Capter 1 Eercise 1.1 Q. 1. (i) = 1 + 5 = 1 (ii) = 1 + 5 = 7 (iii) = 1 0 = 99 (iv) = 41 40 = 9 (v) = 61 11 = 60 (vi) = 65 6 = 16 Q.. (i) = 8 sin 1 = 4.1 5.5 (ii) = cos 68 = 14.7 1 (iii) = sin 49 = 15.9 4 (iv) = tan 4 = 5.9 (v) = 8.5 cos 18 = 8.1 (vi) = 5 tan 5 = 6.6 Q.. (i) A = cos 1 ( 5 9 ) = 56 (ii) A = sin 1 ( 1 ) = 56 (iii) A = sin 1 ( 4 8 ) = 0 (iv) A = tan 1 ( 15 ) = 56 (v) A = tan 1 ( 4 6 ) = 4 (vi) A = cos 1 ( 6 ) = 71 Q. 4. tan 5 = tan 7 = 4 = tan 5 4 = tan 7 = 4.0 = 7.9 Q. 5. (i) = 4 (ii) A = tan 1 ( 4 1 ) = 18 Q. 6. cos 0 = AC AC = cos 0 ( AD = cos 0 ( ( cos 0 ) ) cos 0 80 AE = = cos 0 9 cos 0 ) Q. 7. Q. 8. Q. 9. 7 0 m tan 7 = 0 0 tan 7 = So, use te tangent function as sown above to evaluate te eigt. 9 = sin 9 = sin 9 + 1.5 =.7 metres =,7 cm 14 Active Mats (Strands 1 5): C 1 Solutions (i) sin = 14 14 = sin 6.4 metres (ii) = 6.4 cos metres Q.. (i) No. For a given opposite lengt, doubling te adjacent lengt does not alve te angle. It alves te tangent of te angle. (ii) 0 1 1
Q. 11. (i) tan 0 = 1 = 1 tan 0 = 4 eigt = 4 + 1.4 = 8. metres Q. 1. (i) 0 (ii) 7 81 tan 7 = 87 = 81 tan 7 = 41 metres 68 tan 68 = 0 0 = tan 68 = 1.1 0 q 81 q = tan 1 ( 81 ) q = 5 Q. 1. (i) sin 0 = = sin 0 = 4 metres (ii) = (4 cos 0 ) = 8 cos 0 = 6.9 metres tan = 0 0 = tan = 46.0 BC = 46. 1.1 = 4.08 metres (ii) 4.08 =.4 m/s Q. 14. (i) tan A = 8 6 = 4 Draw a rigt triangle wit opposite lengt 4 units and adjacent lengt units. 4 units A units Use a ruler, set square and compass. (Diagram sown is not to scale.) Active Mats (Strands 1 5): C 1 Solutions
(ii) B Ptagoras, potenuse lengt is 5 units. sin A = 4 5 cos A = 5 Q. 15. (i) Similar metod as to Q.14. A 4 units (ii) Hpotenuse lengt is 5 units sin A = 5 units Q. 17. (i) (iv) = tan 4 = (7 ) tan 48 tan 4 = 7 tan 48 tan 48 tan 4 + tan 48 = 7 tan 48 (tan 4 + tan 48 ) = 7 tan 48 7 tan 48 = tan 4 + tan 48 (v) =.865. = (.865.) tan 4 =.48 m = 48 cm cos A = 4 5 Q. 16. (i) q = 90 4 = 48 q = sin q (ii) 7 metres (iii) (ii) q q M 4 tan 4 = = tan 4 N = cos q l = cos q + + cos q l = + 0 cos q 7 tan 48 = 7 = (7 ) tan 48 48 (iii) Area = 1 ( cos q)( sin q) + ( sin q) + 1 ( cos q)( sin q) = 50 cos q sin q + 0 sin q + 50 cos q sin q A = 0 sin q + 0 sin q cos q (iv) A = 0 sin 60 + 0 sin 60 cos 60 = 75 cm (v) A = 0 sin 45 + 0 sin 45 cos 45 = 50 + 50 cm Active Mats (Strands 1 5): C 1 Solutions
Q. 18. (i) SU is a tangent to te circle and [OU] in te corresponding radius. A tangent and corresponding radius make a 90 angle wit eac oter. (ii) 90 q (iii) 180 in a triangle. So OSU = 180 90 (90 q) = q (iv) UC = sin q (v) OC = cos q Q.. (i) 6 45ʹ (vi) 6 8ʹ (ii) 16 5ʹ (vii) 7.8ʹ (iii) 68 ʹ (viii) 1 40ʹ (iv) 1 17ʹ (i) 7.0ʹʹ (v) 6 15ʹ () 56.7ʹ Q. 4. (i) A = 90 44 44ʹ = 45 16ʹ (ii) B = 180 50 48ʹ 70 = 59 1ʹ (iii) C = 7 14ʹ + 9 51ʹ = 165 5ʹ (iv) D = 1 (180 0 40ʹ) (vi) Ptagoras: sin q + cos q = 1 (vii) 1 O U tan q = 1 1 = tan q SU = Eercise 1. 4 Active Mats (Strands 1 5): C 1 Solutions 1 tan q Q. 1. (i) 0ʹ (vii) 4 45ʹ (ii) 15ʹ (viii) 6 15ʹ (iii) 45ʹ (i) 55 08ʹ (iv) 5 4ʹ () 0 0ʹ (v) 4 1ʹ (i) 0 45ʹ (vi) 15 0ʹ (ii) 0 4ʹ Q.. (i).5 (vi).55 (ii).67 (vii) 4.17 (iii) 5.8 (viii) 58.75 (iv) 70.7 (i) 56.5 (v) 11.6 () 8. q S = 74 40ʹ Q. 5. (i) AC = 9 + 1 = 15 km 15 (ii) tan 6.87 = DC 15 DC = tan 6.87 DC 0 km (iii) AD = 15 + 0 = 5 km Lengt = 9 + 1 + 0 + 5 = 66 km Q. 6. (i) As CBA = 90 a is acute (onl 180 in a triangle) (ii) If 45 < a < 90 For suc values of a, te opposite lengt eceeds te adjacent lengt (i.e. AB > BC ). (iii) sin a: opposite lengt is alwas less tan potenuse lengt. cos a: adjacent lengt is alwas less tan potenuse lengt. (iv) Coose sa a = 0 sin a = sin 60 0.866 sin a = sin 0 = 1 0.866 1 So statement (b tis eample) is false.
Q. 7. (i) Sun q q = 90 89 50ʹ 58ʹʹ = 9ʹ ʹʹ.84 8 (ii) sin 9ʹʹʹ =.84 8 = sin 9ʹ ʺ 80 50 58 Moon.84 8 Eart =,445,718,857 =.445718857 9 in metres (iii) Time = Distance Speed.445718857 9 =.0 8 = 8.15. 8.15 seconds (iv) r =.445718857 9 Circumference = pr 1.54 metres r = sin 60 16ʹ 4ʹʹ + 960 sin 60 16ʹ 4ʹʹ = r r = (sin 60 16ʹ 4ʹʹ 1) = 960 sin60 16ʹ 4ʹʹ 960 sin 60 16ʹ 4ʹʹ r = sin 60 16ʹ 4ʹʹ 1 r = 6,6.11 r 6,6 km (iv) 7,96 km (v) p(7,96) 45,84 km (vi) 45,84 = 1.697 ours 7,000 Eercise 1. 1 r 4 mins Q. 1. (i) Construct equilateral triangle of side lengt units. Divide in two from top verte. (v) 65 1_ 4 (vi) das = 65.5 4 60 60 = 1,557,600 seconds 1.54 1,557,600 488 m/s 0 Q. 8. (i) 90 60 16ʹ 4ʹʹ = 9 4ʹ 6ʹʹ (ii) r + 960 km (iii) r A O sin 60 16ʹ 4ʹʹ = 60 16 4 r + 960 r r + 960 B 60 1 B Ptagoras: = 1 = So: sin 60 = sin 0 = 1 cos 60 = 1 cos 0 = tan 60 = tan 0 = 1 Construct a rigt-angled isosceles triangle wit equal sides lengt one unit. Active Mats (Strands 1 5): C 1 Solutions 5
45 1 tan 60 = z 9 9 tan 60 = z 9 = z Q.. (i) 60. ΔCAB in equilateral. 45 1 B Ptagoras: = 1 + 1 = So: sin 45 = cos 45 = 1 tan 45 = 1 A 0 45 60 sin A 1 1 cos A 1 1 tan A 1 1 (ii) sin 0 = 1 = = 5 = 5 = 75 = 75 = 5 (ii) RHS (iii). B Ptagoras. (iv) 1 (1)( ) = Q.. (i) = DC (ii) tan 0 = 1 = cos 0 = 5 = z tan 60 = 5 = z z = Q. 5. (i) 45 z = (iii) sin 60 = 6 = 6 = 9 = 6 cos 60 (ii), and = 6 = 6 Active Mats (Strands 1 5): C 1 Solutions 4 + 4 + (iii) 4 + = 4 = ( 1) 4 1 = 4( + 1) = 1 ( + 1) = Q. 4. tan 60 = tan 60 = 6 6 = tan 60 = 45
(iii) 0 60 (iv), 8 6, 16 6 (v) 1 = (vi) tan 0 = 8 6 1 = 8 6 = 4 (1+ ) = 4 4 = 1 + 1.4 (vii) 1.4 1.7 =.7 metres 8 6 Q. 6. (i) Diagram. Se s increasing te opposite lengt as te adjacent lengt remains constant. (ii) 4 + 8 = 4 5 (iii) q = 90 a and b = 90 a (iv) Te bot contain a 90 angle and q = b. BC (v) = 1 1 BC = 4 (vi) AB = (4 ) + 1 = 8 (vii) Yes. In Δ ADE, sin a = 1 In ΔABC, sin a = 4 8 = 1 (viii) AB = 8 units BC = units (i) AD = 8 + = 17 AE = 8 + 4 = 4 5 AF = 8 + 6 = AG = 8 + 8 = 8 Q. 7. ( 1 Q. 8. ( Q. 9. ( 1 () B considering ow te ratio adjacent: potenuse canges as a increases. Let a = DAB, ten EAB, ten FAB etc. Te adjacent lengt AB remains te same, wile te potenuse lengts AD, AE, AF, etc increase. Hence, cos a decreases as a increases. ) + ( 1 ) = 1 + 1 = 1 ) + ( 1 ) + ( ) + ( 1 ) + ( 1 Q.. LHS = Q. 11. (i) O 4 M = ) = 4 + 1 4 + = 4 ) = 1 + 1 4 + 4 = 4 1 = RHS Active Mats (Strands 1 5): C 1 Solutions 0 tan 0 = 4 4 = tan 0 MN = 4 (ii) 4 + (4 ) = 64 = 8 (iii) 4 + 8 = 1 (iv) 1 (8 ) (1) = 48 units Revision Eercises Q. 1. (i) = 6 4 = 4.5 (ii) = 4.6 =.0 (iii) tan 6.6 = 4 4 = tan 6.6 = 8.0 N 7
(iv) = 4. sin 56.4 =.6 (v) cos 47. = 6.8 6.8 = cos 47. =.0 (vi) tan.4 =.6 =.6 tan.4 =.4 (vii) = 6 sin 45 = 4. (viii) = + 4 = 5 Q.. (i) tan 1 ( 4 ) = 7 (ii) cos 1 ( 4 ) = 41 (iii) sin 1 ( 4 6 ) = 4 (iv) tan 1 ( 5 ) = 59 (ii) (iii) 45 45 A H AB = + = 4.4 m B B (v) tan 1 ( 4 ) = 7 Q.. Q D 1 H 40 BD = 1 + =.16 m P 4.1 T tan 40 = 4.1 6 R (iv) (4.4) + + () + (.16) +.16 = 6.96 m 4.1 = tan 40 4.886 tan 6 = 4.886 = 4.886 tan 6 6.75 (v) 6.4 6.96 =,99.904 (,99.904) =,666.56 9 Q. 5. (i) B = tan 1 (0.908) B = tan 1 (0.908) tan B = tan [ tan 1 (0.908)] = 0.65 Area = 1 (4.1 + 6.75)(4.886) (ii) B 6.4 units Q. 4. (i) DH = HE as Δ BDH ΔBHE (RHS) AH = HC as AD = EC ΔABH ΔCBH (SAS) 8 Active Mats (Strands 1 5): C 1 Solutions A 5 tan 5 = 97 = 97 tan 5 4.4 m BT = 4.4 m 97 T
(iii) 4.4 B T C sin 80 0ʹ = 4.4 4.4 = sin 80 0ʹ BC = 4.9 m Q. 6. tan ( 1 ) = 60 Q. 7. 80 0 9 AC = 14 + 0 AC = 596 AC = 149 r = 149 cm Q.. (i) tan 51 = tan 51 = tan 0 = 14.75 + (14.75 + ) tan 0 = (ii) = tan 51 = 14.75 tan 0 + tan 0 (tan 51 tan 0 ) = 14.75 tan 0 = 14.75 tan 0 tan 51 tan 0 Q. 8. Q. 9. 0 tan 0 = 9 9 = tan 0 = 9 Perimeter = 6(9 ) = 54 cm 15 6 = 1 8 15 6 = 1 + 15 Area = 8( 1 ) + 1 (6) ( 1 ) = 4 1 + 9 1 = 1 151. cm 1 AB BC = 140 AB BC = 80 AB (14) = 80 AB = 0 cm 6 (iii) = 1.95 m (iv) = 1.95 tan 51 = 15.99 16 m (v) pr = 5.1 r =.999 r 4 m (vi) Te distance can now be calculated once te distance ( 4) is measured directl. One angle of elevation measurement will ten be sufficient to calculate te eigt, b use of te tangent function. (vii) ( 4) in measured as 8.95. tan 51 = tan 51 = 1.95 = 1.95 tan 51 = 15.99 16 m Active Mats (Strands 1 5): C 1 Solutions 9
Q. 11. (i) tan 61.8 = AB 0 0 tan 61.8 = AB AB = 55.94 AB 56 m (ii) sin BDA = 56 BDA = 7.818 BDA 8 Q. 1. (i) tan 51 6 = 15 = 15 tan 51 6ʹ = 18.95 18.9 m (ii) tan 8 0ʹ = 18.9 15 + Q. 1. A 18.9 15 + = tan 8 0ʹ 19.86 m 1 = 4 + 1 = 17 1 cos A = 17 0 Q. 14. tan 0 15 = 0 + d 0 0 + d = tan 0 15ʹ d = 14.9 d 14 m Q. 15. (i) ACD = 17 (ii) sin 17 = AD sin 17 = AD AD = cm 4 (iii) AC = = 91 cm tan (58 + 17 ) = AB 91 AB = 91 tan (75 0 ) = 6.886 7 cm (iv) BD = 7 = 4 cm Q. 16. (i) DAC = 60 (ii) BD = m (iii) 4 C 60 A B 4 sin 60 = BC 4 BC = sin 60 BC 4.6 m (iv) 4.6 4.1 m Q. 17. (i) Incorrect. tan 50 > 1 (ii) Correct. sin 60 sin 0 (iii) Incorrect. cos 1 < cos 0 (iv) Correct cm 60 0 cm cm 60 60 60 cm 1 cm sin 60 = Active Mats (Strands 1 5): C 1 Solutions
Q. 18. Leaning Tower of Pisa θ.9 q = tan 1 ( 55.86.9 ) q 86.006 55.86 Suurusen curc tower.47 a = tan 1 ( 7.7.47 ) a 84.84 As a < q 7.7 Te Suurusen curc tower is more tilted. Q. 19. (a) Ultan s measurement will be used to calculate te radius of te Spire, wic will allow for te eigt of te Spire to be calculated via use of te tangent function. (b) Q. 0. (a) Q. 1. (a) pr = 7.07 r 1.15 tan 60 = 71.15 71.15 tan 60 = = 1.19 = 1.19 + 1.7 15 m 0 5 l = 0 + 5 l 9. cm (b) (i) tan 60 = = tan 60 = (ii) tan 0 = 4 (4 ) tan 0 = 4 = = = 4 = 4 4 = 4 = 6 = 6 metres l 60 1.7 7.68 56.8 1.6 71.15 Active Mats (Strands 1 5): C 1 Solutions 11
(b) tan 7.68 = 56.8 = 56.8 tan 7.68 = 56.8 tan 7.68 + 1.6 194.1750 metres (c) Second: 194.077 metres Tird: 194.1149 metres (d) 194.1750 + 194.077 + 194.1149 = 194.8 19,41 cm 1 Active Mats (Strands 1 5): C 1 Solutions