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Curved surface area of a cylindrical tin = \(2\pi rh\) \(\therefore 2\pi rh = 704cm^2\) \(2 \times \frac{22}{7} \times 8 \times h = 704\) \(h = \frac{704 \times 7}{2 \times 22 \times 8}\) \(h = 14cm\)

1/3 \pir\(^2\) x 9 = 1848 3 x \pir2 = 1848 r2 = 1848/3 x 7/22 r = 14 \(\text{Volume of a cone} = \frac{1}{3} \pi r^2 h\) \(\frac{1}{3} \times \frac{22}{7} \times r^2 \times 9 = 1848\) \(r^2 = \frac{1848 \times 7}{22 \times 3}\) \(r^2 = 196 \therefore r = 14cm\)

Area of parallelogram = base \(\times\) height. \(513 = base \times 19 \implies base = \frac{513}{19}\) = 27 cm

Area of \(\Delta OXY\) = \(12\frac{1}{4} cm^2\) Area of sector OXY = \(\frac{30}{360} \times \frac{22}{7} \times 7 \times 7\) = \(\frac{77}{6} = 12\frac{5}{6} cm^2\) Area of the shaded portion = \(12\frac{5}{6} - 12\frac{1}{4}\) = \(\frac{7}{12} cm^2\)

∠RQT = 180^\circ - (95^\circ + 48^\circ) = 37^\circ ∠OQR = 90^\circ - 37^\circ = 53^\circ ∠QOR = 180^\circ - (53^\circ + 53^\circ) = 74^\circ QSR = 74^\circ/2 = 37^\circ

Reflex < POR = 360^\circ - 126^\circ = 234^\circ \(\therefore\) < PQR = \(\frac{1}{2} \times 234^\circ\) = 117^\circ

∠TRS = ∠RST = 60o ∠PSR + ∠PQR = 180o, 120o + xo = 180o x = 60o

< QOS = 2 < QPS (angle subtended at the centre) \(\therefore\) < QOS = 2 x 36^\circ = 72^\circ

Construction: Join PQ. Then < RSP = 37^\circ = < RQP (angles on the same segment) But < RPQ = 90^\circ (angle in a semi-circle) \(\therefore\) < PRQ = 180^\circ - (90^\circ + 37^\circ) = 53^\circ

Draw line ANB// ML < MNA = 40^\circ (alternate angle AB/ ML) < ANP = 55^\circ - 40^\circ = 15^\circ < PQR = 15^\circ = < ANP (corresponding angles, PN // RQ)

Sum of exterior angles = 360^\circ 2x + 3x + 2x + 3x + 5x = 360^\circ 15x = 360^\circ x = 360^\circ/15 = 24^\circ

\(\frac{\text{t}}{t + 4}\) = \(\frac{7}{7 + 5}\) \(\frac{\text{t}}{t + 4}\) = \(\frac{7}{12}\) 12t = 7t + 28 12t - 7t = 28 5t = 28 t = \(\frac{28}{5}\) = 5.6cm

360/18 = 20

< PTQ = 25^\circ (base angles of an isos. triangle) \(\therefore\) < PQT = 180^\circ - (25^\circ + 25^\circ) = 130^\circ (sum of angles in triangle PQT) \(\therefore\) < RST = 130^\circ - 75^\circ = 55^\circ (exterior angle = sum of 2 opp. interior angles)

In the second quadrant, \(\sin 120 = \sin (180 - 120)\) = \(\sin 60\)