trigonometry

More documents

Recommendations

Info

124 GEOMETRY AND TRIGONOMETRY 9. sin −1 0.2341 10. cos −1 0.8271 11. tan −1 0.8106 12. sec −1 1.6214 13. cosec −1 2.4891 14. cot −1 1.9614 [ 13.54 ◦ ,13 ◦ 32 ′ ] , 0.236 rad [ 34.20 ◦ ,34 ◦ 12 ′ ] , 0.597 rad [ 39.03 ◦ ,39 ◦ 2 ′ ] , 0.681 rad [ 51.92 ◦ ,51 ◦ 55 ′ ] , 0.906 rad [ 23.69 ◦ ,23 ◦ 41 ′ ] , 0.413 rad [ 27.01 ◦ ,27 ◦ 1 ′ ] , 0.471 rad In Problems 15 to 18, evaluate correct to 4 significant figures. 15. 4 cos 56 ◦ 19 ′ − 3 sin 21 ◦ 57 ′ [1.097] 16. 11.5 tan 49 ◦ 11 ′ − sin 90 ◦ 3 cos 45 ◦ [5.805] 17. 5 sin 86 ◦ 3 ′ 3 tan 14 ◦ 29 ′ − 2 cos 31 ◦ 9 ′ [−5.325] 18. 6.4 cosec 29 ◦ 5 ′ − sec 81 ◦ 2 cot 12 ◦ [0.7199] 19. Determine the acute angle, in degrees and minutes, correct ( to the nearest minute, given 4.32 sin 42 by sin −1 ◦ 16 ′ ) 7.86 [21 ◦ 42 ′ ] 20. If tan x = 1.5276, determine sec x, cosec x, and cot x. (Assume x is an acute angle) [1.8258, 1.1952, 0.6546] In Problems 21 to 23 evaluate correct to 4 significant figures 21. (sin 34 ◦ 27 ′ )(cos 69 ◦ 2 ′ ) (2 tan 53 ◦ 39 ′ ) [0.07448] 22. 3 cot 14 ◦ 15 ′ sec 23 ◦ 9 ′ [12.85] 23. cosec 27 ◦ 19 ′ + sec 45 ◦ 29 ′ 1 − cosec 27 ◦ 19 ′ sec 45 ◦ 29 ′ [−1.710] 24. Evaluate correct to 4 decimal places: (a) sine (−125 ◦ ) (b) tan (−241 ◦ ) (c) cos (−49 ◦ 15 ′ ) [ ] (a) −0.8192 (b) −1.8040 (c) 0.6528 25. Evaluate correct to 5 significant figures: (a) cosec (−143 ◦ ) (b) cot (−252 ◦ ) (c) sec (−67[ ◦ 22 ′ ) ] (a) −1.6616 (b) −0.32492 (c) 2.5985 12.7 Sine and cosine rules To ‘solve a triangle’ means ‘to find the values of unknown sides and angles’. If a triangle is right angled, trigonometric ratios and the theorem of Pythagoras may be used for its solution, as shown in Section 12.4. However, for a non-right-angled triangle, trigonometric ratios and Pythagoras’ theorem cannot be used. Instead, two rules, called the sine rule and the cosine rule, are used. Sine rule With reference to triangle ABC of Fig. 12.19, the sine rule states: a sin A = Figure 12.19 b sin B = c sin C The rule may be used only when: (i) 1 side and any 2 angles are initially given, or (ii) 2 sides and an angle (not the included angle) are initially given. Cosine rule With reference to triangle ABC of Fig. 12.19, the cosine rule states:
INTRODUCTION TO TRIGONOMETRY 125 or or a 2 = b 2 + c 2 − 2bc cos A b 2 = a 2 + c 2 − 2ac cos B c 2 = a 2 + b 2 − 2ab cos C The rule may be used only when: (i) 2 sides and the included angle are initially given, or (ii) 3 sides are initially given. 12.8 Area of any triangle The area of any triangle such as ABC of Fig. 12.19 is given by: (i) 2 1 × base × perpendicular height, or (ii) 2 1 ab sin C or 2 1 ac sin B or 2 1 bc sin A,or (iii) √ [s(s − a)(s − b)(s − c)], where s = a + b + c 2 Figure 12.20 Area of triangle XYZ = 2 1 xy sin Z = 2 1 (15.2)(18.00) sin 62◦ = 120.8 cm 2 (or area = 2 1 xz sin Y = 2 1 (15.2)(17.27) sin 67◦ = 120.8 cm 2 ). It is always worth checking with triangle problems that the longest side is opposite the largest angle, and vice-versa. In this problem, Y is the largest angle and XZ is the longest of the three sides. Problem 22. Solve the triangle PQR and find its area given that QR = 36.5 mm, PR = 29.6 mm and ∠Q = 36 ◦ . Triangle PQR is shown in Fig. 12.21. B 12.9 Worked problems on the solution of triangles and finding their areas Problem 21. In a triangle XYZ, ∠X = 51 ◦ , ∠Y= 67 ◦ and YZ = 15.2 cm. Solve the triangle and find its area. The triangle XYZ is shown in Fig. 12.20. Since the angles in a triangle add up to 180 ◦ , then Z = 180 ◦ − 51 ◦ − 67 ◦ = 62 ◦ . Applying the sine rule: Using Using 15.2 sin 51 ◦ = 15.2 sin 51 ◦ = y sin 67 ◦ = y = 15.2 sin 51 ◦ = z = z sin 62 ◦ y and transposing gives: sin 67◦ 15.2 sin 67◦ sin 51 ◦ = 18.00 cm = XZ z and transposing gives: sin 62◦ 15.2 sin 62◦ sin 51 ◦ = 17.27 cm = XY Figure 12.21 Applying the sine rule: 29.6 sin 36 ◦ = 36.5 sin P from which, sin P = 36.5 sin 36◦ 29.6 = 0.7248 Hence P = sin −1 0.7248 = 46 ◦ 27 ′ or 133 ◦ 33 ′ . When P = 46 ◦ 27 ′ and Q = 36 ◦ then R = 180 ◦ − 46 ◦ 27 ′ − 36 ◦ = 97 ◦ 33 ′ . When P = 133 ◦ 33 ′ and Q = 36 ◦ then R = 180 ◦ − 133 ◦ 33 ′ − 36 ◦ = 10 ◦ 27 ′ . Thus, in this problem, there are two separate sets of results and both are feasible solutions. Such a situation is called the ambiguous case.
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: ( ) 1 (c) cot −1 2.1273 = tan −
Page 13 and 14: INTRODUCTION TO TRIGONOMETRY 127 Ap
Page 15 and 16: INTRODUCTION TO TRIGONOMETRY 129 th
Page 17 and 18: from which, AO sin 50◦ sin B = =
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
show all

trigonometry

Create successful ePaper yourself

Delete template?

Save as template?