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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

Number of students scoring at least 50 marks = Number of students scoring 50 and above From the table 53, 70, 84, 59, 90, 60, 81, 73, 50, 37, 67, 68, 64, 52. Hence, 14 students scored at least 50 marks

Mode = a + \(\frac{(b - a)(F_m - F_b)}{2F_m - F_a - F_b}\) = \(L_1 + \frac{\Delta_1 x^\text{c}}{\Delta_1 + \Delta_2}\) = \(10 + \frac{(20 - 10)(27 - 10)}{2(27) - 10 - 19}\) = 10 + \(\frac{170}{25}\) = 10 + 6.8 = 16.8

Average age of 110 students = 16 ∴ Total age = 16 x 10 = 160 years Age of teachers = x, total number of people now = 11 mean age = 19 Total age of new group = 19 x 11 = 209 Age of teachers = x = (209 - 160) = 49 yrs

Median = L1 + (\(\frac{Ef}{fm}\)) - fo \(\frac{\sum f}{2}\) = \(\frac{20}{2}\) = 10, L1 = 10.5, fo = 6, fm = 5 Median = 10.5 + \(\frac{(10 - 6)}{5}\)5 = 10.5 + 4 = 14.5

Possible outcomes are 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30. Prime numbers has only two factors itself and 1 The prime numbers among the group are 23, 29. Probability of choosing a prime number = \(\frac{\text{Number of prime}}{\text{No. of total Possible Outcomes}}\) = \(\frac{2}{11}\)

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

Chance of x = \(\frac{1}{2}\) Change of Y = \(\frac{2}{3}\) Chance of Z = \(\frac{1}{4}\) Chance of Y and Z only occurring = Pr (Y ∩ Z ∩ Xc) where Xc = 1 - Pr(X) 1 - \(\frac{1}{2}\) = \(\frac{1}{2}\) = Pr(Y) x Pr(Z) x Pr(Xc) = \(\frac{2}{3}\) x \(\frac{1}{4}\) x \(\frac{1}{2}\) = \(\frac{1}{12}\)

The two equations are y = x(2 - x) and y = (x - 1)(x - 3) The root of these equations are points where the graph of the equations cuts the x axis; but at these points = 0 put y = 0; 0 = x(2 - x) 0 = (x - 1)(x - 3) x = 0 or x = 2; x = 1 or x = 3 The values of x at P, Q, R are increasing towards the positive direction of x-axis at P, x = 1 at Q, x = 2 at R, x = 3 P, Q, R are respectively (1, 2, 3)

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