Chapter 12 Exercise 12.1
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1 hapter 1 Eercise 1.1 Q. 1. (i) = = 9 (ii) = = 75 = 85 (iii) + 4 = 5 = 9 = 3 (iv) + 10 = 6 = 576 = 4 (v) 5 = + 7 = 576 = 4 (vi) 41 9 = = 40 Q.. (i) = Q. 3. = 10 = 10 (ii) = = = (iii) = = 34 = 34 (iv) 7 = = 4 = = 6 + = 6.5 =.5 m Q. 4. (i) = = 5 y + 5 = 13 y = 13 5 y = 144 y = 1 (ii) = = = 36 + y y = 599 y = 77 (iii) 100 = 8 + = = y 16 = y y = 4 (iv) = = 100 = 10 6 = 10 + y y = 576 y = 4 Q (i) 80 (80) = = 60 cm (ii) = = = 100 (iii) rea = = 4800 cm ctive Maths 3 Book Strands 1 5 h 1 Solutions 1
2 Q. 6. (a) Side a Side b Hypotenuse c (b) a b c Tick if true ü ü ü ü ü ü ü Q. 7. The width of a door frame is 84 cm and its height is 187 cm 84 cm = X 187 cm = 405 = 405 = 05 Q. 8. (a) Side a Side b Hypotenuse c ctive Maths 3 Book Strands 1 5 h 1 Solutions
3 (b) a b c Tick if true ü ü ü ü ü ü ü Q. 9. Triangle with sides: 85, 77, = 75, = = 75 Triangle is right-angled. Q. 10. Triangle with sides: 7, 4, 5 5 = 65, = = 65 Triangle is right-angled. Q. 11. Triangle with sides 11, 60, = 371, = = 371 Triangle is right-angled. Eercise 1. Q. 1. (i) sin = 5 13 (ii) sin = 1 9 Q.. (i) sin = (ii) sin = 13 (iii) sin = 0 1 cos = 13 tan = 5 0 cos = 9 4 cos = 58 cos = 13 4 cos = tan = 0 40 tan = sin B = tan = 3 sin B = 13 tan = 4 sin B = cos B = 58 cos B = 3 13 cos B = 0 4 tan B = 40 tan B = 3 tan B = 4 Q. 3. (i) () (ii) (i) (iii) (ii) (iv) (iii) (v) (iv) (vi) (v) (vii) 0.16 (vi) (viii) 0.5 (vii) (i) (viii) Q. 4. (i) 38.6 (vi) 8.84 (ii) 9.61 (vii) (iii) (viii) 46.3 (iv) 5.94 (i) (v) 0.6 () ctive Maths 3 Book Strands 1 5 h 1 Solutions 3
4 Q. 5. (i) 30ʹ (iv) 5 4ʹ (ii) 15ʹ (v) 1 1ʹ (iii) 45ʹ (vi) 0 0ʹ Q. 6. (i).5 (iv) (ii) (v) 11.6 (iii) 5.83 (vi) Q. 7. (i) 75 57ʹ (iv) 41 34ʹ (ii) 48 14ʹ (v) 17 4ʹ (iii) 43 55ʹ (vi) 6 49ʹ (vii) 73 54ʹ (i) 1ʹ (viii) 13 57ʹ () 58 01ʹ Q. 8. tan B = 9 40 B = tan 1 ( 9 40 ) = Q. 9. cos = = ʹ Eercise 1.3 Q. 1. (i) = tan (ii) = cos = 10 cos 60 = 5 (iii) = cos = 10.3 (iv) = sin 35 0 = (v) = 16 sin 30 = 8 (vi) = cos = 40 cos 45 = 8.8 Q.. (i) y = 7 tan 3 = 4.37 (ii) y = 50 sin 9 = 4.4 (iii) y = 18 cos 35 = (iv) y = 1 tan 30 = 6.93 (v) y = 6 cos 50 = (vi) y = 10 sin 30 = 5 Q. 3. = 100 sin 40 = 64.8 y = 64.8 tan 10 y = Q. 4. = 3 tan 50 z = = z = z 1 y = y = Q. 5. (i) sin = 4 7 = sin = = 35 (ii) cos = 6 = cos 1 6 = 70.5 = 71 (iii) tan = 5 6 = tan = 39.8 = 40 4 ctive Maths 3 Book Strands 1 5 h 1 Solutions
5 (iv) tan = 5 3 = tan = = 59 (v) sin = 7 8 = sin = = 61 (vi) tan = 4 4 = tan 1 1 = 45 Eercise 1.4 Q. 1. (i) sin 67 = 46 = 46 sin 67 Q.. 4 km (ii) 46 = 3 km/hr (iii) 4 = 1 km/hr = tan m = tan 77 = 449. m 00 Q. 3. = tan = tan = tan 30 = 346 m Q. 4. (i) sin 47 = = sin metres (ii) 16 = sin = sin 63 = 18 m Q. 5. tan 69 = 0.4 = 0.4 tan 69 = 57.3 m Q. 6. tan 33 = 5 = 5 tan 33 = 3.5 Liam s height 1.7 m Tree = 4.95 metres Q. 7. tan 78 = 100 = 100 tan 78 = m = 471 m Q. 8. find cos 35 = 6 cos 35 = = 6 (.81915)() = 6 = m = 7.3 m Multiplied by = m 35 6 D 55 z y 15 find y (7.346) = (6) + z z = z = 4. (4.) + (15) = (y) (y) = 4.64 y = m = m Multiplied by = Total length of wire = ( ) = 45.8 m ctive Maths 3 Book Strands 1 5 h 1 Solutions 5
6 Q. 9. (b) DB = = 55 D = tan D = y B 5 (a) BD = 6 5 BD = D Q. 10. (a) tan 5 = h h tan 40 = + 10 (b) h = tan 5 (c) h = ( + 10) tan 40 (d) tan 5 = ( + 10) tan 40 = 19.03m (e) tan 5 tan 40 = 10 tan 40 (tan 5 tan 40) = 10 tan tan 40 = tan 5 tan 40 = m (f) h = (19.03)(tan 5 ) h = 4.36 m BD = 11 BD = 11 BD = feet 3 feet 4 inches (b) sin( DB) = 5 6 DB = sin DB = DB = 56 Eercise 1.5 Q sin 1 3 cos 3 1 tan (c) find tan 50 = 5 tan 50 = = (1.1917)() = 5 find y = = 4. feet Q.. (i) 30 (ii) tan 60 = B 1 3 = B (iii) 1 + ( 3 ) = = = 4 = sin 50 = 5 y sin 50 = y =.766 (.766)(y) = 5 y = 6.57 y = 6.53 feet Total length = = % wastage = 1.03 feet ctive Maths 3 Book Strands 1 5 h 1 Solutions Q. 3. (i) = 45 (ii) cos 45 = B 7 1 = B 7 B = 7 B = 7 Q. 4. (i) cos 45 = 4 = 4 Isoceles Δ so y = 4. (ii) 90
7 Q. 5. ( 1 ) Q. 6. ( 3 ) + ( 1 = ) + = 1 + ( ) + ( 3 ) = = 4 Q. 7. ( ) ( ) Q ( 3 ) = 4 3 sin 30 cos 30 = = 1 3 = tan 30 3 Eercise 1.6 Q. 1. Q.. Q. 3. Q. 4. Q. 5. Q. 6. Q. 7. (1)(15) sin 55 = 73.7 square units (4)(0) sin 30 = 10 sq units (14)(18) sin 40 = sq units (5)(30) sin 5 = sq units (10)(10) sin 70 = sq units (1)(1) sin 80 = sq units (7)(8.4) sin 6 = 5.96 sq units Q. 8. (9)(18.4) sin 8 = 8.8 sq units Q. 9. (i) 6 rea = 1 8 rea = ab sin 1 = (6)(8) sin sin = = sin 1 = 30 Q. 10. (ii) 5 3 rea = 37.5 ctive Maths 3 Book Strands 1 5 h 1 Solutions 10 rea = ab sin 37.5 = ( 5 3 )(10) sin sin = 3 36 = sin = rea = 30 rea = ab sin 30 = (15)()(sin 36) = = 7 Q. 11. (a) D 75 cos X = (60)(75) BD = D X Y B 50 cos Y = (75)(50) DB = (5.89) (b) DB = 180 ( ) = B 7
8 Q. 1. Y 6 4 Q Q Q X (i) cos XYZ = Y 3 Z X 4 Z Q Q Q Q Eercise 1.8 (XZ) = 5 3 (XZ) = 16 (XZ) = 4 sin XYZ = 4 5 (ii) rea = ab sin Eercise 1.7 rea = (6)(4) ( 4 5 ) rea = 48 5 rea = 9.6 units θ θ S θ T 360 θ Q. 1. cos = 45 cos 45 = 1 Q.. sin = 30 = sin , 360 = Q. 3. cos = 60 cos 60 Q Q.. 1 Q Q Q Q = Q. 4. sin = 30 sin 30 = Q. 5. tan = 30 tan 30 = ctive Maths 3 Book Strands 1 5 h 1 Solutions
9 Q. 6. cos = 45 cos 45 = 1 Q. 7. sin = 60 sin 60 = 3 Q. 8. cos = 30 cos 30 = 3 Q. 9. tan = 60 tan 60 = 3 Q. 10. tan 60 = 3 Q Q Q Q Q Q Q Q Q Q Q. 1. sin = sin is positive = 30 OR = Q.. cos = 1 = T S cosine is positive = 45 OR = = 315 Q. 3. tan = tan is positive = 60 OR = = 40 ctive Maths 3 Book Strands 1 5 h 1 Solutions 9
10 Q. 4. sin = = 5 OR = Q. 5. tan = 1 3 = sin is negative tan is negative = 150 OR = = 315 Q. 6. sin = = 40 OR = = 300 sin is negative Eercise 1.9 Q. 1. Q.. sin 50 = 10 sin 40 = 10 sin 50 sin 40 = 11.9 sin 30 = 0 sin 40 = 0 sin 30 sin 40 = Q. 3. Q. 4. Q. 5. sin 15 = 10 sin 5 = 10 sin 15 sin 5 = sin 30 = 0 sin sin 30 = sin 100 = sin 70 = 10 sin 60 = 10 sin 70 sin 60 = ctive Maths 3 Book Strands 1 5 h 1 Solutions
11 Q. 6. Q. 7. Q (i) sin = sin sin =.4596 = sin = 7.36 (ii) = 180 ( ) = (iii) rea = ab sin rea = (15)(5)(sin 10.64) rea = units = units sin 1 = sin sin sin = = sin ( 1 1 sin ) = (i) sin sin (53 8ʹ) = sin =.93 = sin 1.93 = = 68 6ʹ (ii) 180 = ( ʹ) = (iii) rea = ab sin rea = (8)(9.3) (sin ) rea = 9.65 units Q. 9. Q. 10. Q. 11. (iv) rea = ab sin 9.65 = (8)()(sin68.44) sin 16 =.59 sin 44 6ʹ = sin 44 6ʹ sin = 14 = sin ( 1 16 sin 44 6ʹ 14 ) = (i) sin 7 X 1 sin 45 = 1 sin = 7 4 = 4.36 = 4 (ii) rea = ab sin rea = (7)(1)(sin 111) rea = 39.1 units (iii) rea = ab sin 39.1 = (1)()(sin 4) = PQ sin 60 = 80 sin sin 60 PQ = sin 70 = 74 m RR sin 50 = 80 sin sin 50 PR = sin 70 = 65 m ctive Maths 3 Book Strands 1 5 h 1 Solutions 11
12 Q sin sin 40 = sin sin = = sin ( 1 80 sin ) = 5 10 rea = (80)(10) sin 115 = metres Q. 13. (a) sin 50 = sin = 4.6 cm (b) V = (.4)(.98)(.48) (c) (d) V = m 3 98 = (48) + (98) = 11,908 = cm 48 5 The diagonal is greater than the length of the bar so the bar will lie flat on the floor of the bo = 40 + (109.1) = 13, = 116. cm Ma length = 116. cm Eercise 1.10 Q. 1. a = (4)(5) cos 60 a = 1 a = 1 = 4.58 Q.. a = (5)(10) cos 70 a = a = 9.53 Q. 3. a = (9)(1) cos 80 a = a = Q. 4. a = (5)(5) cos 45 34ʹ a = a = 3.87 Q. 5. a = (15)(1) cos 85 a = a = Q. 6. cos = (5)(8) = cos 1 (0.5) = 60 Q. 7. cos = (5)(3) = cos 1 ( 0.5) = 10 Q. 8. cos = (5)(6) = cos 1 (0.75) = 41 Q. 9. cos = (10)(9) = cos 1 (0.9556) = 17 Q. 10. cos = (3)(4) = cos 1 (0) = 90 Q. 11. (i) = (4)(6) cos 60 = 8 = 8 = ctive Maths 3 Book Strands 1 5 h 1 Solutions
13 (ii) B B sin 30 = 5.9 sin 53 8ʹ 5.9 sin 30 B = sin 53 8ʹ B = 3.31 (iii) B = (5.9)(3.31) sin 96 5ʹ = 8.69 D = (4)(6) sin 60 = rea = units squared EG Q. 1. (i) sin 56 = 7 sin 30 7 sin 56 EG = sin 30 = 1 cm (ii) cos EHG = (8)(6) EHG = cos 1 ( ) EHG = 117 (iii) (6)(8) sin 117 = (7)(1)sin 94 = square units Q. 13. (i) 5 cm 7 cm B (iii) Q. 14. (i) 5 9 cos = (5)(7) = cos 1 ( 0.1) = 96 (iv) (5)(7) sin units square (ii) (iii) 7 cm 7 5 cm 3 cm cos = (5)(7) = cos 1 (0.986) = (iv) (5)(7) sin square units Q. 15. (i) = 15 km (ii) = 1.5 km (iii) 150 (iv) = (15)(1.5) cos 150 = = 6.57 km (ii) 96 9 cm Q. 16. (i) = (176)(135) cos = = km 110 km (ii) Journey = = 41 km T = D S = hrs. 80 Total trip = 4.5 hrs. ctive Maths 3 Book Strands 1 5 h 1 Solutions 13
14 (iii) We need both journey times. ar: 17 =.54 hrs = hrs 3 mins 50 km/h 110 Helicopter: = 0.39 hrs Limerick to Waterford 80 km/h = 0.63 hrs Dublin to Limerick Stopover 1 hr Total time to Waterford including stopover =.0 hrs = hrs 1 min The car arrives in Waterford 31 mins after the helicopter. The helicopter must be in the air in hrs from time of landing. Ma. length for meeting is 1hr 9 mins. Q. 17. (i) cos B = (33.07)(33.07) B = cos 1 (0.464) B = (ii) (33.07)(33.07) sin m (iii) = m (iv) = m Q. 18. (i) = (35)(40) cos 10 = 45 = 65 m (ii) = 140 m (iii) Use are of Δ. (35)(40) sin 10 = (35)( height) 40 sin 10 = metres m or sin 60 = 40 = 40 sin 60 B = metres. Eercise 1.11 Q.. (i) p(0) ( ) = 4.44p cm Q. 1. (i) p(10) = 13.89p cm (ii) p(1) (ii) p(5) ( ) = 60p cm = 13.89p cm (iii) p (18) (iii) p(30) ( ) = 189p cm = 8.33p cm 14 ctive Maths 3 Book Strands 1 5 h 1 Solutions
15 Q. 3. (i) q 360 p (6) = 6p cm 6q 360 = 1 q = 60 (ii) q 360 p (1) = 68p q 360 = q = 135 (iii) q 81p = 45p 360 q 360 = q = 00 Q. 4. (i) q p (18) = 7p 360 q 360 = 7 36 q = 70 (ii) q p (6) = 4p 360 q 360 = 4 1 q = 10 (iii) q p (4) = 40p 360 q 360 = q = 300 Q. 5. (i) (7)(7)sin 70 = 3.0 sq units (ii) p(7) = 9.93 sq units (iii) Shaded region = 6.91 sq units Q. 6. Δ (13)(10) sin 40 = Sector 360 p 15 = Shaded = = sq units Revision Eercises Q. 1. (a) sin a = 0 9 sin b = 1 9 (b) (i) sin a 7 sin a = sin 100 = 15 cots a = 1 9 cos b = sin a = sin 1 (0.4596) a = 7 k (ii) sin 53 = 15 sin sin 53 k = sin 100 k = 1.16 cm tan a = 0 1 tan b = 1 0 (iii) y = (1.16)(0) cos 10 y = y = 8.13 cm Q.. (a) sin a cos a tan a p q p q (b) (i) ()( 3 ) sin 80 = = 3 = 3.5 (ii) y = (3.5) ( 3 ) cos 80 = y = 3.64 ctive Maths 3 Book Strands 1 5 h 1 Solutions 15
16 Q. 3. (a) a = = 30 sin a = cos a = 3 tan a = 1 3 sin 10 = cos 10 = 3 tan 10 = 1 3 m = 3 n = = ( 3, ) (b) (i) sin 70 = 6 sin 65 6 sin 70 = sin 65 = 6. cm (ii) (6.+5)(6)sin 45 = = 4 sq units Q. 4. (a) (i) sin = cos 60 sin = = 30 (ii) tan B = 1 B = 45 (iii) sin = cos = 45 (iv) sin D cos 30 = 3 4 sin D ( 3 ) = 3 4 sin D = = 3 = 3 D = 60 (b) (i) tan b = 5 1 (ii) cos b = 1 13 h = h = 13 (iii) sin b = 5 13 (iv) sin b = sin b cos b = ( 5 13 )( 1 13 ) = Q. 5. O = 1 = 4 km OB = 16 = 3 km B = (4)(3) cos 90 B = 1600 B = 40 km Q. 6. (a) PQ sin 30 = 1 sin 30 = 1 1 = sin 30 = 4 PR P 4 R PR = (17) cos 50 = PR = Q 16 ctive Maths 3 Book Strands 1 5 h 1 Solutions
17 (b) 15 cm 50 y Q. 8. (i) 40 = sin y = sin = 15 sin 50 y = 15 sin 40 = y = rea = sq units Q. 7. (i) r = cm (ii) p.09 cm (iii) (10)(10) sin cm (iv) () = 39.1 cm β α d First, John should set up the clinometer at a point,, a distance d from the base of the wall. The value of d can be found using the measuring tape. From, John will measure the angle of elevation, b, of the billboard and the angle of elevation, a, of the bottom of the billboard. h (ii) = tan 3 5 h = 5 tan 3 h = m = 00 cm a + h = tan 45 5 a + h = 5 tan 45 a + h = 5 a + = 5 a = 3 m = 300 cm Q. 9. (i) = (6)(10) cos 47 = = 7.36 m (ii) (iii) α 10 tan a =.4 10 a = tan m a = tan = 6 = 6 tan =.4 m ctive Maths 3 Book Strands 1 5 h 1 Solutions 17
18 Q. 10. (i) (ii) Q. 11. (i) m (ii) cos 15 = B 30 B = 30 cos 15 B = 8.98 m tan 35 = B 8.98 B = 8.98 tan 35 Hill + tower = 0.9 metres Hill = sin 15 = 30 = 7.76 Tower = = 1.53 metres 5 cm cm D? B L D B 5 60 X sin() 5 sin = = sin 60 5 sin 60 = (79.8) (iii) F = 180 ( ) = (40.) (iv) D E sin (40.) = 18 sin sin (40.) = sin 60 = 13.4 DE = 13.4 cm F 0 cm 18 cm D E 18 ctive Maths 3 Book Strands 1 5 h 1 Solutions
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