = \(\frac{6 \times 10^{-2}}{1.5 \times 10^{3}}\) = \(4 \times 10^{-2 - 3} = 4 \times 10^{-5}\)