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If \(sin P = \frac{5}{13}\) from right angled triangle from Pythagoras theorem \(BC^2 = 13^2 - 5^2\) = 169 - 25 BC = \(\sqrt{144}\) = 12 ∴ cos P - tan P = \(\frac{12}{13} - \frac{5}{12}\) =\(\frac{79}{156}\)

\((PR)^2 = (PQ)^2 + (QR)^2\) \(10^2 = 5^2 + (QR)^2\) \((QR)^2 = 100 - 25\) \(QR = \sqrt{75}\) = \(8.660\) \(\approxeq\) 8.7 km

25^\circ 45' = 25^\circ + \(\frac{45}{60}\)^\circ = 25^\circ + 0.75^\circ = 25.75^\circ

Total number of balls = 5 + 3 + 4 = 12 balls P(first ball is red) = \(\frac{5}{12}\) P(second ball is red) = \(\frac{4}{11}\) \(\therefore\) P(both balls are red) = \(\frac{5}{12} \times \frac{4}{11}\) = \(\frac{5}{33}\)

Total number of balls = 5 + 3 + 4 = 12 balls P(one ball is green and the other is blue) = P(first ball is green and second blue) + P(first ball is blue and the second green) = \(\frac{3}{12} \times \frac{4}{11} + \frac{4}{12} \times \frac{3}{11}\) = \(\frac{2}{11}\)

The probability that an event will certainly occur + the probability that an event will certainly not occur = The probability that an event will certainly occur = 1 - the probability that an event will certainly not occur The probability that an event will certainly not occur = 1 + The probability that an event will certainly occur Option C is true.

Range = 30.5 - 10.5 = 20 years

MODE \(\approx\) 20.5 + 3.5 = 24years

The size of an exterior angle = \(\frac{360}{n}\) From the options, all options can perfectly divide 360^\circ, except 21^\circ.

< PQS = 50^\circ (PQ = PR) < PQS = < SPQ = 50^\circ (QS = PS) In triangle PSQ, < PQS + < QPS + < PSQ = 180^\circ 50^\circ + 50^\circ + < PSQ = 180^\circ < PSQ = 180^\circ - 100^\circ = 80^\circ

In \(\Delta\) PRQ, < SRQ = 90^\circ - 58^\circ < SRQ = < QTS = x (angles in the segment) \(\therefore\) x = 32^\circ

From the diagram r\(^2\) = B\(^2\) + 15\(^2\) = 64 + 225 r = \(\sqrt{289}\) = 17cm

Perimeter of the sector = \(2r + \frac{\theta}{360} \times 2\pi r\) = \(2(14) + \frac{135}{360} \times 2 \times \frac{22}{7} \times 14\) = \(28 + 33\) = 61 cm

\(\frac{LM}{LP} = \frac{3}{4}\) \(\frac{Area of LMN}{Area of LPQ} = \frac{3^2}{4^2}\) Area of LPQ = \(\frac{16}{9}\times 18 = 32 cm^2\) Area of quadrilateral MPQN = 32 – 18 = \(14 cm^2\)

In \(\Delta\) RQP, QR\(^2\) + QP\(^2\) = RP\(^2\) 3\(^2\) + QP\(^2\) = 5\(^2\) QP\(^2\) = 5\(^2\) - 3\(^2\) QP = \(\sqrt{16}\) = 4 cm \(\therefore\) Area = \(\frac{1}{2} \times base \times height\) = \(\frac{1}{2} \times 3 \times 4\) = 6 cm\(^2\)