Chapter 5. Exercise 5A. Chapter minor arc AB = θ = 90 π = major arc AB = minor arc AB =

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1 Chapter 5 Chapter 5 Exercise 5. minor arc = cm. major arc = cm. minor arc = cm 4. major arc = = 6 = cm 5. minor arc = = cm 6. major arc = = cm 7. minor sector= = 4cm 8. minor sector= = cm 9. major sector= = 8 cm 0. minor sector= cm. First, from arc length l to angle θ l = θ 60 r θ = 60l r Then from angle θ to sector area a a = θ 60 r = 60l r 60 r = 60lr 70r = lr So for question minor sector= cm. major sector= 40 0 = 00cm. minor segment = minor sector triangle = sin 00 86cm 4. θ = 60l r = 88 minor segment = θ 60 r r sin θ = sin cm ( 88 ) 5. θ = = 4 minor segment = 90 0 sin 4 4cm 6. minor segment = sin 60 = 4 7 = 4 6 = ( )cm 7. minor segment = sin 5 = 7 8 = 7 9 ( = 9 ) cm 8. θ = 60 0 = 50 minor segment = sin(50) 9. 5cm = 5 = 5 = 5 50 sin 50 5 (5 ) cm (a) minor arc= = 9.cm (b) major arc= = cm

2 Exercise 5 Chapter cm r. θ cm 78cm r = 4 60 r = = cm 78 = 60 θ θ = θ θ 0cm. 50 cm 5cm a = = cm θ = cos a = sin cm 5. In half an hour the minute hand sweeps out 80 or half a circle. Its tip travels d = = cm. 40 In half an hour the hour hand sweeps out 4 of a full circle. Its tip travels 8cm d = 4 8 = cm = 0 = = 80 0 = 40 a = sin 40 9cm 6. The ship is travelling through degrees of latitude. = 60 = 80. The ship travels 80 nautical miles. ne nautical mile in kilometres is 60 d = km

3 Exercise 5 Chapter 5 7. The circumference of the base of the cone is equal to the arc length of the sector: r = r = 0 The slant height of the cone is the radius of the sector: h + r = 0 h = = 0 ( ) = = 0 9 = 0 5 Exercise 5. θ = = rads. θ = =.5 rads. θ = 5 = 5 4. θ = 5 =.5 5. θ = 4 = 4 6. θ = 8 = = 5 80 = = 8 80 = = 0 80 = = = = = = = = = = = = = 6. 9 = 9 80 = = = = 0 80 = = 0 80 = = 6 80 = = = = = 6. 6 rads = 6 80 = 0 4. rads = 80 = rads = 8 80 = 50 rads = 0 80 = rads = 5 80 = rads = = rads = rads = 6 80 = 75. rads = 80 = rads = 8 80 = rads = 80 = rads = 5 80 = rads = 6 80 = rads = 4 80 = rads = = 7 rads = 8 80 = = = = = = = = =

4 Exercise 5 Chapter = = = = R = R = R = R = R = R = R = R = For exact value problems, you should work towards knowing these in radians as well as degrees, so you don t have to first convert to degrees. This will come with time and effort. You should deliberately memorise the following table: θ sin θ cos θ tan θ undefined 59. sin 4 = makes an angle of 6 quadrant, so sin 5 6 = 6. 4 makes an angle of 4 quadrant, so cos 4 = 6. sin = 6. makes an angle of quadrant, so sin = makes an angle of 4 quadrant, so sin 4 = 65. cos 4 = 66. makes an angle of quadrant, so tan = 67. cos = tan is undefined 69. makes an angle of quadrant, so cos = makes an angle of 6 quadrant, so tan 5 6 = makes an angle of 6 quadrant, so cos 5 6 = 7. makes an angle of 0 quadrant, so tan = 0 7. cos = 74. makes an angle of 0 quadrant, so sin = 0 Questions are single-step calculator exercises, so there s no point in reproducing the solutions here. Refer to the answers in Sadler. 9. (a) revolutions/second= = 6 radians/second. (b) 5 revolutions/minute= 5 60 = radians/second. (c) 90 degrees/second= 4 radians/second. 9. (a) radians/minute = revolution/minute (b) 4 radians/second = 4 60 = 45 radians/minute = 45 =.5 revolutions/minute (c) radians/second = 60 = 0 radians/minute = 0 = 0 revolutions/minute 9. sin = 6 x x = 6 sin tan. = 8 x 8 x = tan Let h be the perpendicular height in cm. 96. x sin. = 4 sin.8 4 sin. x = sin.8 = θ = sin 0.6 = h 0 h = 0 sin 0.6. x = h x = cos

5 Exercise 5C Chapter = cos x ( 5.0 x = cos ) D E 99. (a) 4 = (b) = 4 (c) 5 6 = 5 (d) = (a) 50 grads= 0.50 = 4 (b) 75 grads= 0.75 = 8 (c) 0 grads= 0.0 = 0 (d) 0 grads= 0.0 = 0 0. (a) C Let d be the diameter of the pipe. D = tan 0.5 D D = D = tan 0.5 d tan 0.5 = d.09 = 0.95d The scale along must be set up so that cm units are 0.95cm apart, starting with 0 at. (b) If C = then D is isosceles so D=D or D = 0.5d. This would be simpler to construct as cm units along would be exactly 0.5cm apart. Exercise 5C. l = rθ = = 4cm. l = rθ = 0.5 = 5cm. l = rθ = 7.8 ( 4.5).9cm 4. a = r θ = 4 = 8cm 5. a = r θ = 6.5 = 45cm 6. a = r θ = 0 ( 4) = 4.cm 7. a = r (θ sin θ) = 59 ( sin ) = 75.9cm 8. a = r (θ sin θ) = 5 ((.5) sin(.5)) = 0.4cm 9. a = r (θ sin θ) = 7.5 (. sin.) = 9.cm 0.. l 5cm l = rθ = 5. = 8cm 5

6 Exercise 5C Chapter cm 4. 8cm θ 0cm (a) a = r θ = = 90cm (b) a = cm θ = l/r = 0/8 =.5 a = r θ. l 8cm = 8.5 = 80cm (a) l = rθ = 8 = 8cm (b) a = r (θ sin θ) 5. = 8 ( sin) 6cm 5.cm. θ l 5cm 5cm r θ = 9 6 θ = 9 8θ = 9 (a) r θ = 5 5 θ = 5 θ = 6 5 l = rθ = = 6cm θ = a = r (θ sin θ) = 6 ( sin 6. a = segment C segment D = R θ r θ = θ (R r ) =.5 ( 6 ) = 8cm ) 0.7cm (b) a = r (θ sin θ) = 5 ( 6 5 sin6 5 ).5cm 7. a = θ (R r ) =.5 (9 5 ) = 4cm 6

7 Exercise 5C Chapter cm 6cm 0cm = C.855 area sector = =.8 area = 48.8 = 4.6cm Consider.. rea CD is equal to the area of the segment created by chord D minus the area of the segment created by chord C. = cos =.87 segment = 8 (.87 sin.87) Similarly 0.46 = cos =.855 segment = 6 (.855 sin.855) 6.0 total area = = 6.57cm 9. area segment C = = 5.6 area D = 5 sin area CD = = 6.60cm 0. The C and C are perpendicular to and respectively, since a tangent is perpendicular to a radius to the same point. This makes calculating the area of the halves of the quadrilateral simple. area C = 6 8 = 4 area C = 48 C = tan a = 5 ( sin ) 5 ( sin ) = 5 ( sin ( sin )) = 5 ( sin + sin ) =.65cm. (a) l = rθ = = 60cm each way, or 0cm total.. (b) C= 75 sin cm rc C exceeds chord C by.6cm. 50mm 70mm Consider. 40mm = cos =.88 segment = 50 (.88 sin.88) Similarly 5.9 = cos =.550 segment = 40 (.550 sin.550) total area = mm 7

8 Exercise 5C Chapter 5 fence 4. 0cm 7cm 5cm θ 6m 0m 6. To cos θ = 6 0 θ = cos = cos = arc = = cos find the major segment, use θ: a = 0 ( θ sin( θ)) 69m =.88 arc = perimeter = cm 7. fence θ 5m m fence It may be simplest to deal with this as the area of the circle minus the area of two identical minor segments. 0cm 5. θ 0+=cm P a = 45.9m cos θ = 5 θ = cos cos θ = θ = cos.0 percentage =.0 00% 5% a seg = r (θ sin θ) = (.8 sin.8) 09.76m a = a a seg m 8

9 Exercise 5C Chapter 5 0cm 6cm θ 6+4+6=6cm 6cm D ngle θ cos θ = 5 9 Minor arc E θ = cos C This is similar to the Situation at the beginning of the chapter. Straight segments and CD ngle θ Major arc C Minor arc D = 6 0 = 4cm cos θ = 0 6 θ = cos C = r( θ) = 6(.5) 6.90 D = rθ = Total length = cm α 5+4+0=9cm 9. θ φ 0+8+=0cm E 5cm 5cm 0cm 8cm F G Straight segments and EF = 9 5 = 6cm Straight segments CD and GH CD = 0 8 = 4cm C D cm H ngle φ E = rθ = cos φ = 8 0 Minor arc DH φ = cos.85 DH = rφ - 8 =.85.7 Minor arcs C and FG 0 α = ( θ φ).04 G = rα = Total length = ( + C + CD) + E + DH = ( ) cm 0. perimeter: r + rθ = 4 subst. for θ: area: r ( 4 r rθ = 4 r θ = 4 r r θ = 0 ) = 0 7r r = 0 r 7r + 0 = 0 (r 5)(r ) = 0 r = 5 or r = for r = 5, θ = 4 5 for r =, θ = 4 = 0.8 (an acute angle) = 5 (a reflex angle) 9

10 Exercise 5C Chapter 5 (a) Radius = 5 cm. (b) Radius = cm.. r is the slant height of the cone, and θ is such that the arc length of the sector equals the circumference of the base of the cone. r = = cm rθ = 8 θ = RDINS. 4. First, find the area of the major segment, then find the capacity:. a = 0 0 sin = 50 = 5 a sector = 5 = 5 6 a shaded = a a sector = = 5 5 5cm = 5 0cm ( ) cm a = 0 5 = 5 θ = tan a sector = a shaded = a a sector = 5.84.cm cos θ = 0 60 θ = cos a = 60 ( θ sin( θ)) 6849cm V = al = cm 8L 5. (a) a I = bh = 5 40 = 00cm a II = a I = 00cm a III = a I + a II = 600cm for segment IV: r = = θ = tan a IV = r θ a III = cm (b) l = cm

11 Exercise 5C Chapter 5 C = cos cm C 5cm 5cm CD = 5 5 = 0 E a DC = CD(r + r ) D = 0 (0 + 5) = cos C E = 5 5 C E. a CE = 0. 5cm a = ab sin C = 5 6 sin a sector = a sector = a sector C = a shaded = a a sector a sector a sector C = a shaded =.7 a 4.70 = 8.6% 6.55 sin D E 4 = 5 5 D E h x R a DE = a shaded = a DC a CE a de cm P u S 7. 4cm cm C cm = cos = cos v Q The lowest position of the piston is r above the top of the wheel. Therefore it is r above the bottom of the wheel. Thus the fixed length of the drive rod PR= r. It should be clear that h = r since at the low position the arm goes straight down from the piston to the bottom of the wheel, and at the high position the arm goes straight down from the piston to the top of the wheel. If arc length PQ is equal to r, the angle it subtends is PQ= RDIN. The height of point P relative to point is u = r cos. Similarly the horizontal position of point P is v = r sin. The length from point S to R can be determined

12 Miscellaneous Exercise 5 Chapter 5 by Pythagoras Theorem: SR = (r) (r sin ) = 9r r sin = r 9 sin =.88r x = SR r u =.88r r r cos = r(.88 cos ) = 0.r x h = 0.r r 7% Miscellaneous Exercise 5. No worked solution necessary.. No worked solution necessary.. (a) x <.5: x 5 < 0; 4 x > 0 (x 5) = 4 x x + 5 = 4 x x = (which is in the part of the domain we are considering, so we do not reject it.).5 x < 4: x 5 0; 4 x > 0 x 5 = 4 x x = 9 x = x 4: x 5 0; 4 x 0 x 5 = (4 x) x = We don t really need to do the third part (which yields a solution outside the part of the domain we re considering), as there can only be two solutions. (You can see that if you graph the left and right hand sides of the equation and see where they intersect.) Solution: x = or x =. (b) x <.5: x 5 < 0; 4 x > 0 (x 5) < 4 x x + 5 < 4 x x > (which is in the part of the domain we are considering, so we have < x <.5.).5 x < 4: x 5 0; 4 x > 0 x 5 < 4 x x < 9 x < (which is in the part of the domain we are considering, so we have.5 x < 4.) Combining the two parts of solution we get: Solution: < x <. (c) This inequality must be true where the previous inequality is false, i.e.: Solution: x or x. There is another way we could look at this. We know the left and right hand sides are equal at and, so the inequality must be true either between and or less than /greater than. We can decide which by testing a suitable value: substitute (for example) for x and decide whether the inequality holds true. Here 5 = and 4 = so LHS<RHS and the second inequality holds for the interval < x <. 4. l = rθ = 40 = 60cm/s 5. (a) C = C = 0 6 = 8cm (b) D = D + = = 6 = 9cm (c) cos C = C C = cos - C = cos

13 Miscellaneous Exercise 5 Chapter 5 (d) We want FE. T tan FE = F E E FE = tan - F E E = tan N h 50m C (i + 5j) 6. P (7i j) (a) tan 4 = h 50 h = 50 tan 4 m P = 4 5 P = 4 5 ( ) P = 4 5 ( ) + = (b) C = = 40 cos 40 = 50 C C = 50 cos 40 65m = 0.8(7i j) + 0.(i + 5j) = (5.6i 0.8j) + (0.i + j) = 5.8i + 0.j (c) tan 40 = 50 = 50 tan 40 4m N 70 (d) tan T = h -.6 T = tan θ 9. = 45 sin θ sin 0 = sin 0 θ = sin The plane must fly on a bearing of 70 4 = 66. = sin 45 ( = 0 ) 7.65m y the similarity of the triangles and JI: JI = m

14 Miscellaneous Exercise 5 Chapter 5 0. a = kc k = b c = + 4 = 5 5 = 5 a = 5(i 4j) = 5i 0j a + b = 5i 0j 7i + 4j = 8i + 4j a + b = = 80 = 4 5 ( h k ) ( a = k h ) c h k = 0 h + k = h + k = and k h = 0 h + k = 0 7k = (+ ) k = ( ) 7 h + = 0 (subst into ) 7 h = 6 7. a M E c C D E = a + c M = h E = ha + h c M = a + M = a ha + h c = ( h)a + h c D = c + a M = k D = k(c + a) = k a + kc ( h)a + h c = k a + kc ( h)a k a = kc h c. (a) = (9i 6j) + t( i + 5j) = (9 t)i + ( 6 + 5t)j = 5 (9 t) + ( 6 + 5t) = t + 4t t + 5t = 65 (b) = 9t 6t + 97 = 65 9t 6t 8 = 0 (t 6)(9t + 8) = 0 t = 6s = (5i + j) ((9 t)i + ( 6 + 5t)j) = (5 9 + t)i + ( + 6 5t)j = ( 4 + t)i + (7 5t)j = 9t 6t + 97 = ( 4 + t) + (7 5t) 9t 6t + 97 = 96 56t + 4t t + 5t 9t 6t + 97 = 9t 6t t = 5 t = 5.s 4