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Tan \(x^o = \frac{100m}{80}\) Tan \(x^o = Tan^{-1} 1.25\) x = 51.34\(^o\)

\(60^2 + x^2 + x^2\) \(360^2 = 2x^2\) \(x^2\) = 1800 x = \(\sqrt{1800}\) x = 42.4264 x = 42.4264 perimeter = 4x = 4 x 42.4264 = 169.7056 = 120\(\sqrt{2}\) = 120\(\sqrt{2}\)

Find the minimum value of y is the point where the base of the curve corresponds to the value of y i.e. at -2.1

RSP = 90 < substance in semi a circle RPS = 180 - (90 + 24) = 180 - (114) = 66 TPS = 180 - 66 = 114 RST = 24 < STR = 180 - (114 + 24) = 180 - 138 = 42\(^o\)

(a) \((y - 1)\log_{10} 4 = y\log_{10} 16\) \((y - 1)\log_{10} 4 = y \log_{10} 4^{2}\) \((y - 1)\log_{10} 4 = 2y\log_{10} 4\) Equating both sides, we have \(y - 1 = 2y \implies -1 = 2y - y\) \(\therefore y = -1\) (b) Let the distance from my house to the office = c. At 4km/h, the time taken to get to the office from the house = \(\frac{c}{4} hr\) At 5km/h, the time taken to get to the office from the house = \(\frac{c}{5} hr\) \(\frac{c}{4} = \frac{c}{5} + \frac{30}{60}\) \(\frac{c}{4} - \frac{c}{5} = \frac{1}{2}\) \(\frac{c}{20} = \frac{1}{2} \implies c = 10km\)

(a) \(\frac{2}{3}(3x - 5) - \frac{3}{5}(2x - 3) = 3\) \(\frac{10(3x - 5) - 9(2x - 3)}{15} = 3\) \(30x - 50 - 18x + 27 = 3 \times 15 = 45\) \(12x = 45 + 23 = 68\) \(x = \frac{68}{12} = 5\frac{2}{3}\) (b) \(n + r = 80^\circ .... (1)\) \(180^\circ - m + r = 88^\circ .... (2)\) From (1), r = 80^\circ - n. Putting into (2), we have \(180 - m + 80 - n = 88 \implies - m - n = 88 - 260 = - 172^\circ\) \(-(m + n) = - 172 \implies m + n = 172^\circ\).

(a) \(< P = 23.6^\circ\) (Alternate angle) \(\tan 23.6 = \frac{x}{50}\) \(x = 50\tan 23.6\) = \(50 \times 0.4369\) = \(21.845 m \approxeq 22 m\) (to the nearest metre) (b) Given that \(\Delta TRU = 45cm^{2}\) \(\implies \frac{1}{2}(bh) = 45\) \(10h = 45 \times 2 = 90\) \(h = 9cm\) Area of trapezium = \(\frac{1}{2}(a + b)h\), where \(a = 6cm; b = 10 + 6 = 16cm; h = 9cm\) = \(\frac{1}{2}(6 + 16) \times 9\) = \(11 \times 9\) = \(99 cm^{2}\)

(a) \(T_{n} = a + (n - 1)d\) (For an AP series) \(T_{6} = 37 = a + (6 - 1) d = a + 5d\) \(a + 5d = 37 .... (1)\) \(S_{n} = \frac{n}{2}(2a + (n - 1) d)\) or \(S_{n} = \frac{n}{2}(a + l)\) where a = first term; and l = last term. Since, we are given the sum of the first 6 terms and also given the sixth term which is the last term in the sum, we can use the second formula where the sixth term is the last term in this case. \(S_{6} = 147 = \frac{6}{2} (a + 37)\) \(147 = 3(a + 37) \implies 147 = 3a + 111 \) \(3a = 147 - 111 = 36 \implies a = 12\) (b) From equation (1) above, \(a + 5d = 37\) \(12 + 5d = 37\) \(5d = 37 - 12 = 25\) \(d = 5\) \(S_{15} = \frac{15}{2} (2(12) + (15 - 1)(5))\) \(\frac{15}{2} (24 + 70) = \frac{15}{2} (94)\) = \(15 \times 47\) = \(705\).