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2 - a2 + c2 - 2a Cos B Cos B = \(\frac{a^2 + c^2 - b^2}{2ac}\) \(\frac{1}{5}\) = \(\frac{x^2 + ?(x - 4)^2 - (x + 4)^2}{2x (x - 4)}\) \(\frac{1}{5}\)= \(\frac{x(x - 16)}{2x(x - 4)}\) \(\frac{1}{5}\) = \(\frac{x - 16}{2x - 8}\) = 5(x - 16) = 2x - 8 3x = 72 x = \(\frac{72}{3}\) = 24

\(\frac{1}{tan\theta}\) = \(\frac{cos\theta}{sin\theta}\) sin\(\theta\) = \(\frac{x}{y}\) cos\(\theta\) = \(\frac{\sqrt{y^2 - x^2}}{y}\)

Asector = \(\frac{60}{360} \times \pi r^2\) = \(\frac{1}{6} \pi r^2\) A\(\bigtriangleup\) = \(\frac{1}{2}r^2 \sin 60^o\) \(\frac{1}{2} r^2 \times \frac{\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}\) A\(\text{shaded portion}\) = Asector - A\(\bigtriangleup\) = (\(\frac{1}{6} \pi r^2 - \frac{r^2\sqrt{3}}{4})^3\) = \(\frac{\pi r^2}{2} - \frac{3r^2\sqrt{3}}{4}\) = \(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)

From the diagram, PQRSTW is a regular hexagon. Hexagon is a six sided polygon. Sum of interior angles of polygon = (2n - 4)90o = [2 x 6 - 4] x 90 = 8 x 90 = 720o each angle = \(\frac{720^o}{6} = 120^o\) and TVS = \(\frac{120}{2} = 60^o\)

< PSQ = 35o, < SPR = 30o O is the centre of the circle, PRQ = 65o < OQP = 90 - 65 = 25o

From the diagram, o, o o - (70 + 44o) = 66o

\(\frac{PQ}{6 + RT} = \frac{6}{12} = \frac{6}{PS}\)(Similar triangles) \(\frac{6}{12} = \frac{8}{PS}\) PS = \(\frac{12 \times 8}{6}\) = 16cm

< R = 180o - 45o (sum of angles on a straight line) < R = < P = 45o (corresponding angles) < PSO = < P = 45o (\(\bigtriangleup\)PSO is a right angle)

by Pythagoras's theorem, XW = 3cm Also by Pythagoras theorem, XY2 = 62 + 32 XY2 = 36 + 9 = 45 XY = \(\sqrt{45} = 3 \sqrt{5}\)

O\(\hat{P}\)T = 90º (The radius is perpendicular to the tangent at the point of contact.) Sum of angles in \(\triangle\) PQT = 30 + ( 90 + x ) + (x + 2x) = 180 = 120 + 4x = 180 4x = 180 - 120 = 60º x = \(\frac{60}{4}\) = 15^\circ Therefore, P\(\hat{T}\)O = 2x = 2 x 15 = 30º

Let the bearing of H from P be represented by x^\circ. \(\tan \theta = \frac{30}{60} = 0.5\) \(\theta = \tan^{-1} (0.5) = 26.565^\circ\) \(x = 180^\circ + (90 - 26.565)\) \(x = 180 + 63.435 = 243.435^\circ\) = \(243^\circ 26'\)

Sum of interior angles of polygon = (2n - 4)rt o = (12 - 4) x 90o = 8 x 90o = 720o each interior angle will have \(\frac{720^o}{6}\) = 120o

From the diagram, sin 60o = \(\frac{PR}{8}\) PR = 8 sin 60 = \(\frac{8\sqrt{3}}{2}\) = 4\(\sqrt{3}\) Cos 45o = \(\frac{PR}{PS}\) = \(\frac{4 \sqrt{3}}{PS}\) PS Cos45o = 4\(\sqrt{3}\) PS = 4\(\sqrt{3}\) x 2 = 4\(\sqrt{6}\)

Given Cos z = L, z is an acute angle \(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = cos z = \(\frac{\text{cos z}}{\text{sin z}}\) cosec z = \(\frac{1}{\text{sin z}}\) cot z - cosec z = \(\frac{\text{cos z}}{\text{sin z}}\) - \(\frac{1}{\text{sin z}}\) cot z - cosec z = \(\frac{L - 1}{\text{sin z}}\) sec z = \(\frac{1}{\text{cos z}}\) tan z = \(\frac{\text{sin z}}{\text{cos z}}\) sec z = \(\frac{1}{\text{cos z}}\) + \(\frac{\text{sin z}}{\text{cos z}}\) = \(\frac{1}{l}\) + \(\frac{\text{sin z}}{L}\) the original eqn. becomes \(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = \(\frac{L - \frac{1}{\text{sin z}}}{1 + sin \frac{z}{L}}\) = \(\frac{L(L - 1)}{\text{sin z}(1 + \text{sin z})}\) = \(\frac{L(L - 1)}{\text{sin z} + 1 - cos^2 z}\) = sin z + 1 = 1 + \(\sqrt{1 - L^2}\) = \(\frac{L(L - 1)}{1 - L + 1 \sqrt{1 - L^2}}\)

Using the method of difference of two cubes, \(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\) \((0.03)^{3} - (0.02)^{3} = (0.03 - 0.02)((0.03)^{2} + (0.03)(0.02) + (0.02)^{2}\) = \((0.01)(0.0009 + 0.0006 + 0.0004)\) = \(0.01 \times 0.0019\) = \(0.000019\)