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130 GEOMETRY AND TRIGONOMETRY Figure 12.30 6. A laboratory 9.0 m wide has a span roof which slopes at 36 ◦ on one side and 44 ◦ on the other. Determine the lengths of the roof slopes. [6.35 m, 5.37 m] 12.12 Further practical situations involving <strong>trigonometry</strong> Problem 28. A vertical aerial stands on horizontal ground. A surveyor positioned due east Figure 12.31 DC 2 = 30.02 0.261596 = 3440.4 AB sin 50 ◦ = AO sin B of the aerial measures the elevation of the top as 48 ◦ . He moves due south 30.0 m and measures the elevation as 44 ◦ . Determine the height of the aerial. Hence, height of aerial, DC = √ 3440.4 = 58.65 m In Fig. 12.31, DC represents the aerial, A is the initial position of the surveyor and B his final position. Problem 29. A crank mechanism of a petrol From triangle ACD, tan 48 ◦ = DC AC , engine is shown in Fig. 12.32.Arm OA is 10.0 cm long and rotates clockwise about O. The connecting rod AB is 30.0 cm long and end B is from which AC = DC tan 48 ◦ constrained to move horizontally. Similarly, from triangle BCD, BC = DC tan 44 ◦ For triangle ABC, using Pythagoras’ theorem: BC 2 = AB 2 + AC 2 Figure 12.32 ( ) DC 2 ( ) DC 2 tan 44 ◦ = (30.0) 2 (a) For the position shown in Fig. 12.32 determine the angle between the connecting rod + tan 48 ◦ AB and the horizontal and the length of OB. ( ) (b) How far does B move when angle AOB DC 2 1 tan 2 44 ◦ − 1 tan 2 48 ◦ = 30.0 2 changes from 50 ◦ to 120 ◦ ? DC 2 (1.072323 − 0.810727) = 30.0 2 (a) Applying the sine rule:
from which, AO sin 50◦ sin B = = AB = 0.2553 10.0 sin 50◦ 30.0 Hence B = sin −1 0.2553 = 14 ◦ 47 ′ (or 165 ◦ 13 ′ , which is impossible in this case). Hence the connecting rod AB makes an angle of 14 ◦ 47 ′ with the horizontal. Angle OAB = 180 ◦ − 50 ◦ − 14 ◦ 47 ′ = 115 ◦ 13 ′ . Applying the sine rule: 30.0 sin 50 ◦ = OB sin 115 ◦ 13 ′ from which, OB = 30.0 sin 115◦ 13 ′ sin 50 ◦ = 35.43 cm (b) Figure 12.33 shows the initial and final positions of the crank mechanism. In triangle OA ′ B ′ , applying the sine rule: 30.0 sin 120 ◦ = 10.0 sin A ′ B ′ O from which, sin A ′ B ′ 10.0 sin 120◦ O = = 0.2887 30.0 INTRODUCTION TO TRIGONOMETRY 131 Hence B moves 11.71 cm when angle AOB changes from 50 ◦ to 120 ◦ . Problem 30. The area of a field is in the form of a quadrilateral ABCD as shown in Fig. 12.34. Determine its area. Figure 12.34 A diagonal drawn from B to D divides the quadrilateral into two triangles. Area of quadrilateral ABCD = area of triangle ABD + area of triangle BCD = 2 1 (39.8)(21.4) sin 114◦ + 2 1 (42.5)(62.3) sin 56◦ = 389.04 + 1097.5 = 1487 m 2 B Figure 12.33 Hence A ′ B ′ O = sin −1 0.2887 = 16 ◦ 47 ′ (or 163 ◦ 13 ′ which is impossible in this case). Angle OA ′ B ′ = 180 ◦ − 120 ◦ − 16 ◦ 47 ′ = 43 ◦ 13 ′ . Applying the sine rule: 30.0 sin 120 ◦ = OB ′ sin 43 ◦ 13 ′ from which, OB ′ = 30.0 sin 43◦ 13 ′ sin 120 ◦ = 23.72 cm Since OB = 35.43 cm and OB ′ = 23.72 cm then BB ′ = 35.43 − 23.72 = 11.71 cm. Now try the following exercise. Exercise 60 Further problems on practical situations involving <strong>trigonometry</strong> 1. PQ and QR are the phasors representing the alternating currents in two branches of a circuit. Phasor PQ is 20.0A and is horizontal. Phasor QR (which is joined to the end of PQ to form triangle PQR) is 14.0A and is at an angle of 35 ◦ to the horizontal. Determine the resultant phasor PR and the angle it makes with phasor PQ. [32.48A, 14 ◦ 19 ′ ] 2. Three forces acting on a fixed point are represented by the sides of a triangle of dimensions 7.2 cm, 9.6 cm and 11.0 cm. Determine the angles between the lines of action and the three forces. [80 ◦ 25 ′ ,59 ◦ 23 ′ ,40 ◦ 12 ′ ]
Page 1 and 2: Geometry and trigonometry 12 Introd
Page 3 and 4: INTRODUCTION TO TRIGONOMETRY 117 Fi
Page 5 and 6: INTRODUCTION TO TRIGONOMETRY 119 5.
Page 7 and 8: Hence 129.9 + x = 75 tan 20 ◦ = 7
Page 9 and 10: ( ) 1 (c) cot −1 2.1273 = tan −
Page 11 and 12: INTRODUCTION TO TRIGONOMETRY 125 or
Page 13 and 14: INTRODUCTION TO TRIGONOMETRY 127 Ap
Page 15: INTRODUCTION TO TRIGONOMETRY 129 th
Page 19 and 20: Geometry and trigonometry 13 Cartes
Page 21 and 22: 4. (−5.4, 3.7) 5. (−7, −3) 6.
Page 23 and 24: Geometry and trigonometry 14 The ci
Page 25 and 26: (a) Since π rad = 180 ◦ then 1 r
Page 27 and 28: For example, Fig. 14.8 shows a circ
Page 29 and 30: The unit of angular velocity is rad
Page 31 and 32: THE CIRCLE AND ITS PROPERTIES 145 P
Page 33 and 34: 60° 12 cm 12 cm h ASSIGNMENT 4 147
Page 35 and 36: 180° X Figure 15.2 90° Y Quadrant
Page 37 and 38: θ = tan −1 1.7629 = 60 ◦ 26
Page 39 and 40: TRIGONOMETRIC WAVEFORMS 153 B Figur
Page 41 and 42: where α is a phase displacement co
Page 43 and 44: TRIGONOMETRIC WAVEFORMS 157 Figure
Page 45 and 46: (c) When t = 10 ms ( then v = 340 s
Page 51 and 52: Now try the following exercise. Exe
Page 53 and 54: tan x + sec x LHS = ( sec x 1 + tan
Page 55 and 56: TRIGONOMETRIC IDENTITIES AND EQUATI
Page 57 and 58: 3. 2 cosec 2 t − 5 cosec t = [ 12
Page 59 and 60: Geometry and trigonometry 17 The re
Page 61 and 62: THE RELATIONSHIP BETWEEN TRIGONOMET
Page 63 and 64: Hence = ( tan x + π ) ( tan x −
Page 65 and 66: COMPOUND ANGLES 179 B Figure 18.3 H
Page 67 and 68:
COMPOUND ANGLES 181 ◦ ′ ◦ ′
Page 69 and 70:
COMPOUND ANGLES 183 Now try the fol
Page 71 and 72:
From equation (7), cos 6x + cos 2x
Page 73 and 74:
COMPOUND ANGLES 187 p i v p v + i B
Page 75 and 76:
Geometry and Trigonometry Assignmen
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