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Now, t + 60o + 50o = 180o t = 180o - 110o t = 70o

The bearing of X from Y = 360 - 120 = 240

Required distance = x where \(\frac{1}{50m}\) = sin 60 x 50m x 0.866 x = 43.3m

2(30 + x) = 360 - 200 60 + 2x = 160 2x = 160 - 60 2x = 100 x = \(\frac{100}{2}\) x = 50

No expanded working has been attached for this question yet. The verified answer is Option B.

(a) (i) Mode = 2 (ii) Third quartile : Sum of frequency = 24 \(Q_{3} = \frac{3}{4} \times 24 = 18\) The 18th position falls at 3. (iii) P(at least 2 children) = \(\frac{7 + 4 + 3 + 2}{24}\) = \(\frac{16}{24}\) = \(\frac{2}{3}\) (b) The sector for 1 child = \(\frac{5}{24} \times 360\) = \(75^\circ\).

(a) From the diagram, \(|AC|^{2} = 24^{2} + 7^{2}\) = \(576 + 49\) \(|AC|^{2} = 625 \implies |AC| = \sqrt{625} = 25 cm\) \(Area = \frac{1}{2} \times b \times h\) = \(\frac{7 \times 24}{2}\) = \(84 cm^{2}\). (b) Total Surface Area of prism = Area of two right angles ( ABC and EFD) + Area of 3 rectangles (ACDF, ABEF and BCDE) = \(( 2 \times 84 cm^{2}) + (25 \times 40)cm^{2} + (24 \times 40)cm^{2} + (7 \times 40)cm^{2}\) = \((168 + 1000 + 960 + 280) cm^{2}\) = \(2408 cm^{2}\)

(a) (i) From \(\Delta\) TFO, \(\tan x = \tan \theta\) (alternate angles) \(\tan x = \frac{25}{30} = 0.8333\) \(x = \tan^{-1} (0.8333)\) = \(39.80^\circ \approxeq 40^\circ\) Angle of depression of the pole from the woman = 90^\circ - 40^\circ = 50^\circ. (ii) From \(\Delta\) TPQ, \(\tan 46 = \frac{25}{TQ}\) \(TQ = \frac{25}{\tan 46}\) \(TQ = \frac{25}{1.0355}\) = \(24.1429 m\) Height of the pole : \(TF - TQ\) = \(30 m - 24.1429 m\) = \(5.8571 m\) \(\approxeq 6m\)

(a) (b) /RS/ = 4.4 cm.

(a) Perimeter of a circle = \(2\pi r\). Length of a side of square = 2r in terms of r. Sum of perimeter of circle and square = \((2 \pi r) + (4 \times 2r)\) \(2r (\pi + 4) = 100\) \(r ( \frac{22}{7} + 4 ) = 50\) \(r = 50 \div \frac{50}{7}\) \(r = 7 cm\) (b) (i) L = \(4 \times breadth\) B = breadth. 60 cm = 2(4B + B) 2(5B) = 60 cm \(B = \frac{60}{10} = 6 cm\) \(L = 4 \times 6 = 24 cm\). (ii) Let the diagonal be d. \(d^{2} = 6^{2} + 24^{2}\) \(d^{2} = 36 + 576\) \(d^{2} = 612\) \(d = \sqrt{612}\) = \(24.74 cm \approxeq 24.7 cm\).

(4x+3y)\(^2\) - (3x-2y)\(^2\) Using the differences of two squares: (4x+3y+3x-2y)(4x+3y-(3x-2y)) (4x+3y+3x-2y)(4x+3y-3x+2y) (x+5y)(7x+y)

Exterior angle = sum of two interior opposite angles y = 50^\circ (vert. opp ∠s are equal) X = y + 30 X = 50 + 30 X = 80^\circ

Exterior angle = 360 / n = 360 / 12 = 30o

Area of the figure = Area of the rectangle + area of the semicircle \(=L \times h + \frac{1}{2}\pi r^2\) 5 \(\times 15 + \frac{1}{2} \times \frac{22}{7} \times \left(\frac{5}{2}\right)^2\) =75+\(\frac{(22\times 25)}{8\times7}\) =75 + \(\frac{550}{56}\) =84.8cm\(^2\)

If a length of a chord is equal to the length of a radius in the same circle, the triangle formed y the chord and the radii is an equilateral triangle ∴each angle = 60o