Directions for completing the IB Math SL Presumed Knowledge unit

Transcription

1 Directions for completing the IB Mth SL Presumed Knowledge unit Plese complete the Bckground Knowledge unit nd do not wit to strt completing it! Go through ech section nd mke sure you cn do ech problem. These re ll skills tht you should hve in order to be successful in Mth SL. Mke sure your work is net nd orgnized for ech problem you complete. As you re working through ech problem, mke note of ny problems you do not understnd s well s ny questions you my hve. I will spend some time ech dy during the first couple weeks of school nswering your questions nd ddressing your concerns. I will not be collecting your pcket. However, your first test is over the contents of this unit. You will notice some different terminology used in this book. For instnce, the book uses fctorise where we re used to sying fctor. Look t the exmples in the blue boxes if you re unsure wht term mens, Google it, or emil me. You cn emil me t rtillmn@district70.org. I will check my emil periodiclly throughout the summer. Topics Included: 1. Rdicls 2. Scientific nottion. Inequlities nd set nottion 4. Liner equtions nd inequlities 5. Absolute vlue 6. Opertions with polynomils 7. Fctoring polynomils 8. Bsic lgebr 9. Rtionl expressions 10. Pythgoren Theorem 11. Trigonometry Hve gret summer nd I look forwrd to working with you this coming school yer.

2 Bckground knowledge Contents: A B C D E F G H I J K L M N Surds nd rdicls Scientific nottion (stndrd form) Number systems nd set nottion Algebric simplifiction Liner equtions nd inequlities Modulus or bsolute vlue Product expnsion Fctoristion Formul rerrngement Adding nd subtrcting lgebric frctions Congruence nd similrity Pythgors theorem Coordinte geometry Right ngled tringle trigonometry

3 2 BACKGROUND KNOWLEDGE This chpter contins mteril tht is ssumed knowledge for the course. It does not cover ll ssumed knowledge, s other necessry work is revised within the chpters. A SURDS AND RADICALS A rdicl is ny number which is written with the rdicl sign p. A surd is rel, irrtionl rdicl such s p 2, p, p 5 or p 6. Surds re present in solutions to some qudrtic equtions. p 4 is rdicl but is not surd s it simplifies to 2. p is the non-negtive number such tht p p =. Properties: ² p is never negtive, so p > 0. ² p is meningful only for > 0. ² p b = p p b for > 0 nd b > 0. r p ² b = p for > 0 nd b>0. b Exmple 1 Write s single surd: p 2 p b p 18 p 6 p 2 p = p 2 = p 6 b = p p 18 6 q 18 6 = p or p 18 p 6 = p 6 p p 6 = p EXERCISE A 1 Write s single surd or rtionl number: p p 5 b ( p ) 2 c 2 p 2 p 2 d p 2 2 p 2 e p 7 2 p 7 f p p 12 2 g p p12 6 h p p18 Exmple 2 Simplify: p +5 p b 2 p 2 5 p 2 Compre with 2x 5x = x p +5 p = ( + 5) p =8 p b 2 p 2 5 p 2 = (2 5) p 2 = p 2

4 BACKGROUND KNOWLEDGE 2 Simplify the following mentlly: 2 p 2+ p 2 b 2 p 2 p 2 c 5 p 5 p 5 d 5 p 5+ p 5 e p 5 5 p 5 f 7 p +2 p g 9 p 6 12 p 6 h p 2+ p 2+ p 2 Exmple Write p 18 in the form p b where nd b re integers nd is s lrge s possible. p 18 = p 9 2 f9 is the lrgest perfect squre fctor of 18g = p 9 p 2 = p 2 Write the following in the form p b where nd b re integers nd is s lrge s possible: p 8 b p 12 c p 20 d p 2 e p p p p 27 f 45 g 48 h 54 p p p p i 50 j 80 k 96 l 108 Exmple 4 Simplify: 2 p 75 5 p 27 2 p 75 5 p 27 =2 p 25 5 p 9 =2 5 p 5 p = 10 p 15 p = 5 p 4 Simplify: 4 p p 12 b p 2+ p 50 c p 6+ p 24 d 2 p p 12 e p 75 p p p p 12 f Exmple 5 Write denomintor. 9 p without rdicl in the 9p = 9 p p p = 9p = p

5 4 BACKGROUND KNOWLEDGE 5 Write without rdicl in the denomintor: 1 p2 b 6 p c 7 p2 d 10 p 5 e 10 p 2 f 18 p 6 g 12 p h 5 p7 i 14 p 7 j 2 p p 2 B SCIENTIFIC NOTATION (STANDARD FORM) Scientific nottion (or stndrd form) involves writing ny given number s number between 1 nd 10, multiplied by power of 10, i.e., 10 k where 1 6 <10 nd k 2 Z. Exmple 6 Write in stndrd form: b 0: = : fshift deciml point 4 plces to the =: left nd g b 0: = 8: fshift deciml point 4 plces to the =8: right nd g EXERCISE B 1 Express the following in scientific nottion: 259 b c 2:59 d 0:259 e 0: f 40:7 g 4070 h 0:0407 i j k 0: Express the following in scientific nottion: The distnce from the Erth to the Sun is m. b Bcteri re single cell orgnisms, some of which hve dimeter of 0:000 mm. c d e A speck of dust hs width smller thn 0:001 mm. The core temperture of the Sun is 15 million degrees Celsius. A single red blood cell lives for bout four months. During this time it will circulte round the body times.

6 BACKGROUND KNOWLEDGE 5 Exmple 7 Write s n ordinry number: : b 5: : =: = 20 b 5: = : =0: Write s n ordinry deciml number: 4 10 b c 2:1 10 d 7: e : f 8: g 4: 10 7 h Write s n ordinry deciml number: 4 10 b c 2:1 10 d 7: e : f 8: g 4: 10 7 h Write s n ordinry deciml number: The wvelength of light is m. b The estimted world popultion for the yer 2000 ws 6: c d The dimeter of our glxy, the Milky Wy, is light yers. The smllest viruses re mm in size. 6 Find, correct to 2 deciml plces: (: ) (4: ) b (6: ) 2 c : d (9: ) (7: ) e 1 : f (1:2 10 ) 7 If missile trvels t 5400 km h 1, how fr will it trvel in: 1 dy b 1 week c 2 yers? Give your nswers in scientific nottion correct to 2 deciml plces, nd ssume tht 1 yer ¼ 65:25 dys. istockphoto.com 8 Light trvels t speed of 10 8 metres per second. How fr will light trvel in: 1 minute b 1 dy c 1 yer? Give your nswers with deciml prt correct to 2 deciml plces, nd ssume tht 1 yer ¼ 65:25 dys.

7 6 BACKGROUND KNOWLEDGE C NUMBER SYSTEMS AND SET NOTATION NUMBER SYSTEMS We use: ² R to represent the set of ll rel numbers. These include ll of the numbers on the number line. negtives 0 positives ² N to represent the set of ll nturl numbers. N = f0, 1, 2,, 4, 5,...g ² Z to represent the set of ll integers. Z = f0, 1, 2,, 4,...g ² Z + is the set of ll positive integers. Z + = f1, 2,, 4,...g ² Q to represent the set of ll rtionl numbers which re ny numbers of the form p where p nd q re integers, q 6= 0. q SET NOTATION fx j <x<2g reds the set of ll vlues tht x cn be such tht x lies between nd 2. the set of ll such tht Unless stted otherwise, we ssume tht x is rel. EXERCISE C 1 Write verbl sttements for the mening of: fx j x>5, x 2 R g b fx j x 6, x 2 Z g c fy j 0 <y<6g d fx j 2 6 x 6 4, x 2 Z g e ft j 1 <t<5g f fn j n<2 or n > 6g Exmple 8 Write in set nottion: 0 4 x b included not included - 4 x fx j 1 6 x 6 4, x 2 N g or fx j 1 6 x 6 4, x 2 Z g b fx j 6 x<4, x 2 R g 2 Write in set nottion: b c d e f

8 BACKGROUND KNOWLEDGE 7 Sketch the following number sets: fx j 4 6 x<10, x 2 N g b fx j 4 <x6 5, x 2 Z g c fx j 5 <x6 4, x 2 R g d fx j x> 4, x 2 Z g e fx j x 6 8, x 2 R g D ALGEBRAIC SIMPLIFICATION To nswer the following questions, you will need to remember: ² the distributive lw (b + c) =b + c ² power nottion 2 =, =, 4 =, nd so on. EXERCISE D 1 Simplify if possible: x +7x 10 b x +7x x c 2x +x +5y d 8 6x 2x e 7b +5b f x 2 +7x 2 Remove the brckets nd then simplify: (2x + 5) + 4(5 + 4x) b 6 2(x 5) c 5(2 b) 6( 2b) d x(x 2 7x + ) (1 2x 5x 2 ) Simplify: 2x(x) 2 b 2 b 9b 4 c p 16x 4 d (2 2 ) 4 E LINEAR EQUATIONS AND INEQUALITIES When deling with inequlities: ² multiplying or dividing both sides by negtive reverses the inequlity sign. ² do not multiply or divide both sides by the unknown or term involving the unknown. EXERCISE E 1 Solve for x: 2x + 5 = 25 b x 7 > 11 c 5x + 16 = 20 d x 7 = 10 e 6x + 11 < 4x 9 f x 2 =8 5 g 1 2x > 19 h 1 2 x +1= 2 x 2 i 2 x 4 = 1 2 (2x 1)

9 8 BACKGROUND KNOWLEDGE 2 Solve simultneously for x nd y: ½ x +2y =9 b x y = d ½ 5x 4y = 27 x +2y =9 e ½ 2x +5y = 28 x 2y =2 ½ x +2y =5 2x +4y =1 c f ½ 7x +2y = 4 x +4y = 14 8 x >< 2 + y =5 x >: + y 4 =1 F MODULUS OR ABSOLUTE VALUE The modulus or bsolute vlue of rel number is its size, ignoring its sign. For exmple: the modulus of 7 is 7, nd the modulus of 7 is lso 7. GEOMETRIC DEFINITION The modulus of rel number is its distnce from zero on the number line. Becuse the modulus is distnce, it cnnot be negtive We denote the modulus of x s jxj. jxj is the distnce of x from 0 on the rel number line. x If x>0 If x<0 0 x x x 0 jx j cn be considered s the distnce of x from. ALGEBRAIC DEFINITION jxj = x if x > 0 x if x<0 or jxj = p x 2 MODULUS EQUATIONS It is cler tht jxj = 2 hs two solutions, x = 2 nd x = 2, since j2j = 2 nd j 2j =2. If jxj = where > 0, then x =.

10 BACKGROUND KNOWLEDGE 9 EXERCISE F 1 Find the vlue of: 5 ( 11) b j5j j 11j c j5 ( 11)j d ( 2) ( 2) e j 6j j 8j f j 6 ( 8)j 2 If = 2, b =, nd c = 4 find the vlue of: jj b jbj c jjjbj d jbj e j bj f jj jbj g j + bj h jj + jbj i jj 2 j 2 k c jcj l jj Solve for x: jxj = b jxj = 5 c jxj =0 d jx 1j = e j xj =4 f jx +5j = 1 g jx 2j =1 h j 2xj = i j2 5xj = 12 G PRODUCT EXPANSION y = 2(x 1)(x + ) cn be expnded into the generl form y = x 2 + bx + c. Likewise, y = 2(x ) 2 +7 cn lso be expnded into this form. Following is list of expnsion rules you cn use: ² ( + b)(c + d) =c + d + bc + bd ² ( + b)( b) = 2 b 2 fdifference of two squresg ² ( + b) 2 = 2 +2b + b 2 fperfect squresg Exmple 9 Expnd nd simplify: (2x + 1)(x + ) b (x 2)(x + ) (2x + 1)(x + ) =2x 2 +6x + x + =2x 2 +7x + b (x 2)(x + ) =x 2 +9x 2x 6 =x 2 +7x 6

11 10 BACKGROUND KNOWLEDGE EXERCISE G 1 Expnd nd simplify using ( + b)(c + d) =c + d + bc + bd: (2x + )(x + 1) b (x + 4)(x + 2) c (5x 2)(2x + 1) d (x + 2)(x 5) e (7 2x)(2 + x) f (1 x)(5 + 2x) g (x + 4)(5x ) h (1 x)(2 5x) i (7 x)( 2x) j (5 2x)( 2x) k (x + 1)(x + 2) l 2(x 1)(2x + ) Exmple 10 Expnd using the rule ( + b)( b) = 2 b 2 : (5x 2)(5x + 2) b (7 + 2x)(7 2x) (5x 2)(5x + 2) = (5x) = 25x 2 4 b (7 + 2x)(7 2x) =7 2 (2x) 2 = 49 4x 2 2 Expnd using the rule ( + b)( b) = 2 b 2 : (x + 6)(x 6) b (x + 8)(x 8) c (2x 1)(2x + 1) d (x 2)(x + 2) e (4x + 5)(4x 5) f (5x )(5x + ) g ( x)( + x) h (7 x)(7 + x) i (7 + 2x)(7 2x) j (x + p 2)(x p 2) k (x + p 5)(x p 5) l (2x p )(2x + p ) Exmple 11 Expnd using the perfect squre expnsion rule: (x + 2) 2 b (x 1) 2 (x + 2) 2 = x 2 + 2(x)(2) = x 2 +4x +4 b (x 1) 2 = (x) 2 + 2(x)( 1) + ( 1) 2 =9x 2 6x +1 Expnd nd simplify using the perfect squre expnsion rule: (x + 5) 2 b (x + 7) 2 c (x 2) 2 d (x 6) 2 e ( + x) 2 f (5 + x) 2 g (11 x) 2 h (10 x) 2 i (2x + 7) 2 j (x + 2) 2 k (5 2x) 2 l (7 x) 2 4 Expnd the following into the generl form y = x 2 + bx + c: y = 2(x + 2)(x + ) b y = (x 1) 2 +4 c y = (x + 1)(x 7) d y = (x + 2) 2 11 e y = 4(x 1)(x 5) f y = 1 2 (x + 4)2 6 g y = 5(x 1)(x 6) h y = 1 2 (x + 2)2 6 i y = 5 2 (x 4)2

12 BACKGROUND KNOWLEDGE 11 Exmple 12 The use of Expnd nd simplify: 1 2(x + ) 2 b 2( + x) (2 + x)( x) brckets is essentil! 1 2(x + ) 2 =1 2[x 2 +6x + 9] =1 2x 2 12x 18 = 2x 2 12x 17 b 2( + x) (2 + x)( x) =6+2x [6 2x +x x 2 ] =6+2x 6+2x x + x 2 = x 2 + x 5 Expnd nd simplify: 1 + 2(x + ) 2 b 2 + (x 2)(x + ) c ( x) 2 d 5 (x + 5)(x 4) e 1 + 2(4 x) 2 f x 2 x (x + 2)(x 2) g (x + 2) 2 (x + 1)(x 4) h (2x + ) 2 + (x + 1) 2 i x 2 +x 2(x 4) 2 j (x 2) 2 2(x + 1) 2 H FACTORISATION Algebric fctoristion is the reverse process of expnsion. For exmple, (2x + 1)(x ) 2x 2 5x is expnded to 2x 2 5x, wheres is fctorised to (2x + 1)(x ). Notice tht 2x 2 5x = (2x + 1)(x ) hs been fctorised into two liner fctors. Flow chrt for fctorising: Expression Difference of two squres 2 b 2 =( + b)( b) Perfect squre 2 +2b + b 2 =( + b) 2 Tke out ny common fctors Recognise type Sum nd Product type x 2 + bx + c, 6= 0 ² find c ² find the fctors of c which dd to b ² if these fctors re p nd q, replce bx by px + qx ² complete the fctoristion Sum nd Product type x 2 + bx + c x 2 + bx + c =(x + p)(x + q) where p + q = b nd pq = c

13 12 BACKGROUND KNOWLEDGE Exmple 1 Fully fctorise: x 2 12x b 4x 2 1 c x 2 12x + 6 Remember tht ll fctoristions cn be checked by expnsion! x 2 12x =x(x 4) b 4x 2 1 = (2x) = (2x + 1)(2x 1) c x 2 12x + 6 = x 2 + 2(x)( 6) + ( 6) 2 =(x 6) 2 fx is common fctorg fdifference of two squresg fperfect squre formg EXERCISE H 1 Fully fctorise: x 2 +9x b 2x 2 +7x c 4x 2 10x d 6x 2 15x e 9x 2 25 f 16x 2 1 g 2x 2 8 h x 2 9 i 4x 2 20 j x 2 8x + 16 k x 2 10x + 25 l 2x 2 8x +8 m 16x x + 25 n 9x x +4 o x 2 22x Exmple 14 Fully fctorise: x x +9 b x 2 +x + 10 x x +9 = (x 2 +4x + ) = (x + 1)(x + ) b x 2 +x + 10 = [x 2 x 10] = (x 5)(x + 2) f is common fctorg fsum =4, product =g fremoving 1 s common fctor to mke the coefficient of x 2 be 1g fsum =, product = 10g 2 Fully fctorise: x 2 +9x +8 b x 2 +7x + 12 c x 2 7x 18 d x 2 +4x 21 e x 2 9x + 18 f x 2 + x 6 g x 2 + x +2 h x 2 42x + 99 i 2x 2 4x 2

14 BACKGROUND KNOWLEDGE 1 j 2x 2 +6x 20 k 2x 2 10x 48 l 2x x 12 m x 2 +6x n x 2 2x 1 o 5x x + 40 FACTORISATION BY SPLITTING THE x-term Using the distributive lw to expnd we see tht: (2x + )(4x + 5) =8x x + 12x + 15 =8x x + 15 We will now reverse the process to fctorise the qudrtic expression 8x x + 15: 8x x + 15 Step 1: Split the middle term =8x x + 12x + 15 Step 2: Group in pirs = (8x x) + (12x + 15) Step : Fctorise ech pir seprtely =2x(4x + 5) + (4x + 5) Step 4: Fctorise fully = (4x + 5)(2x + ) The trick in fctorising these types of qudrtic expressions is in Step 1. The middle term is split into two so the rest of the fctoristion cn proceed smoothly. Rules for splitting the x-term: The following procedure is recommended for fctorising x 2 + bx + c : ² Find c: ² Find the fctors of c which dd to b: ² If these fctors re p nd q, replce bx by px + qx: ² Complete the fctoristion. Exmple 15 Fully fctorise: 2x 2 x 10 b 6x 2 25x x 2 x 10 hs c =2 10 = 20: The fctors of 20 which dd to 1 re 5 nd +4: ) 2x 2 x 10 = 2x 2 5x +4x 10 = x(2x 5) + 2(2x 5) = (2x 5)(x + 2) b 6x 2 25x + 14 hs c =6 14 = 84: The fctors of 84 which dd to 25 re 21 nd 4: ) 6x 2 25x + 14 = 6x 2 21x 4x + 14 =x(2x 7) 2(2x 7) = (2x 7)(x 2)

15 14 BACKGROUND KNOWLEDGE Fully fctorise: 2x 2 +5x 12 b x 2 5x 2 c 7x 2 9x +2 d 6x 2 x 2 e 4x 2 4x f 10x 2 x g 2x 2 11x 6 h x 2 5x 28 i 8x 2 +2x j 10x 2 9x 9 k x 2 + 2x 8 l 6x 2 +7x +2 m 4x 2 2x +6 n 12x 2 16x o 6x 2 9x + 42 p 21x 10 9x 2 q 8x 2 6x 27 r 12x 2 + 1x + s 12x x + t 15x 2 22x +8 u 14x 2 11x 15 Exmple 16 Fully fctorise: (x + 2) + 2(x 1)(x + 2) (x + 2) 2 (x + 2) + 2(x 1)(x + 2) (x + 2) 2 =(x + 2)[ + 2(x 1) (x + 2)] fs (x + 2) is common fctorg =(x + 2)[ + 2x 2 x 2] =(x + 2)(x 1) 4 Fully fctorise: (x + 4) + 2(x + 4)(x 1) b 8(2 x) (x + 1)(2 x) c 6(x + 2) 2 + 9(x + 2) d 4(x + 5) + 8(x + 5) 2 e (x + 2)(x + ) (x + )(2 x) f (x + ) 2 + 2(x + ) x(x + ) g 5(x 2) (2 x)(x + 7) h (1 x) + 2(x + 1)(x 1) Exmple 17 Fully fctorise using the difference of two squres : (x + 2) 2 9 b (1 x) 2 (2x + 1) 2 (x + 2) 2 9 =(x + 2) 2 2 = [(x + 2) + ][(x + 2) ] =(x + 5)(x 1) b (1 x) 2 (2x + 1) 2 = [(1 x) + (2x + 1)][(1 x) (2x + 1)] = [1 x +2x + 1][1 x 2x 1] = x(x + 2) 5 Fully fctorise: (x + ) 2 16 b 4 (1 x) 2 c (x + 4) 2 (x 2) 2 d 16 4(x + 2) 2 e (2x + ) 2 (x 1) 2 f (x + h) 2 x 2 g x 2 (x + 2) 2 h 5x 2 20(2 x) 2 i 12x 2 27( + x) 2

16 BACKGROUND KNOWLEDGE 15 INVESTIGATION ANOTHER FACTORISATION TECHNIQUE Wht to do: 2 If x 2 + bx + c = 1 By expnding, show tht Using 2 on 8x x + 15, we hve 8x x + 15 = (x + p)(x + q) = x 2 +[p+q]x+ (x + p)(x + q), show tht p + q = b nd pq = c. (8x + p)(8x + q) 8 ) p = 12, q = 10, or vice vers ) 8x 2 (8x + 12)(8x + 10) + 22x + 15 = 8 4(2x + )2(4x + 5) = 8 1 = (2x + )(4x + 5) b where ( p + q = 22 pq =8 15 = 120 Use the method shown to fctorise: i x x +8 ii 12x x +6 iii 15x x 8 Check your nswers to by expnsion. h pq i : I FORMULA REARRANGEMENT In the formul D = xt + p we sy tht D is the subject. This is becuse D is expressed in terms of the other vribles x, t nd p. We cn rerrnge the formul to mke one of the other vribles the subject. We do this using the usul rules for solving equtions. Whtever we do to one side of the eqution we must lso do to the other side. Exmple 18 Mke x the subject of D = xt + p. If D = xt + p ) xt + p = D ) xt + p p = D p fsubtrcting p from both sidesg ) xt = D p ) xt t = D p t ) x = D p t fdividing both sides by tg

17 16 BACKGROUND KNOWLEDGE EXERCISE I 1 Mke x the subject of: + x = b b x = b c 2x + = d d c + x = t e 5x +2y = 20 f 2x +y = 12 g 7x +y = d h x + by = c i y = mx + c Exmple 19 Mke z the subject of c = m z. c = m z c z = m z z ) cz = m cz ) c = m c fmultiplying both sides by zg fdividing both sides by cg ) z = m c 2 Mke z the subject of: Mke: z = b c b z = d c d = 2 z d z 2 = z the subject of F = m b r the subject of C =2¼r c d the subject of V = ldh d K the subject of A = b K : Exmple 20 Mke t the subject of s = 1 2 gt2 where t> gt2 = s fwriting with t 2 on LHSg ) gt2 = 2 s fmultiplying both sides by 2g ) gt 2 =2s gt 2 ) g = 2s g ) t 2 = 2s g r 2s ) t = g fdividing both sides by gg fs t>0g

18 4 Mke: BACKGROUND KNOWLEDGE (Chpter 1) 17 r the subject of A = ¼r 2 if r>0 b x the subject of N = x5 c r the subject of V = 4 ¼r d x the subject of D = n x : 5 Mke: p the subject of d = b l the subject of T = 1 n 5p l c the subject of c = p r l 2 b 2 d l the subject of T =2¼ g e the subject of P = 2( + b) f h the subject of A = ¼r 2 +2¼rh g r the subject of I = E R + r 6 Mke the subject of the formul k = d2 2b. b Find the vlue for when k = 112, d = 24, b =2. h q the subject of A = B p q : 7 The formul for determining the volume of sphere of rdius r is V = 4 ¼r. Mke r the subject of the formul. b Find the rdius of sphere which hs volume of 40 cm. 8 The distnce trvelled by n object ccelerting from sttionry position is given by the formul S = 1 2 t2 cm where is the ccelertion in cm s 2 nd t is the time in seconds. Mke t the subject of the formul. Consider t>0 only. b Find the time tken for n object ccelerting t 8 cm s 2 to trvel 10 m. 9 The reltionship between the object nd imge distnces (in cm) for concve mirror cn be written s 1 f = 1 u + 1 v where f is the focl length, u is the object distnce nd v is the imge distnce. Mke v the subject of the formul. b Given focl length of 8 cm, find the imge distnce for the following object distnces: i 50 cm ii 0 cm. 10 According to Einstein s theory of reltivity, the mss of prticle is given by the m 0 formul m = r ³, where m 0 is the mss of the prticle t rest, v 2 1 v is the speed of the prticle, nd c c is the speed of light in vcuum. Mke v the subject of the formul given v>0. b Find the speed necessry to increse the mss of prticle to three times its rest mss, i.e., m =m 0. Give the vlue for v s frction of c. c A cyclotron incresed the mss of n electron to 0m 0 : With wht velocity must the electron hve been trvelling, given c = 10 8 ms 1?

19 18 BACKGROUND KNOWLEDGE J To dd or subtrct lgebric frctions, we combine them into single frction with the lest common denomintor (LCD). For exmple, x 1 x + 2 ADDING AND SUBTRACTING ALGEBRAIC FRACTIONS hs LCD of 6, so we write ech frction with denomintor 6. Exmple 21 Write s single frction: 2+ x b x 1 x x ³ x =2 + x x 2x + = x b x 1 x + 2 = 2 µ x 1 µ x (x 1) (x + ) = 6 2x 2 x 9 = 6 x 11 = 6 EXERCISE J 1 Write s single frction: + x 5 d x 2 4 Exmple 22 Write x +1 x 2 2 s single frction. b 1+ x e 2+x + x 4 5 x +1 x 2 2 µ µ x +1 x 2 = 2 x 2 x 2 (x + 1) 2(x 2) = x 2 x +1 2x +4 = x 2 = x +5 x 2 c + x 2 2 2x +5 f x flcd =(x 2)g

20 BACKGROUND KNOWLEDGE 2 Write s single frction: 1+ x+2 d 2x 1 + x+1 b 2 + e x 4 c x x+1 f x x Write s single frction: c 2x + 5 x + 2x 5 x 2 x 2 5x + x 4 x+4 K b d 1 1 x 2 x 2x + 1 x+4 x 2x + 1 CONGRUENCE AND SIMILARITY CONGRUENCE Two tringles re congruent if they re identicl in every respect prt from position. The tringles hve the sme shpe nd size. There re four cceptble tests for the congruence of two tringles. Two tringles re congruent if one of the following is true: ² corresponding sides re equl in length (SSS) ² two sides nd the included ngle re equl (SAS) ² two ngles nd pir of corresponding sides re equl (AAcorS) ² for right ngled tringles, the hypotenuse nd one other pir of sides re equl (RHS). If congruence cn be proven then ll corresponding lengths, ngles nd res must be equl.

21 BACKGROUND KNOWLEDGE 21 Exmple 24 Estblish tht pir of tringles is similr, then find x given BD = 20 cm. 12 cm B A x cm C E (x + 2) cm D ² - x +2 x smll lrge 12 cm B A E x cm C 20 cm (x + 2) cm D The tringles re equingulr nd hence similr. ) x +2 = x fsides in the sme rtiog ) 12(x + 2) = 20x ) 12x + 24 = 20x ) 24 = 8x ) x = EXERCISE K.2 1 In ech of the following, estblish tht pir of tringles is similr, nd hence find x: b c A P U 2 cm B C 5 cm X x cm 6 cm cm x cm Q S x cm 7 cm R 2 cm D E 6 cm T V 5 cm Y 6 cm Z d D e f X A B 4 cm x cm cm 5 cm E C Y 2 cm U 5 cm x cm Z V S P 8 cm x cm 10 cm 50 Q 50 9 cm R T 2 A fther nd son re stnding side-by-side. The fther is 1:8 m tll nd csts shdow :2 m long, while his son s shdow is 2:4 m long. How tll is the son?

22 22 BACKGROUND KNOWLEDGE L PYTHAGORAS THEOREM The hypotenuse is the longest side of right ngled tringle. It is opposite the right ngle. Pythgors Theorem is: In right ngled tringle, the squre of the length of the hypotenuse is equl to the sum of the squres of the lengths of the other two sides. c 2 = 2 + b 2 c b This theorem, known to the ncient Greeks, is vluble becuse: ² if we know the lengths of ny two sides of right ngled tringle then we cn clculte the length of the third side ² if we know the lengths of the three sides then we cn determine whether or not the tringle is right ngled. The second sttement here relies on the converse of Pythgors Theorem, which is: If tringle hs sides of length, b nd c units nd 2 + b 2 = c 2 then the tringle is right ngled nd its hypotenuse is c units long. Exmple 25 Find the unknown length in: 1 m 1 m 08. m x m 08. m x m x 2 =0: ) x = p (0: ) ) x ¼ 1:2806 So, the length is bout 1:28 m. Exmple 26 Find the unknown length in: 5 m x m 17. m 5 m x m 17. m x 2 +1:7 2 =5 2 ) x 2 =5 2 1:7 2 ) x = p (5 2 1:7 2 ) ) x ¼ 4:7021 So, the length is bout 4:70 m.

23 BACKGROUND KNOWLEDGE 2 EXERCISE L.1 1 Find, correct to significnt figures, the vlue of x in: b c 1 m x m 12. m 8. m x m x m 1. 2 m 21. m 18. m 2 How high is the roof bove the wlls in the following roof structures? b 92. m m 16 m m Bob is bout to tee off on the sixth, pr 4 t the Royl Golf Club. He chooses to hit over the lke, directly t the flg. If the pin is 15 m from the wter s edge, how fr must he hit the bll to cler the lke? 208m 142m lke 15m 4 Tee A siling ship sils 46 km north then 74 km est. Drw fully lbelled digrm of the ship s course. b How fr is the ship from its strting point? PYTHAGORAS THEOREM IN -D PROBLEMS The theorem of Pythgors is often used twice in -D problem solving. Exmple 27 The floor of room is 6 m by 4 m, nd the floor to ceiling height is m. Find the distnce from corner point on the floor to the opposite corner point on the ceiling.

24 24 BACKGROUND KNOWLEDGE The required distnce is [AD]. We join [BD]. A m B 4 m x m C 6 m y m D In BCD, x 2 = fpythgorsg ) x 2 = = 52 In ABD, y 2 = x ) y 2 = = 61 ) y = p 61 ¼ 7:81 So, the distnce is bout 7:81 m. EXERCISE L.2 1 A pole [AB] is 16 m tll. At point 5 m below B, four wires re connected from the pole to the ground. Ech wire is pegged to the ground 5 m from the bse of the pole. Wht is the totl length of wire needed if totl of 2 m extr is needed for tying? B 5 m A 5 m 2 A cube hs sides of length 10 cm. Find the length of digonl of the cube. digonl A room is 7 m by 4 m nd hs height of m. Find the distnce from corner point on the floor to the opposite corner of the ceiling. 4 A pyrmid of height 40 m hs squre bse with edges of length 50 m. Determine the length of the slnt edges. 5 An eroplne P is flying t n ltitude of P m. The pilot spots two ships A nd km B. Ship A is due south of P nd 22:5 km 10 km wy in direct line. Ship B is due est km nd 40:8 km from P in direct line. Find B b the distnce between the two ships. A M COORDINATE GEOMETRY THE NUMBER PLANE The position or loction of ny point in the number plne cn be specified in terms of n ordered pir of numbers (x, y), where x is the horizontl step from fixed point O, nd y is the verticl step from O.

25 THE THREE BASIC TRIGONOMETRIC RATIOS BACKGROUND KNOWLEDGE 5 sin µ = OPP HYP, ADJ cos µ = HYP, OPP tn µ = ADJ HYP ADJ OPP sin µ, cos µ nd tn µ re bbrevitions for sine µ, cosine µ nd tngent µ: The three formule bove re clled the trigonometric rtios nd re the tools we use for finding side lengths nd ngles of right ngled tringles. However, before doing this we will clculte the trigonometric rtios in right ngled tringles where we know two of the sides. Exmple 40 2 cm Find, without using clcultor, sin µ, cos µ nd tn µ: cm So, HYP x cm OPP sin µ = OPP HYP = p 1, ADJ If the hypotenuse is x cm long x 2 = ) x 2 = 1 ) x = p 1 cos µ = ADJ HYP = 2 p 1, fpythgorsg fs x>0g tn µ = OPP ADJ = 2. Exmple 41 If µ is n cute ngle nd clcultor. sin µ = 1, find cos µ nd tn µ without using We drw right ngled tringle nd mrk on ngle µ so tht OPP =1unit nd HYP =units. Now x = 2 fpythgorsg ) x 2 +1=9 ) x 2 =8 ) x = p 8 fs x>0g ) cos µ = ADJ HYP = p 8 nd tn µ = OPP ADJ = 1 p 8 : x 1

26 PROBLEM SOLVING USING TRIGONOMETRY BACKGROUND KNOWLEDGE 4 Trigonometry is very useful brnch of mthemtics. Heights nd distnces which re very difficult or even impossible to mesure cn often be found using trigonometry. ngle of elevtion ngle of depression horizontl ngle of elevtion Exmple 48 Find the height of tree which csts shdow of 12:4 m when the sun mkes n ngle of 52 o to the horizon m h m Let h m be the tree s height. For the 52 o ngle, OPP = h, ADJ = 12:4 ) tn 52 o = h 12:4 ) 12:4 tn 52 o = h ) h ¼ 15:9 ) the tree is 15:9 m high. EXERCISE N.5 1 Find the height of flgpole which csts shdow of 9:2 m when the sun mkes n ngle of 6 o to the horizontl. 2 A hill is inclined t 18 o to the horizontl. It runs down to the bech so its bse is t se level. b If I wlk 1.2 km up the hill, wht is my height bove se level? If I m 500 metres bove se level, how fr hve I wlked up the hill? B 120 m C A surveyor stnding t A notices two posts B nd C on the opposite side of cnl. The posts 7 re 120 m prt. If the ngle of sight between cnl the posts is 7 o, how wide is the cnl? A 4 A trin must climb constnt grdient of 5:5 m for every 200 m of trck. Find the ngle of incline.

27 44 BACKGROUND KNOWLEDGE 5 Find the ngle of elevtion to the top of 56 m high building from point A which is 11 m from its bse. Wht is the ngle of depression from the top of the building to A? A 6 The ngle of depression from the top of 120 m high verticl cliff to bot B is 16 o. Find how fr the bot is from the bse of the cliff. 11 m 56 m 120 m se B 7 Srh mesures the ngle of elevtion to the top of tree s 2:6 o from point which is 250 m from its bse. Her eye level, from which the ngle mesurement is tken, is 1:5 m bove the ground. Assuming the ground is horizontl, find the height of the tree. 8 Kylie mesures the ngle of elevtion from point on level ground to the top of building 120 metres high to be 2 o. She wlks towrds the building until the ngle of elevtion is 45 o. How fr did she wlk? 9 A circulr trck of rdius r m is bnked t n ngle of µ to the horizontl. The idel speed for the bend is given by the formul s = p gr tn µ where g =9:8 ms 2. Wht is the idel speed for vehicle trvelling on circulr trck of rdius 100 m which is bnked t n ngle of 15 o? b At wht ngle should trck of rdius 200 m be bnked if it is designed for vehicle trvelling t 20 ms 1? Exmple 49 A builder hs designed the roof structure illustrted. The pitch of the roof is the ngle tht the roof mkes with the horizontl. Find the pitch of this roof. 87. m 1 m 1 m 1 m 87. m 75. m 15 m x m By constructing n ltitude of the isosceles tringle, we form two right ngled tringles. For ngle µ : ADJ =7:5, HYP =8:7 ) cos µ = 7:5 8:7 µ 7:5 ) µ = cos 1 8:7 ) µ ¼ 0:450 o ) the pitch is pproximtely o.

28 BACKGROUND KNOWLEDGE Find µ, the pitch of the roof. 8. m 06. m 15 m 06. m 11 If the pitch of the given roof is 2 o, find the B length of the timber bem [AB]. 2 A 08. m 16 m 08. m 12 An open right-circulr cone hs verticl ngle mesuring 40 o nd bse rdius of 0 cm. Find the cpcity of the cone in litres A A refrigertor is tipped ginst verticl wll so it P cn be serviced. It mkes n ngle of 70 o with the 1 m horizontl floor. How high is point A bove the floor? 2 m 70 Q N 14 From n observer O, the ngles of elevtion to the bottom nd the top of flgpole re 6 o nd 8 o respectively. Find the height of the flgpole. C B 6 O m A 15 The ngle of depression from the top of 150 m high cliff to bot t se is 7 o. How much closer to the cliff must the bot move for the ngle of depression to become 19 o? 16 A helicopter flies horizontlly t 100 km h 1. An observer notices tht it tkes 20 seconds for the helicopter to fly from directly overhed to being t n ngle of elevtion of 60 o. Find the height of the helicopter bove the ground. 17 [AC] is stright shore line nd B is bot out t se. Find the shortest distnce from the bot to the shore if A nd C re 5 km prt. A 5 km B C

29 46 BACKGROUND KNOWLEDGE 18 A regulr pentgonl grden plot is to be constructed with sides of length 20 m. d m Find the width of lnd d m required for the plot. 20 m