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In terms of distance, a circle has a total distance or perimeter of 2\pir or \pid Where r is radius and d is the diameter So perimeter = \(\frac{22}{7}\) x 100 = 314.2857m To cover a distance of 1000m, he is going to round the circular track for \(\frac{1000}{314.2857}\) = 3.18 \(\approxeq\) 3 (to the nearest whole number)

Given length of the room = 12m; breadth = 9m and height = 8m. The room is a cuboid in shape, therefore the length of the diagonal = \(\sqrt{l^2 + b^2 + h^2}\) = \(\sqrt{12^2 + 9^2 + 8^2}\) =\(\sqrt{289}\) = 17m. The diagonal makes an angle with the diagonal of the floor: \(\sqrt{12^2 + 9^2}\) = \(\sqrt{225}\) = 15m The cosine of the angle that the diagonal makes with the floor (\(\theta\)) = \(\frac{15}{17}\).

From the figure, ∠ XMN = ∠ XZY Angle X is common So, ∠ XNM = ∠ XYZ Then from the angle relationship \(\frac{XM}{XZ}\) = \(\frac{XN}{XY}\) = \(\frac{MN}{ZY}\) XM = 8, XZ = 12 + 4 = 16, XN = 12, XY = 8 + YM \(\frac{8}{16}\) = \(\frac{12}{(8 + YM) }\) Cross multiply 8(8 + YM) = 192 64 + 8YM = 192 8YM = 128 YM = \(\frac{128}{8}\) = 16cm

First quadrant: Sin, Cos and Tan are all positive Second quadrant: Sin is positive, Cos is negative and Tan is negative Third quadrant: Tan is positive, Sin is negative and Cos is negative Fourth quadrant: Cos is positive, Sin is negative and Tan is negative The correct option is the third quadrant only where Tan\theta is positive and Sin\theta is negative

Locus is the path traced at by a point which moves in accordance with a certain law. It is also the set of all possible position occupied by an object The path traced from all possible location of 4cm from a given point P form a circle of radius 4cm with centre P.

Sin(5x - 28) = Cos(3x - 50)………..i But Sin\alpha = Cos(90 - \alpha) So Sin(5x - 28) = Cos(90 - [5x - 28]) Sin(5x - 28) = Cos(90 - 5x + 28) Sin(5x - 28) = Cos(118 - 5x)………ii Combining i and ii Cos(3x - 50) = Cos(118 - 5x) 3x - 50 = 118 - 5x Collecting the like terms 3x + 5x = 118 + 50 8x = 168 x = \(\frac{168}{8}\) x = 21\(^o\) Answer is B

∫\(\frac{5}{x}\) dx = 5 ∫\(\frac{1}{x}\) = 5Inx Since the integral of \(\frac{1}{x}\) is Inx ∫\(^2\) \(_1\)∫ \(\frac{5}{x}\) dx = 5 dx = 5 (In<2 – InIn1) = 3.4657 = 3.47

P( tail on a coin) = \(\frac{1}{2}\) Even numbers on a care 2, 4 and 6 P( even number on a die) = \(\frac{3}{6}\) = \(\frac{1}{2}\) P( tail on a coin and even number on a die) = \(\frac{1}{2}\) x \(\frac{1}{2}\) = \(\frac{1}{4}\)

Using Pythagoras theorem PR\(^2\) = 5\(^2\) + 12\(^2\) 25 + 144 = 169 PR = \sqrt(169)= 13km

An equilateral triangle has all sides equal and all angles equal as 600 Area = \(\frac{1}{2}\) absin\theta Area = \(\frac{1}{2}\) x 8 x 8 x sin60 = \(\frac{1}{2}\) x 64 x \(\sqrt{\frac{3}{2}}\) = 16\sqrt3 cm\(^2\)

Cos \theta = \(\frac{adj}{hyp}\) = \(\frac{300}{600}\) = 0.5 \theta = Cos - 10.5 = 60 ∠ RPQ = ∠ PQs So the bearing of P from Q is 180 + 60 = 240\(^o\) Answer is D