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Let A = (1, -3), B = (6, 2) and C = (0,4) \(d =\sqrt{ (y_2 - y_1)^2 + (x_2 - x_1)^2}\) |AB| = \(\sqrt{(2 - (-3))^2 + (6 - 1)^2}\) = \(\sqrt{(2 + 3)^2 + (6 - 1)^2}\) = \(\sqrt{5^2 + 5^2}\) = \(\sqrt{25 + 25}\) = \(\sqrt50\) = 5\(\sqrt2\) units |BC| = \(\sqrt{(4 - 2)^2 + (0 - 6)^2}\) = \(\sqrt{2^2 + (-6)^2}\) = \(\sqrt{4 + 36}\) = \(\sqrt40\) = 2\(\sqrt10\) units |AC| = \(\sqrt{(4 - (-3)^2 + (0 - 1)^2}\) = \(\sqrt{(4 + 3)^2 + (0 - 1)^2}\) = \(\sqrt{7^2 + (-1)^2}\) = \(\sqrt{49 + 1}\) = \(\sqrt50\) = 5\(\sqrt{2}\) units Since two sides are equal (|AB| = |AC|), then the triangle is Isosceles.

Gradient(slope) m = \(\frac{ y_2 - y_1}{ x_2 - x_1}\) the points are \((\frac{1}{2}, \frac{- 1}{3}) and ( 3 , \frac{2}{3})\) m = \(\frac{\frac{2}{3} - (\frac{-1}{3})}{3 - \frac{1}{2}}\) = \(\frac{\frac{2}{3} + \frac{1}{3}}{3 - \frac{1}{2}}\) = \(1 \div\frac{5}{2}\) = \(1\times\frac{2}{5}\) Therefore, m = \(\frac{2}{5}\)

cos (a + 14)^\circ = sin (4a + 6)^\circ Recall, cos \theta = sin (90 - \theta) ⇒ sin (90 - (a + 14)) = sin (4a + 6) ⇒ sin (90 - a - 14) = sin (4a + 6) ⇒ sin (76 - a) = sin (4a + 6) ⇒ 76 - a = 4a + 6 ⇒ 76 - 6 = 4a + a ⇒ 70 = 5a \(a = \frac{70}{5}\) therefore, a = 14

The height of the second building H = h + 24 tan \theta = \(\frac {opp}{adj}\) tan 30\(^o = \frac {h}{x}\) \(\implies\frac{\sqrt 3}{3} = \frac {h}{x}\) \(\implies x = \sqrt 3 = 3h\) \(\implies x = \frac {3h}{\sqrt 3}\) ....(i) tan 60\(^o = \frac {24}{x}\) \(\implies\sqrt 3 = \frac {24}{x}\) \(\implies x\sqrt 3 = 24\) \(\implies x = \frac {24}{\sqrt 3}\) ....(ii) Equate equation (i) and (ii) \(\implies \frac {3h}{\sqrt 3} = \frac {24}{\sqrt 3}\) \(\implies\) 3h = 24 \(\implies h = \frac {24}{3}\) = 8m ∴The height of the second building = 8 + 24 = 32m

From the diagram above; Tan\(\theta = \frac{opp}{adj}\) tan50^\circ = \(\frac{124}{d}\) d = \(\frac{124}{tan50}\) therefore, d = 104.05m

Tan\(\theta = \frac{opp}{adj}\) Tan\(\theta = \frac{56}{110}\) \(\theta = tan^1(0.5091)\) therefore, \(\theta\) = 27^\circ ( to the nearest degree)

a. solution above bi. From ∆PMT the height of the building is PT \(Tan\theta = \frac{opposite}{adjacent}\) tan66^\circ = \(\frac{PT}{50}\) Therefore, PT = \(tan 66^\circ\times50 = 112.3m\) ( to 1 d.p) bii. From ∆PCT \(Tan53^\circ = \frac{PT}{CT}\) \(Tan53^\circ = \frac{112.3}{CT}\) \(Tan53^\circ\times CT = 112.3\) \(CT = \frac{112.3}{Tan53^\circ}\) = 84. 62m. MC = CT - MT = 84.62 - 50 = 34.6m Therefore, MC = 34.6m ( to 1 d.p ).

tan \theta = \(\frac {opp}{adj} = \frac {RQ}{QP} = \frac {6}{8}\) tan \theta = 0.75 \theta = tan\(^{-1} (0.75) = 36.87^o\) ∴ The bearing of R from P = 360\(^o\) - 36.87\(^o\) = 323\(^o\) (to the nearest degree)

Speed = \(\frac{Distance}{Time}\) ⇒ Time = \(\frac{Distance}{Time}\) Let D = distance between the two airports ∴ Time taken to get to the airport = \(\frac{D}{120}\) and Time taken to return =\( \frac{D}{150}\) Since total time of flight= 3hours, ⇒ \(\frac{D}{120} + \frac{D}{150}\) = 3 ⇒ \(\frac{15D + 12D}{1800}\) = 3 ⇒ \(\frac{27D}{1800}\) = 3 ⇒ \(\frac{3D}{200} = \frac{3}{1}\) ⇒ 3D = 200 x 3 ∴ D =\(\frac{ 200\times3}{3}\)= 200km

AB = \(\frac{\theta}{360}\times 2\pi Rcos\propto\) (distance on small circle) = 64 - 56 = 8\(^o\) \(\propto = 86^o\) ⇒ AB = \(\frac{8}{360}\) x 2 x 3.142 x 6370 x cos 86 ⇒ AB = \(\frac{22,338.29974}{360}\) ∴ AB = 62km (to the nearest km)

Consider |NS| \theta = 180^\circ - (90^\circ + 35^\circ) (sum of angles on a straight line is 180^\circ) = 180^\circ - 90^\circ - 35^\circ = 90^\circ - 35^\circ = 55^\circ ∴ His new bearing is N55^\circE.

a. the diagram above solution to question a. bi. Using the sine rule \(\frac{sinP}{XY} = \frac{sinY}{PX}\) = \(\frac{sin93^\circ}{180} = \frac{sinY}{60}\) cross multiply \(sinY\times180 = sin93^\circ\times60\) sinY = 0.3329 Y = \(sin^{-1}0.3329\) Y = 19.44^\circ the angle between 30^\circ and 19.44^\circ at Y = 30^\circ - 19.44^\circ = 10.56^\circ Therefore, the bearing X from Y = 360^\circ - 10.56^\circ = 349^\circ( to 3 sig. figure) bii. Using cosine rule Let a = |PY|, b = |PX| and c = |XY| \(a^2 = b^2 + c^2 - 2bcCos A\) where A = 67.56^\circ \(a^2 = 60^2 + 180^2 - 2(60)(180)cos 67.56^0\) \(a2 = 3600 + 32400 - 8245.06\) \(a^2 = 27754.94^0\) a = \(\sqrt{27754.94}\) a = 166.6 m ∴ |PY| = 167 m (to 3 s.f)

\(lim_{x\to2} \frac {x^2 + 4x - 12}{x^2 - 2x}\) = \(lim_{x\to2} \frac {(x - 2)(x + 6)}{x(x - 2)}\) \(lim_{x\to2} \frac {x + 6}{x}\) \(\frac {2 + 6}{2} = \frac {8}{2}\) = 4

y = \(\sqrt[3]{x^2(2x - x^2)} = x^{2/3} (2x - x^2)\) = \(2x^{5/3} - x^{8/3}\) Now, we can differentiate the function \(\therefore \frac {dy}{dx} = \frac {10x^{2/3}}{3} - \frac {8x^{5/3}}{3}\)

Area of a circle (A) = \(\pi r^2\) Given \(\frac{dA}{dt} = 1.5cm^2s^{-1}\) \(\frac{dr}{dt}\) = ? A = 2cm\(^2\) Now 2 = \(\pi r^2\) = r\(^2 = \frac {2}{\pi}\) r = \(\sqrt \frac {2}{\pi}\) cm = 0.798cm \(\frac {dr}{dt} = \frac {dA}{dt} \times \frac {dr}{dt}\) \(\frac {dA}{dr} = 2\pi r\) (differentiating A = \(\pi r^2)\) \(\frac {dr}{dA} = \frac {1}{2\pi r}\) \(\frac {dr}{dt} = 1.5 \times \frac {1}{2\times \pi \times 0.798} = 1.5 \times 0.199\) \(\frac {dr}{dt} = 0.299cms^{-1}\) (to 3 s.f)