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\(\frac{\cot (90 - \theta)}{sin^2\theta}\) \(\cot (90 - \theta) = \tan \theta\) \(\therefore \frac{\cot (90 - \theta)}{\sin^{2} \theta} = \frac{\tan \theta}{\sin^{2} \theta}\) \(\tan \theta = \frac{\sqrt{3}}{3}\) \(\sin \theta = \frac{1}{2} \implies \sin^{2} \theta = \frac{1}{4}\) \(\frac{\cot(90 - \theta)}{\sin^{2} \theta} = \frac{\sqrt{3}}{3}\div\frac{1}{4}\) = \(\frac{4}{\sqrt{3}}\)

A solid sphere of radius \( 4\, \text{cm} \) has a mass of \( 64\, \text{kg} \). What will be the mass of a shell of the same metal whose internal and external radii are \( 2\, \text{cm} \) and \( 3\, \text{cm} \) respectively? Calculate the density of the solid sphere: The volume \( V \) of a solid sphere is given by the formula: \(V = \frac{4}{3} \pi r^3\) For the solid sphere with radius \( r = 4\, \text{cm} \): \(V = \frac{4}{3} \pi (4)^3 = \frac{4}{3} \pi (64) = \frac{256}{3} \pi \, \text{cm}^3\) To find the density \( \rho \): \(\rho = \frac{\text{mass}}{\text{volume}} = \frac{64\, \text{kg}}{\frac{256}{3} \pi \, \text{cm}^3}\) Converting mass to grams (since \( 1\, \text{kg} = 1000\, \text{g} \)): \(\rho = \frac{64000\, \text{g}}{\frac{256}{3} \pi} = \frac{64000 \times 3}{256 \pi} = \frac{192000}{256 \pi} = \frac{750}{\pi} \, \text{g/cm}^3\) Calculate the volume of the shell: The volume \( V_{\text{shell}} \) of the shell is the difference between the volume of the outer sphere and the inner sphere. Outer sphere = (radius \( 3\, \text{cm} \)): \(V_{\text{outer}} = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi (27) = 36 \pi \, \text{cm}^3\) Inner sphere = (radius \( 2\, \text{cm} \)): \(V_{\text{inner}} = \frac{4}{3} \pi (2)^3 = \frac{4}{3} \pi (8) = \frac{32}{3} \pi \, \text{cm}^3\) Calculate the volume of the shell: \(V_{\text{shell}} = V_{\text{outer}} - V_{\text{inner}} = 36 \pi - \frac{32}{3} \pi\) Converting \( 36 \) to a fraction with a denominator of 3: \(V_{\text{outer}} = \frac{108}{3} \pi\) Now subtract: \(V_{\text{shell}} = \left(\frac{108}{3} \pi - \frac{32}{3} \pi\right) = \frac{76}{3} \pi \, \text{cm}^3\) Calculate the mass of the shell: Using the density \( \rho = \frac{750}{\pi} \, \text{g/cm}^3 \): \(\text{Mass of shell} = \text{Density} \times \text{Volume} = \frac{750}{\pi} \times \frac{76}{3} \pi\) The \( \pi \) cancels out: \(\text{Mass of shell} = \frac{750 \times 76}{3} = \frac{57000}{3} \approx 19000 \, \text{g} \approx 19 \, \text{kg}\)

Area of trapezium = 14cm2 Area of trapezium = \(\frac{1}{2}\)(a + b)h 14 = \(\frac{1}{2}\)(4 + 3)h 14 = \(\frac{7}{2}\)h h = \(\frac{14 \times 2}{7}\) = 4 Area of SQR = \(\frac{1}{2}\)(3 x 4) = \(\frac{12}{2}\) = 6.0

P(R1) = \(\frac{6}{10}\) = \(\frac{2}{3}\) P(R1 n R11) = P(both red) \(\frac{3}{5}\) x \(\frac{5}{9}\) = \(\frac{1}{3}\)

Length of an area = \(\frac{\theta}{360}\) x 2\(\pi\)r Longitude difference = 147 + 153 = 30NoN distance between xy = \(\frac{\theta}{360}\) x 2\(\pi\)R cos60o = \(\frac{300}{360}\) x 4 x 104 x \(\frac{1}{2}\) = 1.667 x 104km(1667 km)

Total students = 120 grade = 18 Zo = \(\frac{18}{120}\) x \(\frac{360}{1}\) = 54o

a{\(\frac{x + 1}{x - 2} - \frac{x - 1}{x + 2}\)} = a{\(\frac{(x + 1)(x + 2)- (x - 1)(x - 2)}{(x - 2)(x + 2)}\)} = 6 \(\frac{6x}{x^2 - 4}\) = 6x a = x2 - 4

By Pythagoras, KZ2 = 252 - 52 KZ2 = (25 + 15)(25 - 15) = 400 KZ = \(\sqrt{400}\) = 20 area of XYZ = \(\frac{1}{2} \times 28 \times 15\) = 210 sq. cm

yo = 180o - (86o + 58o) 180 - 144 = 36o xo = 180 - (94 + 58) 180 -152 = 28 (xo, yo) = (28o, 36o)

< PRQ = \(\frac{1}{2}\) < POQ = 90o < PSR + < PQR = 180o < PQR = 180o - 145o = 35o \(\bigtriangleup\)PQR is a right angled triangle x = 90 - < PQR = 90o - 35o = 55o

HJ2 = a2 + a2 = 2a2 HJ = \(\sqrt{2a^2} = a \sqrt{2}\) HN2 = a2 + (a\(\sqrt{2}\))2 = a2 + 2a2 = 3a2 HN = \(\sqrt{3a^2}\) = a\(\sqrt{3}\)cm

180 - \(\phi\)o = \(\theta\)o + \(\phi\)o (Sum of opposite interior angle equal to its exterior angle) 180 = 2\(\phi\) + \(\theta\)o

Area of sector = \(\frac{120}{360} \times \pi \times (3)^2 = 3 \pi\) Area of triangle = \(\frac{1}{2} \times 3 \times 3 \times \sin 120^o\) = \(\frac{9}{2} \times \frac{\sqrt{3}}{2} = \frac{9\sqrt {3}}{4}\) Area of shaded portion = 3\(\pi - \frac{9\sqrt {3}}{4}\) = 3 (\(\pi - 3 \frac{\sqrt{3}}{4}\))

In the figure, angle x = 20o + 80o + 40o = 140o

1 - \(\sqrt{13}\) and 1 + \(\sqrt{13}\) sum of roots - \(1 + \sqrt{13} + 1 - \sqrt{13} = 2\) Product of roots = (1 - \(\sqrt{13}\)) (1 + \(\sqrt{13}\)) = -12 x2 - (sum of roots) x + (product of roots) = 0 x2 - 2x - 12 = 0