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Solutions to Unit 3A Specialist Mathematics by A.J. Sadler Prepared by: Glen Prideaux 2009
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12Exercise 2A20. x y y2y1tan = xy1y1 = xtantan = xy2y2 = xtany = y1 y2= xtan xtan= x_ 1tan 1tan_= x_ tantan tan tantan tan_= x_tan tantan tan_

21.xzysin = xzz = xsintan = yzy = z tan=_ xsin_tan= xtansin

22.xzycos = zxz = xcos cos = yzy = z cos = (xcos ) cos = xcos cos

64.422 60y45 wxw2= 42+ (22)2= 16 + 4 2= 24w = 24= 4 6= 26xsin = wsin60x = wsinsin60= 26 sin32= 26 sin1 23= 432 sin3= 42 sinysin45 = xsiny = xsin45sin= x 12sin= 42 sin 12sin= 4 sinsin

= 12 10 10 sin 3= 5032= 253asector = 12 52

3asector= 253 3 256= 253 252= 25_3 2_cm232.OAB5cm10cma

= 12 10 5= 25 = tan1 105 1.11asector = 12 521.11 13.84ashaded = a

asector= 25 13.84 11.2cm233. r is the slant height of the cone, and is suchthat the arc length of the sector equals the cir-cumference of the base of the cone.r =_282+ 82= 453 29.1cmr = 2 8 = 16453 1.73RADIANS34. First, nd the area of the major segment, thennd the capacity:cos 2 = 1060 = 2 cos1 16 2.81a = 12 602(2 sin(2 )) 6849cm2V = al= 6849 120 821915cm3 822L35. (a) aI = 12bh= 12 15 40= 300cm2aII = aI = 300cm2aIII = aI +aII = 600cm2for segment IV:r =_402+ 152= 573 42.72 = 2 tan1 1540 0.718aIV = 12r2 aIII= 12 1825 0.718 600 55cm2(b) l = 40 2 +30 2+ 573 2 +573 0.718256cm62Exercise 5C36.O1O2C DA B5cm15cm10cm5cmECD =_15252= 102aO1O2DC = 12CD(r1 +r2)= 12 102(10 + 5)= 752 106.07cos CO1E = 515CO1E 1.23aCO1E = 12 1021.23 61.55sinDO2E 4 = 515DO2E 0.34 + 4 1.91aDO2E = 12 521.91 23.88ashaded = aO1O2DC aCO1E adO2E 106.07 61.55 23.88 20.64cm237.A BC4cm 3cm2cmA = cos1 62+ 72522 6 7 0.775B = cos1 52+ 72622 5 7 0.997C = cos1 62+ 52722 6 5 1.369a

= 12ab sinC= 12 5 6 sin1.369 14.70asector A = 12 420.775 6.20asector B = 12 320.997 4.49asector C = 12 221.369 2.74ashaded = a

asector A asector Basector C= 14.70 6.20 4.49 2.74 1.27ashadeda

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