Background knowledge

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1 Bckground knowledge Contents: B C D E F G H I J K L M N Surds nd rdicls Scientific nottion (stndrd form) Numer systems nd set nottion lgeric simplifiction Liner equtions nd inequlities Modulus or solute vlue Product expnsion Fctoristion Formul rerrngement dding nd sutrcting lgeric frctions Congruence nd similrity Pythgors theorem Coordinte geometry Right ngled tringle trigonometry

2 BCKGROUND 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. SURDS ND RDICLS rdicl is ny numer which is written with the rdicl sign p. surd is rel, irrtionl rdicl such s p, p 3, p 5 or p 6. Surds re present in solutions to some qudrtic equtions. p 4 is rdicl ut is not surd s it simplifies to. p is the non-negtive numer such tht p p =. Properties: ² p is never negtive, so p > 0. ² p is meningful only for > 0. ² p = p p for > 0 nd > 0. r p ² = p for > 0 nd >0. Exmple 1 Write s single surd: p p 3 p 18 p 6 p p 3 = p 3 = p 6 = p p 18 6 q 18 6 = p 3 or p 18 p 6 = p 6 p 3 p 6 = p 3 EXERCISE 1 Write s single surd or rtionl numer: p 3 p 5 ( p 3) c p p d 3 p p e 3 p 7 p 7 f p p 1 g p p1 6 h p p18 3 Exmple Simplify: 3 p 3+5 p 3 p 5 p Compre with x 5x = 3x 3 p 3+5 p 3 =(3+5) p 3 =8 p 3 p 5 p =( 5) p = 3 p

3 BCKGROUND KNOWLEDGE 3 Simplify the following mentlly: p +3 p p 3 p c 5 p 5 3 p 5 d 5 p 5+3 p 5 e 3 p 5 5 p 5 f 7 p 3+ p 3 g 9 p 6 1 p 6 h p + p + p Exmple 3 Write p 18 in the form p where nd re integers nd is s lrge s possile. p 18 = p 9 f9 is the lrgest perfect squre fctor of 18g = p 9 p =3 p 3 Write the following in the form p where nd re integers nd is s lrge s possile: p 8 p 1 c p 0 d p 3 e p p p p 7 f 45 g 48 h 54 p p p p i 50 j 80 k 96 l 108 Exmple 4 Simplify: p 75 5 p 7 p 75 5 p 7 = p p 9 3 = 5 p p 3 =10 p 3 15 p 3 = 5 p 3 4 Simplify: 4 p 3 p 1 3 p + p 50 c 3 p 6+ p 4 d p 7 + p 1 e p 75 p p p p 1 f Exmple 5 Write p 9 3 without rdicl in the denomintor. 9p 3 = p 9 3 p p3 3 = 9p 3 3 =3 p 3

4 4 BCKGROUND KNOWLEDGE 5 Write without rdicl in the denomintor: 1 p 6 p3 c 7 p d 10 p 5 e 10 p f 18 p 6 g 1 p 3 h 5 p7 i 14 p 7 j p 3 p B SCIENTIFIC NOTTION (STNDRD FORM) Scientific nottion (or stndrd form) involves writing ny given numer s numer etween 1 nd 10, multiplied y power of 10, i.e., 10 k where 1 6 <10 nd k Z. Exmple 6 Write in stndrd form: : = 3: fshift deciml point 4 plces to the =3: left nd g 0: = 8: fshift deciml point 4 plces to the =8: right nd g EXERCISE B 1 Express the following in scientific nottion: c :59 d 0:59 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. Bcteri re single cell orgnisms, some of which hve dimeter of 0:0003 mm. c speck of dust hs width smller thn 0:001 mm. d e The core temperture of the Sun is 15 million degrees Celsius. single red lood cell lives for out four months. During this time it will circulte round the ody times.

5 BCKGROUND KNOWLEDGE 5 Exmple 7 Write s n ordinry numer: 3: 10 5: : 10 =3:0 100 = 30 5: = : =0: Write s n ordinry deciml numer: c : d 7: e 3: f 8: g 4: h Write s n ordinry deciml numer: c : d 7: e 3: f 8: g 4: h Write s n ordinry deciml numer: The wvelength of light is m. The estimted world popultion for the yer 000 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 deciml plces: (3: ) (4: ) (6:4 10 ) c 3: d (9: ) (7: 10 6 ) e 1 3: f (1: 10 3 ) 3 7 If missile trvels t 5400 km h 1, how fr will it trvel in: 1 dy 1 week c yers? Give your nswers in scientific nottion correct to deciml plces, nd ssume tht 1 yer ¼ 365:5 dys. istockphoto.com 8 Light trvels t speed of metres per second. How fr will light trvel in: 1 minute 1 dy c 1 yer? Give your nswers with deciml prt correct to deciml plces, nd ssume tht 1 yer ¼ 365:5 dys.

6 6 BCKGROUND KNOWLEDGE C NUMBER SYSTEMS ND SET NOTTION NUMBER SYSTEMS We use: ² R to represent the set of ll rel numers. These include ll of the numers on the numer line. negtives 0 positives ² N to represent the set of ll nturl numers. N = f0, 1,, 3, 4, 5,...g ² Z to represent the set of ll integers. Z = f0, 1,, 3, 4,...g ² Z + is the set of ll positive integers. Z + = f1,, 3, 4,...g ² Q to represent the set of ll rtionl numers which re ny numers of the form p where p nd q re integers, q 6= 0. q SET NOTTION fx j 3 <x<g reds the set of ll vlues tht x cn e such tht x lies etween 3 nd. the set of ll such tht Unless stted otherwise, we ssume tht x is rel. EXERCISE C 1 Write verl sttements for the mening of: fx j x>5, x R g fx j x 6 3, x Z g c fy j 0 <y<6g d fx j 6 x 6 4, x Z g e ft j 1 <t<5g f fn j n< or n > 6g Exmple 8 Write in set nottion: 0 4 x included not included -3 4 x fx j 1 6 x 6 4, x N g or fx j 1 6 x 6 4, x Z g fx j 3 6 x<4, x R g Write in set nottion: c d e f

7 BCKGROUND KNOWLEDGE 7 3 Sketch the following numer sets: fx j 4 6 x<10, x N g fx j 4 <x6 5, x Z g c fx j 5 <x6 4, x R g d fx j x> 4, x Z g e fx j x 6 8, x R g D LGEBRIC SIMPLIFICTION To nswer the following questions, you will need to rememer: ² the distriutive lw ( + c) = + c ² power nottion =, 3 =, 4 =, nd so on. EXERCISE D 1 Simplify if possile: 3x +7x 10 3x +7x x c x +3x +5y d 8 6x x e 7 +5 f 3x +7x 3 Remove the rckets nd then simplify: 3(x +5)+4(5+4x) 6 (3x 5) c 5( 3) 6( ) d 3x(x 7x +3) (1 x 5x ) 3 Simplify: x(3x) c p 16x 4 d ( ) E LINER EQUTIONS ND INEQULITIES When deling with inequlities: ² multiplying or dividing oth sides y negtive reverses the inequlity sign. ² do not multiply or divide oth sides y the unknown or term involving the unknown. EXERCISE E 1 Solve for x: x +5=5 3x 7 > 11 c 5x +16=0 d x 3 7=10 e 6x +11< 4x 9 f 3x =8 5 g 1 x > 19 h 1 x +1= 3 x i 3 3x 4 = 1 (x 1)

8 8 BCKGROUND KNOWLEDGE Solve simultneously for x nd y: ½ ½ x +y =9 x +5y =8 x y =3 x y = d ½ 5x 4y =7 3x +y =9 e ½ x +y =5 x +4y =1 c f ½ 7x +y = 4 3x +4y =14 8 x >< + y 3 =5 x >: 3 + y 4 =1 F MODULUS OR BSOLUTE VLUE The modulus or solute vlue of rel numer 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 numer is its distnce from zero on the numer line. Becuse the modulus is distnce, it cnnot e negtive We denote the modulus of x s jxj. jxj is the distnce of x from 0 on the rel numer line. x If x>0 If x<0 0 x x x 0 jx j cn e considered s the distnce of x from. LGEBRIC DEFINITION jxj = x if x > 0 x if x<0 or jxj = p x MODULUS EQUTIONS It is cler tht jxj = hs two solutions, x = nd x =, since jj = nd j j =. If jxj = where > 0, then x =.

9 BCKGROUND KNOWLEDGE 9 EXERCISE F 1 Find the vlue of: 5 ( 11) j5j j 11j c j5 ( 11)j d ( ) + 11( ) e j 6j j 8j f j 6 ( 8)j If =, =3, nd c = 4 find the vlue of: jj jj c jjjj d jj e j j f jj jj g j + j h jj + jj i jj j k c jcj l jj 3 Solve for x: jxj =3 jxj = 5 c jxj =0 d jx 1j =3 e j3 xj =4 f jx +5j = 1 g j3x j =1 h j3 xj =3 i j 5xj =1 G PRODUCT EXPNSION y =(x 1)(x +3) cn e expnded into the generl form y = x + x + c. Likewise, y =(x 3) +7 cn lso e expnded into this form. Following is list of expnsion rules you cn use: ² ( + )(c + d) =c + d + c + d ² ( + )( ) = fdifference of two squresg ² ( + ) = + + fperfect squresg Exmple 9 Expnd nd simplify: (x + 1)(x +3) (3x )(x +3) (x + 1)(x +3) =x +6x + x +3 =x +7x +3 (3x )(x +3) =3x +9x x 6 =3x +7x 6

10 10 BCKGROUND KNOWLEDGE EXERCISE G 1 Expnd nd simplify using ( + )(c + d) =c + d + c + d: (x + 3)(x +1) (3x + 4)(x +) c (5x )(x +1) d (x + )(3x 5) e (7 x)( + 3x) f (1 3x)(5 + x) g (3x + 4)(5x 3) h (1 3x)( 5x) i (7 x)(3 x) j (5 x)(3 x) k (x + 1)(x +) l (x 1)(x +3) Exmple 10 Expnd using the rule ( + )( ) = : (5x )(5x +) (7 + x)(7 x) (5x )(5x +) =(5x) =5x 4 (7 + x)(7 x) =7 (x) =49 4x Expnd using the rule ( + )( ) = : (x + 6)(x 6) (x + 8)(x 8) c (x 1)(x +1) d (3x )(3x +) e (4x + 5)(4x 5) f (5x 3)(5x +3) g (3 x)(3 + x) h (7 x)(7 + x) i (7 + x)(7 x) j (x + p )(x p ) k (x + p 5)(x p 5) l (x p 3)(x + p 3) Exmple 11 Expnd using the perfect squre expnsion rule: (x +) (3x 1) (x +) = x +(x)() + = x +4x +4 (3x 1) =(3x) + (3x)( 1) + ( 1) =9x 6x +1 3 Expnd nd simplify using the perfect squre expnsion rule: (x +5) (x +7) c (x ) d (x 6) e (3 + x) f (5 + x) g (11 x) h (10 x) i (x +7) j (3x +) k (5 x) l (7 3x) 4 Expnd the following into the generl form y = x + x + c: y =(x + )(x +3) y =3(x 1) +4 c y = (x + 1)(x 7) d y = (x +) 11 e y =4(x 1)(x 5) f y = 1 (x +4) 6 g y = 5(x 1)(x 6) h y = 1 (x +) 6 i y = 5 (x 4)

11 BCKGROUND KNOWLEDGE 11 Exmple 1 The use of Expnd nd simplify: 1 (x +3) (3 + x) ( + x)(3 x) rckets is essentil! 1 (x +3) =1 [x +6x +9] =1 x 1x 18 = x 1x 17 (3 + x) ( + x)(3 x) =6+x [6 x +3x x ] =6+x 6+x 3x + x = x + x 5 Expnd nd simplify: 1+(x +3) +3(x )(x +3) c 3 (3 x) d 5 (x + 5)(x 4) e 1 + (4 x) f x 3x (x + )(x ) g (x +) (x + 1)(x 4) h (x +3) +3(x +1) i x +3x (x 4) j (3x ) (x +1) H FCTORISTION lgeric fctoristion is the reverse process of expnsion. For exmple, (x + 1)(x 3) x 5x 3 is expnded to x 5x 3, wheres is fctorised to (x + 1)(x 3). Notice tht x 5x 3=(x + 1)(x 3) hs een fctorised into two liner fctors. Flow chrt for fctorising: Expression Difference of two squres =( + )( ) Perfect squre + + =( + ) Tke out ny common fctors Recognise type Sum nd Product type x + x + c, 6= 0 ² find c ² find the fctors of c which dd to ² if these fctors re p nd q, replce x y px + qx ² complete the fctoristion Sum nd Product type x + x + c x + x + c =(x + p)(x + q) where p + q = nd pq = c

12 1 BCKGROUND KNOWLEDGE Exmple 13 Fully fctorise: 3x 1x 4x 1 c x 1x +36 Rememer tht ll fctoristions cn e checked y expnsion! 3x 1x =3x(x 4) 4x 1 =(x) 1 =(x + 1)(x 1) c x 1x +36 = x +(x)( 6) + ( 6) =(x 6) f3x is common fctorg fdifference of two squresg fperfect squre formg EXERCISE H 1 Fully fctorise: 3x +9x x +7x c 4x 10x d 6x 15x e 9x 5 f 16x 1 g x 8 h 3x 9 i 4x 0 j x 8x +16 k x 10x +5 l x 8x +8 m 16x +40x +5 n 9x +1x +4 o x x + 11 Exmple 14 Fully fctorise: 3x +1x +9 x +3x +10 3x +1x +9 =3(x +4x +3) =3(x + 1)(x +3) x +3x +10 = [x 3x 10] = (x 5)(x +) f3 is common fctorg fsum =4, product =3g fremoving 1 s common fctor to mke the coefficient of x e 1g fsum = 3, product = 10g Fully fctorise: x +9x +8 x +7x +1 c x 7x 18 d x +4x 1 e x 9x +18 f x + x 6 g x + x + h 3x 4x +99 i x 4x

13 BCKGROUND KNOWLEDGE 13 j x +6x 0 k x 10x 48 l x +14x 1 m 3x +6x 3 n x x 1 o 5x +10x +40 FCTORISTION BY SPLITTING THE x-term Using the distriutive lw to expnd we see tht: (x + 3)(4x +5) =8x +10x +1x +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 +10x +1x +15 Step : Group in pirs =(8x +10x) + (1x + 15) Step 3: Fctorise ech pir seprtely =x(4x + 5) + 3(4x +5) Step 4: Fctorise fully =(4x + 5)(x + 3) 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 + x + c : ² Find c: ² Find the fctors of c which dd to : ² If these fctors re p nd q, replce x y px + qx: ² Complete the fctoristion. Exmple 15 Fully fctorise: x x 10 6x 5x +14 x x 10 hs c = 10 = 0: The fctors of 0 which dd to 1 re 5 nd +4: ) x x 10 = x 5x +4x 10 = x(x 5) + (x 5) =(x 5)(x +) 6x 5x +14 hs c =6 14 = 84: The fctors of 84 which dd to 5 re 1 nd 4: ) 6x 5x +14=6x 1x 4x +14 =3x(x 7) (x 7) =(x 7)(3x )

14 14 BCKGROUND KNOWLEDGE 3 Fully fctorise: x +5x 1 3x 5x c 7x 9x + d 6x x e 4x 4x 3 f 10x x 3 g x 11x 6 h 3x 5x 8 i 8x +x 3 j 10x 9x 9 k 3x +3x 8 l 6x +7x + m 4x x +6 n 1x 16x 3 o 6x 9x +4 p 1x 10 9x q 8x 6x 7 r 1x +13x +3 s 1x +0x +3 t 15x x +8 u 14x 11x 15 Exmple 16 Fully fctorise: 3(x +)+(x 1)(x +) (x +) 3(x +)+(x 1)(x +) (x +) =(x + )[3 + (x 1) (x + )] fs (x +) is common fctorg =(x + )[3 + x x ] =(x + )(x 1) 4 Fully fctorise: 3(x +4)+(x + 4)(x 1) 8( x) 3(x + 1)( x) c 6(x +) +9(x +) d 4(x +5)+8(x +5) e (x + )(x +3) (x + 3)( x) f (x +3) +(x +3) x(x +3) g 5(x ) 3( x)(x +7) h 3(1 x)+(x + 1)(x 1) Exmple 17 Fully fctorise using the difference of two squres : (x +) 9 (1 x) (x +1) (x +) 9 =(x +) 3 =[(x + ) + 3][(x +) 3] =(x + 5)(x 1) (1 x) (x +1) = [(1 x)+(x + 1)][(1 x) (x + 1)] =[1 x +x + 1][1 x x 1] = 3x(x +) 5 Fully fctorise: (x +3) 16 4 (1 x) c (x +4) (x ) d 16 4(x +) e (x +3) (x 1) f (x + h) x g 3x 3(x +) h 5x 0( x) i 1x 7(3 + x)

15 BCKGROUND KNOWLEDGE 15 INVESTIGTION NOTHER FCTORISTION TECHNIQUE Wht to do: 1 By expnding, show tht (x + p)(x + q) = x +[p+q]x+ If x (x + p)(x + q) + x + c =, show tht p + q = nd pq = c. 3 Using on 8x +x +15, we hve ( 8x (8x + p)(8x + q) p + q = +x +15= where 8 pq =8 15 = 10 ) p =1, q =10, or vice vers ) 8x (8x + 1)(8x + 10) +x +15= 8 4(x + 3)(4x +5) = 8 1 =(x + 3)(4x +5) Use the method shown to fctorise: i 3x +14x +8 ii 1x +17x +6 iii 15x +14x 8 Check your nswers to y expnsion. h pq i : I FORMUL RERRNGEMENT In the formul D = xt + p we sy tht D is the suject. This is ecuse D is expressed in terms of the other vriles x, t nd p. We cn rerrnge the formul to mke one of the other vriles the suject. 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 suject of D = xt + p. If D = xt + p ) xt + p = D ) xt + p p = D p fsutrcting p from oth sidesg ) xt = D p ) xt t = D p t ) x = D p t fdividing oth sides y tg

16 16 BCKGROUND KNOWLEDGE EXERCISE I 1 Mke x the suject of: + x = x = c x + = d d c + x = t e 5x +y =0 f x +3y =1 g 7x +3y = d h x + y = c i y = mx + c Exmple 19 Mke z the suject of c = m z. c = m z c z = m z z ) cz = m cz ) c = m c fmultiplying oth sides y zg fdividing oth sides y cg ) z = m c Mke z the suject of: 3 Mke: z = c z = d c 3 d = z d z = z the suject of F = m r the suject of C =¼r c d the suject of V = ldh d K the suject of = K : Exmple 0 Mke t the suject of s = 1 gt where t>0. 1 gt = s fwriting with t on LHSg ) 1 gt = s fmultiplying oth sides y g ) gt =s gt ) g = s g ) t = s g r s ) t = g fdividing oth sides y gg fs t>0g

17 4 Mke: BCKGROUND KNOWLEDGE (Chpter 1) 17 r the suject of = ¼r if r>0 x the suject of N = x5 c r the suject of V = 4 3 ¼r3 d x the suject of D = n x 3 : 5 Mke: p the suject of d = l the suject of T = 1 n 5p l c the suject of c = p r l d l the suject of T =¼ g e the suject of P =( + ) f h the suject of = ¼r +¼rh g r the suject of I = E R + r 6 Mke the suject of the formul k = d. Find the vlue for when k = 11, d =4, =. h q the suject of = B p q : 7 The formul for determining the volume of sphere of rdius r is V = 4 3 ¼r3. Mke r the suject of the formul. Find the rdius of sphere which hs volume of 40 cm 3. 8 The distnce trvelled y n oject ccelerting from sttionry position is given y the formul S = 1 t cm where is the ccelertion in cm s nd t is the time in seconds. Mke t the suject of the formul. Consider t>0 only. Find the time tken for n oject ccelerting t 8 cm s to trvel 10 m. 9 The reltionship etween the oject nd imge distnces (in cm) 1 for concve mirror cn e written s f = 1 u + 1 v where f is the focl length, u is the oject distnce nd v is the imge distnce. Mke v the suject of the formul. Given focl length of 8 cm, find the imge distnce for the following oject distnces: i 50 cm ii 30 cm. 10 ccording to Einstein s theory of reltivity, the mss of prticle is given y the m 0 formul m = r ³, where m 0 is the mss of the prticle t rest, v 1 v is the speed of the prticle, nd c c is the speed of light in vcuum. c Mke v the suject of the formul given v>0. Find the speed necessry to increse the mss of prticle to three times its rest mss, i.e., m =3m 0. Give the vlue for v s frction of c. cyclotron incresed the mss of n electron to 30m 0 : With wht velocity must the electron hve een trvelling, given c = ms 1?

18 18 BCKGROUND KNOWLEDGE J To dd or sutrct lgeric frctions, we comine them into single frction with the lest common denomintor (LCD). For exmple, x 1 3 x +3 DDING ND SUBTRCTING LGEBRIC FRCTIONS hs LCD of 6, so we write ech frction with denomintor 6. Exmple 1 Write s single frction: + 3 x x 1 3 x x ³ x = + x 3 x x +3 = x x 1 x +3 3 = µ x 1 3 µ x (x 1) 3(x +3) = 6 x 3x 9 = 6 x 11 = 6 EXERCISE J 1 Write s single frction: 3+ x 5 d 3 x 4 Exmple 3x +1 Write x s single frction x e +x 3 = 3x +1 x µ 3x +1 x + x 4 5 µ x x (3x +1) (x ) = x 3x +1 x +4 = x = x +5 x c 3+ x x +5 f x flcd =(x )g

19 BCKGROUND KNOWLEDGE 19 Write s single frction: 1+ 3 x x 4 d x 1 x e 3 x x +1 c 3 x 1 f x 3 Write s single frction: c 3x x +5 + x 5 x 5x x 4 + 3x x +4 d 1 x 1 x 3 x +1 x 3 x +4 x +1 K CONGRUENCE ND SIMILRITY 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 cceptle 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 (SS) ² two ngles nd pir of corresponding sides re equl (cors) ² for right ngled tringles, the hypotenuse nd one other pir of sides re equl (RHS). If congruence cn e proven then ll corresponding lengths, ngles nd res must e equl.

20 0 BCKGROUND KNOWLEDGE Exmple 3 Explin why BC nd DBC re congruent: B C B C D D s BC nd DBC re congruent (SS) s: ² C = DC ² CB = DCB, nd ² [BC] is common to oth. EXERCISE K.1 1 Tringle BC is isosceles with C = BC. [BC] nd [C] re produced to E nd D respectively so tht CE = CD. Prove tht E = BD. E C D B Point P is equidistnt from oth [B] nd [C]. Use congruence to show tht P lies on the isector of BC. P B C 3 Two concentric circles re drwn. t P on the inner circle, tngent is drwn which meets the other circle t nd B. Use tringle congruence to prove tht P is the midpoint of [B]. O B P SIMILRITY Two tringles re similr if one is n enlrgement of the other. Similr tringles re equingulr, nd hve corresponding sides in the sme rtio.

21 BCKGROUND KNOWLEDGE 1 Exmple 4 Estlish tht pir of tringles is similr, then find x given BD =0cm. 1 cm B x cm C E (x + ) cm D ² - x + x smll lrge 1 cm B E x cm C 0 cm (x + ) cm D The tringles re equingulr nd hence similr. ) x + = x 0 1 fsides in the sme rtiog ) 1(x +)=0x ) 1x +4=0x ) 4 = 8x ) x =3 EXERCISE K. 1 In ech of the following, estlish tht pir of tringles is similr, nd hence find x: c P U cm B C 5 cm X x cm 6 cm 3 cm x cm Q S x cm 7 cm R cm D E 6 cm T V 5 cm Y 6 cm Z d D e f X B 4 cm x cm E 3 cm 5 cm C Y cm U 5 cm x cm Z V S P 8 cm x cm 10 cm 50 Q 50 9 cm R T fther nd son re stnding side-y-side. The fther is 1:8 m tll nd csts shdow 3: m long, while his son s shdow is :4 m long. How tll is the son?

22 BCKGROUND KNOWLEDGE L PYTHGORS 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 = + c This theorem, known to the ncient Greeks, is vlule ecuse: 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, nd c units nd + = c then the tringle is right ngled nd its hypotenuse is c units long. Exmple 5 Find the unknown length in: 1 m 1 m 08. m x m 08. m x m x =0:8 +1 ) x = p (0:8 +1 ) ) x ¼ 1:806 So, the length is out 1:8 m. Exmple 6 Find the unknown length in: 5 m x m 17. m 5 m x m 17. m x +1:7 =5 ) x =5 1:7 ) x = p (5 1:7 ) ) x ¼ 4:701 So, the length is out 4:70 m.

23 BCKGROUND KNOWLEDGE 3 EXERCISE L.1 1 Find, correct to 3 significnt figures, the vlue of x in: c 1 m x m 1. m 38. m x m x m 13. m 1. m 18. m How high is the roof ove the wlls in the following roof structures? 9. m m 16 m m 3 Bo is out to tee off on the sixth, pr 4 t the Royl Golf Clu. 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 ll to cler the lke? 08m 14m lke 15m 4 Tee siling ship sils 46 km north then 74 km est. Drw fully lelled digrm of the ship s course. How fr is the ship from its strting point? PYTHGORS THEOREM IN 3-D PROBLEMS The theorem of Pythgors is often used twice in 3-D prolem solving. Exmple 7 The floor of room is 6 my4 m, nd the floor to ceiling height is 3 m. Find the distnce from corner point on the floor to the opposite corner point on the ceiling.

24 4 BCKGROUND KNOWLEDGE The required distnce is [D]. We join [BD]. 3 m B 4 m x m C 6 m y m D In BCD, x =4 +6 fpythgorsg ) x =16+36=5 In BD, y = x +3 ) y =5+9=61 ) y = p 61 ¼ 7:81 So, the distnce is out 7:81 m. EXERCISE L. 1 pole [B] is 16 m tll. t point 5 m elow B, four wires re connected from the pole to the ground. Ech wire is pegged to the ground 5 m from the se of the pole. Wht is the totl length of wire needed if totl of m extr is needed for tying? B 5 m 5 m cue hs sides of length 10 cm. Find the length of digonl of the cue. digonl 3 room is 7 my4 m nd hs height of 3 m. Find the distnce from corner point on the floor to the opposite corner of the ceiling. 4 pyrmid of height 40 m hs squre se with edges of length 50 m. Determine the length of the slnt edges. 5 n eroplne P is flying t n ltitude of P m. The pilot spots two ships nd km B. Ship is due south of P nd :5 km 10 km wy in direct line. Ship B is due est. 5 km nd 40:8 km from P in direct line. Find B the distnce etween the two ships. M COORDINTE GEOMETRY THE NUMBER PLNE The position or loction of ny point in the numer plne cn e specified in terms of n ordered pir of numers (x, y), where x is the horizontl step from fixed point O, nd y is the verticl step from O.

25 BCKGROUND KNOWLEDGE 5 The point O is clled the origin. Once O hs een specified, we drw two perpendiculr xes through it. DEMO The x-xis is horizontl nd the y-xis is verticl. The numer plne is lso known s either: ² the -dimensionl plne, or ² the Crtesin plne, nmed fter René Descrtes. (, ) is clled n ordered pir, where nd re the coordintes of the point. is clled the x-coordinte. is clled the y-coordinte. THE DISTNCE FORMUL y-xis If (x 1, y 1 ) nd B(x, y ) re two points in plne, then the distnce etween these points is given y B = p (x x 1 ) +(y y 1 ). P (, ) x-xis Exmple 8 Find the distnce etween (, 1) nd B(3, 4). (, 1) B(3, 4) B = p (3 ) +(4 1) x 1 y 1 x y = p 5 +3 = p = p 34 units THE MIDPOINT FORMUL If M is hlfwy etween points nd B then M is the midpoint of [B]. M B If (x 1, y 1 ) nd B(x, y ) re two points then µ x1 + x the midpoint M of [B] hs coordintes, y 1 + y : Exmple 9 Find the coordintes of the midpoint of [B] for ( 1, 3) nd B(4, 7): The x-coordinte of the midpoint = 1+4 = 3 =11 The y-coordinte of the midpoint = 3+7 =5 ) the midpoint of [B] is (1 1, 5).

26 6 BCKGROUND KNOWLEDGE THE GRDIENT OR SLOPE OF LINE When looking t line segments drwn on set of xes, it is cler tht different line segments re inclined to the horizontl t different ngles. Some pper to e steeper thn others. The grdient or slope of line is mesure of its steepness. If is (x 1, y 1 ) nd B is (x, y ) then the grdient of [B] is y y 1 x x 1 : Exmple 30 Find the grdient of the line through (3, ) nd (6, 4). (3, ) (6, 4) grdient = y y 1 = 4 x x = x 1 y 1 x y PROPERTIES OF GRDIENT ² horizontl lines hve grdient of 0 (zero) ² verticl lines hve n undefined grdient ² forwrd sloping lines hve positive grdients ² ckwrd sloping lines hve negtive grdients ² prllel lines hve equl grdients ² the grdients of two perpendiculr lines re negtive reciprocls of ech other. If the grdients re m 1 nd m then m = 1 or m 1 m = 1. m 1 This is true except when the lines re prllel to the xes. EQUTIONS OF LINES The eqution of line sttes the connection etween the x nd y vlues for every point on the line, nd only for points on the line Equtions of lines hve vrious forms: m1 m ² ll verticl lines hve equtions of the form x = where is constnt. ² ll horizontl lines hve equtions of the form y = c where c is constnt. ² If stright line hs grdient m nd psses through (, ) then it hs eqution y = m or y = m(x ) fpoint-grdient formg x which cn e rerrnged into y = mx + c fgrdient-intercept formg

27 BCKGROUND KNOWLEDGE 7 ² If stright line hs grdient B nd psses through (x 1, x ) then it hs eqution x By = x 1 By 1 or x By = C fgenerl formg Exmple 31 Find, in grdient-intercept form, the eqution of the line through ( 1, 3) with grdient of 5. To find the eqution of line we need to know its grdient nd point on it. The eqution of the line is y 3 x 1 =5 y 3 ) x +1 =5 ) y 3=5(x +1) ) y =5x +8 Exmple 3 Find, in generl form, the eqution of the line through (1, 5) nd (5, ). The grdient = = y So, the eqution is x 5 = 3 4 y + ) x 5 = 3 4 ) 4y +8=3x 15 ) 3x 4y =3 XES INTERCEPTS xes intercepts re the x- nd y-vlues where grph cuts the coordinte xes. The x-intercept is found y letting y =0. The y-intercept is found y letting x =0. y y-intercept x-intercept x Exmple 33 For the line with eqution x 3y =1, find the xes intercepts. When x =0, 3y =1 ) y = 4 When y =0, x =1 ) x =6 So, the y-intercept is 4 nd the x-intercept is 6:

28 8 BCKGROUND KNOWLEDGE DOES POINT LIE ON LINE? point lies on line if its coordintes stisfy the eqution of the line. Exmple 34 Does (3, ) lie on the line with eqution 5x y =0? Sustituting (3, ) into 5x y =0 gives LHS = 5(3) ( ) = 19 ) LHS 6= RHS ) (3, ) does not lie on the line. FINDING WHERE GRPHS MEET Exmple 35 Use grphicl methods to find where the lines x + y =6 nd x y =6 meet. For x + y =6: when x =0, y =6 when y =0, x =6 For x y =6: when x =0, y =6 ) y = 6 when y =0, x =6 ) x =3 The grphs meet t (4, ). Check: 4+=6 X 4 =6 X x 0 6 y 6 0 x 0 3 y y x-y=6 (4,) x+y=6 x When determining whether two lines meet, there re three possile situtions which my occur. These re: Cse 1: Cse : Cse 3: The lines meet in single point of intersection. The lines re prllel nd never meet. There is no point of intersection. The lines re coincident. There re infinitely mny points of intersection.

29 BCKGROUND KNOWLEDGE 9 EXERCISE M 1 Use the distnce formul to find the distnce etween the following pirs of points: (1, 3) nd B(4, 5) O(0, 0) nd C(3, 5) c P(5, ) nd Q(1, 4) d S(0, 3) nd T( 1, 0). Find the midpoint of [B] for: (3, 6) nd B(1, 0) (5, ) nd B( 1, 4) c (7, 0) nd B(0, 3) d (5, ) nd B( 1, 3). 3 By finding y-step nd n x-step, determine the grdient of ech of the following lines: c d e f 4 Find the grdient of the line pssing through: (, 3) nd (4, 7) (3, ) nd (5, 8) c ( 1, ) nd ( 1, 5) d (4, 3) nd ( 1, 3) e (0, 0) nd ( 1, 4) f (3, 1) nd ( 1, ): 5 Clssify the following pirs of lines s prllel, perpendiculr, or neither. Give resons for your nswers. c d e f 6 Stte the grdient of the line which is perpendiculr to the line with grdient: c 4 d 1 3 e 5 f 0 7 Find, in grdient-intercept form, the eqution of the line through: (4, 1) with grdient (1, ) with grdient c (5, 0) with grdient 3 d ( 1, 7) with grdient 3 e (1, 5) with grdient 4 f (, 7) with grdient 1:

30 30 BCKGROUND KNOWLEDGE 8 Find, in generl form, the eqution of the line through: (, 1) with grdient 3 (1, 4) with grdient 3 c (4, 0) with grdient 1 3 d (0, 6) with grdient 4 e ( 1, 3) with grdient 3 f (4, ) with grdient 4 9 : 9 Find the equtions of the lines through: (0, 1) nd (3, ) (1, 4) nd (0, 1) c (, 1) nd ( 1, 4) d (0, ) nd (5, ) e (3, ) nd ( 1, 0) f ( 1, 1) nd (, 3) 10 Find the equtions of the lines through: (3, ) nd (5, ) (6, 7) nd (6, 11) c ( 3, 1) nd ( 3, 3) 11 Copy nd complete: Eqution of line Grdient x-intercept y-intercept x 3y =6 4x +5y =0 c y = x +5 d x =8 e y =5 f x + y =11 g 4x + y =8 h x 3y =1 1 Does (3, 4) lie on the line with eqution 3x y =1? Does (, 5) lie on the line with eqution 5x +3y = 5? c Does (6, 1 ) lie on the line 3x 8y =? If line hs eqution y = mx + c then the grdient of the line is m nd the y-intercept is c. 13 Use grphicl methods to find where the following lines meet: x +y =8 y =x 6 d x 3y =8 3x +y =1 Exmple 36 y = 3x 3 3x y = 1 e x +3y =10 x +6y =11 c 3x + y = 3 x 3y = 4 f 5x +3y =10 10x +6y =0 GRPHING PCKGE stright rod is to pss through the points (5, 3) nd B(1, 8). c Find where this rod meets the rod given y: i x =3 ii y =4 If we wish to refer to the points on the rod (B) etween nd B, how cn we indicte this? Does C(3, 0) lie on the rod (B)? B(1' 8) y=4 x=3 (5' 3)

31 BCKGROUND KNOWLEDGE 31 The line representing the rod hs grdient m = = 5 4 : So, its eqution is y 3 x 5 = 5 4 ) 4(y 3) = 5(x 5) ) 4y 1 = 5x +5 ) 5x +4y =37 i When x =3, 5(3) + 4y =37 ) y =37 ) 4y = ) y =5 1 ) they meet t (3, 5 1 ). We restrict the possile x-vlues to 1 6 x 6 5. c ii When y =4, 5x + 4(4) = 37 ) 5x +16=37 ) 5x =1 ) x = 1 5 ) they meet t ( 1 5, 4). If C(3, 0) lies on the line, its coordintes must stisfy the line s eqution. Now LHS = 5(3) + 4( 0) = =356= 37 ) C does not lie on the rod. 14 Find the eqution of the: horizontl line through (3, 4) verticl line with x-intercept 5 c verticl line through ( 1, 3) d horizontl line with y-intercept e x-xis f y-xis. 15 Find the eqution of the line: through ( 1, 4) which hs grdient 3 4 through P(, 5) nd Q(7, 0) c prllel to the line with eqution y =3x nd pssing through (0, 0) d prllel to the line with eqution x +3y =8 nd pssing through ( 1, 7) e perpendiculr to the line with eqution y = x + 5 nd pssing through (3, 1) f perpendiculr to the line with eqution 3x y =11 nd pssing through (, 5). 16 is the town hll on Scott Street nd D is Post Office on Kech venue. Digonl Rod intersects Scott Street t B nd Kech venue t C. Find the eqution of Kech venue. Find the eqution of Pecock Street. c Find the eqution of Digonl Rod. (Be creful!) d Plunkit Street lies on the mp reference line x =8. Where does Plunkit Street intersect Kech venue? (3, 17) E Pecock St Scott St Digonl Rd C (5, 11) B (7, 0) Kech v D (13, 1)

32 3 BCKGROUND KNOWLEDGE Exmple 37 Find the eqution of the tngent to the circle with centre (, 3) t the point ( 1, 5). C (, 3) P (-1, 5) The grdient of [CP] is 3 5 ( 1) = 3 = 3 ) the grdient of the tngent t P is 3. Since the tngent psses through ( 1, 5), y 5 its eqution is x 1 = 3 ) (y 5) = 3(x +1) ) y 10 = 3x +3 ) 3x y = 13 The tngent is perpendiculr to the rdius t the point of contct. 17 Find the eqution of the tngent to the circle with centre: (0, ) t the point ( 1, 5) (3, 1) t the point ( 1, 1) c (, ) t the point (5, ). Exmple 38 Mining towns re situted t B(1, 6) nd (5, ). Where should the rilwy siding S e locted so tht ore trucks from either or B would trvel equl distnces to rilwy line with eqution x =11? B (1, 6) (5, ) S x=11 rilwy line Suppose S hs the coordintes (11, ). Now BS = S ) p (11 1) +( 6) = p (11 5) +( ) ) 10 +( 6) =6 +( ) fsquring oth sidesg ) = ) 1 +4 =4 100 ) 8 = 96 ) =1 So, the rilwy siding should e locted t (11, 1).

33 BCKGROUND KNOWLEDGE (5, 5) nd B(7, 10) re houses nd y =8 is gs pipeline. Where should the one outlet from the pipeline e plced so tht it is the sme distnce from oth houses? B (7, 10) (5, 5) y=8 19 (CD) is wter pipeline. nd B re two towns. pumping sttion is to e locted on the pipeline to pump wter to nd B. Ech town is to py for their own service pipes nd they insist on equlity of costs. Where should C e locted to ensure equlity of costs? Wht is the totl length of service pipe required? c If the towns gree to py equl mounts, would it e cheper to instll the service pipeline from D to B to? (, 3) C D B (5, 4) Scle: 1 unit 1 km y=7 Exmple 39 Q (, 4) x= P x=6 N tunnel through the mountins connects town Q (, 4) to the port t P. P is on grid reference x=6 nd the distnce etween the town nd the port is 5km. ssuming the digrm is resonly ccurte, find the horizontl grid reference of the port. Scle: ech grid unit is 1 km. Suppose P is t (6, ). Now PQ =5 ) p (6 ) +( 4) =5 ) p 16 + ( 4) =5 ) 16 + ( 4) =5 ) ( 4) =9 ) 4= 3 ) =4 3=7or 1 From the digrm, P is further north thn Q, nd so >4 So, P is t (6, 7) nd the horizontl grid reference is y =7.

34 34 BCKGROUND KNOWLEDGE 0 y=8 Clifton Json s girlfriend lives in house on Clifton Highwy Highwy which hs eqution y = 8. The distnce s the crow flies from Json s house to his girlfriend s house is 11:73 km. If Json lives t (4, 1), wht re the coordintes of his girlfriend s house? Scle: 1 unit 1 km. J (4, 1) 1 circle hs centre (, ) nd rdius r units. P(x, y) moves on the circle. Show tht (x ) +(y ) = r. Find the eqution of the circle with: i centre (4, 3) nd rdius 5 units ii centre ( 1, 5) nd rdius units iii centre (0, 0) nd rdius 10 units iv ends of dimeter ( 1, 5) nd (3, 1). Find the centre nd rdius of the circle: (x 1) +(y 3) =4 x +(y +) =16 c x + y =7 3 Consider the circle with eqution (x ) +(y +3) =0: Stte the circle s centre nd rdius. Show tht (4, 1) lies on the circle. c Find the eqution of the tngent to the circle t the point (4, 1). 4 The perpendiculr isector of chord of circle psses through the centre of the circle. Find the centre of circle pssing through points P(5, 7), Q(7, 1) nd R( 1, 5) y finding the perpendiculr isectors of [PQ] nd [QR] nd solving them simultneously. N RIGHT NGLED TRINGLE TRIGONOMETRY LBELLING RIGHT NGLED TRINGLES C The hypotenuse (HYP) is the side which is opposite the right ngle. It is the longest side of the tringle. HYP OPP For the ngle mrked µ: ² [BC] is the side opposite (OPP) ngle µ ² [B] is the side djcent (DJ) ngle µ. For the ngle mrked Á: ² [B] is the side opposite (OPP) ngle Á ² [BC] is the side djcent (DJ) ngle Á. DJ HYP OPP B C DJ B

35 THE THREE BSIC TRIGONOMETRIC RTIOS BCKGROUND KNOWLEDGE 35 sin µ = OPP HYP, DJ cos µ = HYP, OPP tn µ = DJ HYP DJ OPP sin µ, cos µ nd tn µ re revitions for sine µ, cosine µ nd tngent µ: The three formule ove re clled the trigonometric rtios nd re the tools we use for finding side lengths nd ngles of right ngled tringles. However, efore doing this we will clculte the trigonometric rtios in right ngled tringles where we know two of the sides. Exmple 40 cm Find, without using clcultor, sin µ, cos µ nd tn µ: 3 cm So, HYP x cm OPP sin µ = OPP HYP = 3 p 13, DJ If the hypotenuse is x cm long x = +3 ) x =13 ) x = p 13 cos µ = DJ HYP = p 13, fpythgorsg fs x>0g tn µ = OPP DJ = 3. Exmple 41 If µ is n cute ngle nd clcultor. sin µ = 1 3, find cos µ nd tn µ without using We drw right ngled tringle nd mrk on ngle µ so tht OPP =1unit nd HYP =3units. Now x +1 =3 fpythgorsg ) x +1=9 ) x =8 ) x = p 8 fs x>0g ) cos µ = DJ HYP = p 8 3 nd tn µ = OPP DJ = 1 p 8 : 3 x 1

36 36 BCKGROUND KNOWLEDGE EXERCISE N.1 1 For the following tringles, find the length of the third side nd hence find sin µ, cos µ nd tn µ: c d e f If µ is n cute ngle nd cos µ = 1, find sin µ nd tn µ. If is n cute ngle nd sin = 3, find cos nd tn. c If is n cute ngle nd tn = 4 3, find sin nd cos. 3 For the tringle given, write down expressions for sin µ, cos µ nd tn µ. c sin µ Write in terms of, nd c nd cos µ sin µ hence show tht = tn µ. cos µ 4 The ngle mrked 90 o µ is the complement of µ. Find: i sin µ ii cos µ iii sin(90 o µ) iv cos(90 o µ) Use your results of to complete the following sttements: i The sine of n ngle is the cosine of its... ii The cosine of n ngle is the sine of its c 5 1 Find the length of the remining side. 45 Find sin 45 o, cos 45 o, tn 45 o using the figure. 1 c Use your clcultor to check your nswers. 45

37 BCKGROUND KNOWLEDGE 37 6 Tringle BC is equilterl. [N] is the ltitude corresponding to side [BC]. Stte the mesures of BN nd BN. Find the lengths of [BN] nd [N]. c Without using clcultor, find: i sin 60 o, cos 60 o nd tn 60 o ii sin 30 o, cos 30 o nd tn 30 o. B N C COMMON TRIGONOMETRIC RTIOS We cn summrise the rtios for specil ngles in tle form. Try to lern them. µ 0 o 30 o 45 o 60 o 90 o sin µ 0 1 cos µ 1 p 3 tn µ 0 1 p3 1 1p p 3 1 1p 1 0 p 3 undefined FINDING SIDES ND NGLES Before commencing clcultions, check tht the MODE on your clcultor is set on degrees. In this chpter ll ngle mesure is in degrees. In right ngled tringle, if we wish to find the length of side, we first need to know one ngle nd one other side. Exmple 4 Find, correct to 3 significnt figures, the vlue of x in: 1 cm 54 x cm 56 x 3. 4 For the 54 o ngle, HYP =1, DJ = x. 1 cm 54 x cm ) cos 54 o = x 1 ) 1 cos 54 o = x ) x ¼ 7:05 For the 3:4 o ngle, OPP =1, DJ = x: x 3. 4 So, tn 3:4 o = 56 x 56 ) x = 56 tn 3:4 o ¼ 403:4 ) x ¼ 403

38 38 BCKGROUND KNOWLEDGE In right ngled tringle, if we wish to find the size of n cute ngle we need to know the lengths of two sides. We then need to find the pproprite inverse trigonometric rtio: ² If sin µ = ² If cos µ = ² If tn µ = ³ then µ = sin 1 ³ then µ = cos 1 ³ then µ = tn 1 which reds the ngle with sine of : which reds the ngle with cosine of : which reds the ngle with tngent of. n lterntive nottion for the three inverse trigonometric functions is: ² rcsin µ for sin 1 µ ² rccos µ for cos 1 µ ² rctn µ for tn 1 µ Find help using your clcultor to find inverse trigonometric rtios, consult the grphics clcultor instructions chpter. Exmple 43 Find in degrees, correct to 3 significnt figures: 13 cm 11 cm 11 cm 13 cm For ngle, OPP =11, HYP =13: ) sin = ) = sin 1 ( ) ) ¼ 57:8 o EXERCISE N. 1 Find, correct to 3 significnt figures, the vlue of the unknown in ech of the following: c 35 x cm m 30 cm x m 87. cm 60 x cm d e f x m 65 3 m x m 413 m x m 369 m 53. 9

39 c Find ll unknown side lengths nd ngles of the following tringles: C BCKGROUND KNOWLEDGE 39 Use your clcultor to find the cute ngle µ, to 3 significnt figures, if: sin µ =0:9364 cos µ =0:381 c tn µ =1:731 d cos µ = 7 e sin µ = 1 3 f tn µ = 14 3 g sin µ = p 3 11 h cos µ = 5 p 37 3 Find, correct to 3 significnt figures, the mesure of the unknown ngle in ech of the following: P 51 Q m 10 m B 5 m 5 Find ll unknown sides nd ngles in: 4 x R 10 x y ISOSCELES TRINGLES To use trigonometry with isosceles tringles we invrily drw the perpendiculr from the pex to the se. This ltitude isects the se. Exmple 44 Find the unknowns in the following digrms: 10 cm x cm 5. m 83. m 67 5 cm 67 x cm In the shded right ngled tringle, cos 67 o = 5 x 5 cm 67 ) x = 5 cos 67 o ¼ 1:8

40 40 BCKGROUND KNOWLEDGE 6. m 6. m 83. m In the shded right ngled tringle, ³ sin = :6 8:3 ) = sin 1 µ :6 8:3 ) = sin 1 µ :6 8:3 ¼ 36:5 o CHORDS ND TNGENTS Right ngled tringles occur in chord nd tngent prolems. centre rdius point of contct tngent tngent chord centre tngent Exmple 45 chord of circle sutends n ngle of 11 o t its centre. Find the length of the chord if the rdius of the circle is 6:5 cm. x cm cm We complete n isosceles tringle nd drw the line from the pex to the se. For the 56 o ngle, HYP =6:5, OPP = x ) sin 56 o = x 6:5 ) 6:5 sin 56 o = x ) x ¼ 5:389 ) x ¼ 10:78 ) the chord is out 10:8 cm long. EXERCISE N.3 1 Find, correct to 4 significnt figures, the unknowns in the following: c 6 cm 486. cm 61 4 cm x cm 3 cm 694. cm

41 Find the vlue of the unknown in: c BCKGROUND KNOWLEDGE 41 5 cm 8 cm r cm 14 0 cm 6 cm 10 cm 3 chord of circle sutends n ngle of 89 o t its centre. Find the length of the chord given tht the circle s dimeter is 11:4 cm. 4 chord of circle is 13: cm long nd the circle s rdius is 9:4 cm. Find the ngle sutended y the chord t the centre of the circle. 5 Point P is 10 cm from the centre of circle of rdius 4 cm. Tngents re drwn from P to the circle. Find the ngle etween the tngents. OTHER FIGURES Right ngled tringles cn lso e found in other geometric figures such s rectngles, rhomi, nd trpezi. C P digonl rectngle rhomus trpezium or trpezoid Exmple 46 rhomus hs digonls of length 10 cm nd 6 cm respectively. Find the smller ngle of the rhomus. 5 cm B 3 cm M D C The digonls isect ech other t right ngles, so M =5cm nd BM =3cm. In BM, µ will e the smllest ngle s it is opposite the shortest side. tn µ = 3 5 ) µ = tn 1 ( 3 5 ) ) µ ¼ 30:964 o The required ngle is µ s the digonls isect the ngles t ech vertex. So, the ngle is out 61:9 o :

42 4 BCKGROUND KNOWLEDGE EXERCISE N.4 1 rectngle is 9: my3:8 m. Wht ngle does its digonl mke with its longer side? The digonl nd the longer side of rectngle mke n ngle of 43: o. If the longer side is 1:6 cm, find the length of the shorter side. 3 rhomus hs digonls of length 1 cm nd 7 cm respectively. Find the lrger ngle of the rhomus. 4 The smller ngle of rhomus mesures 1:8 o nd the shorter digonl hs length 13:8 cm. Find the lengths of the sides of the rhomus. Exmple 47 Find x given: 10 cm x cm B 10 cm x cm y cm y cm D M N C We drw perpendiculrs [M] nd [BN] to [DC], creting right ngled tringles nd the rectngle BNM. In DM, sin 65 o = y 10 ) y = 10 sin 65 o In BCN, sin 48 o = y 10 sin 65o = x x 10 sin 65o ) x = sin 48 o ¼ 1: 5 Find the vlue of x in: Find the unknown ngle in: 3 m x m stormwter drin is to hve the shpe illustrted. Determine the ngle the left hnd side mkes with the ottom of the drin. 5 m 100 m 3 m

43 PROBLEM SOLVING USING TRIGONOMETRY BCKGROUND KNOWLEDGE 43 Trigonometry is very useful rnch of mthemtics. Heights nd distnces which re very difficult or even impossile to mesure cn often e found using trigonometry. ngle of elevtion ngle of depression horizontl ngle of elevtion Exmple 48 Find the height of tree which csts shdow of 1:4 m when the sun mkes n ngle of 5 o to the horizon m h m Let h m e the tree s height. For the 5 o ngle, OPP = h, DJ =1:4 ) tn 5 o = h 1:4 ) 1:4 tn 5 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 9:3 m when the sun mkes n ngle of 63 o to the horizontl. hill is inclined t 18 o to the horizontl. It runs down to the ech so its se is t se level. If I wlk 1. km up the hill, wht is my height ove se level? If I m 500 metres ove se level, how fr hve I wlked up the hill? 3 B 10 m C surveyor stnding t notices two posts B nd C on the opposite side of cnl. The posts 37 re 10 m prt. If the ngle of sight etween cnl the posts is 37 o, how wide is the cnl? 4 trin must clim constnt grdient of 5:5 m for every 00 m of trck. Find the ngle of incline.

44 44 BCKGROUND KNOWLEDGE 5 Find the ngle of elevtion to the top of 56 m high uilding from point which is 113 m from its se. Wht is the ngle of depression from the top of the uilding to? 6 The ngle of depression from the top of 10 m high verticl cliff to ot B is 16 o. Find how fr the ot is from the se of the cliff. 113 m 56 m 10 m se B 7 Srh mesures the ngle of elevtion to the top of tree s 3:6 o from point which is 50 m from its se. Her eye level, from which the ngle mesurement is tken, is 1:5 m ove the ground. ssuming 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 uilding 10 metres high to e 3 o. She wlks towrds the uilding until the ngle of elevtion is 45 o. How fr did she wlk? 9 circulr trck of rdius r m is nked t n ngle of µ to the horizontl. The idel speed for the end is given y the formul s = p gr tn µ where g =9:8 ms. Wht is the idel speed for vehicle trvelling on circulr trck of rdius 100 m which is nked t n ngle of 15 o? t wht ngle should trck of rdius 00 m e nked if it is designed for vehicle trvelling t 0 ms 1? Exmple 49 uilder 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 13 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 : DJ =7:5, HYP =8:7 ) cos µ = 7:5 8:7 µ 7:5 ) µ = cos 1 8:7 ) µ ¼ 30:450 o ) the pitch is pproximtely 30 1 o.

45 BCKGROUND KNOWLEDGE Find µ, the pitch of the roof. 83. m 06. m 15 m 06. m 11 If the pitch of the given roof is 3 o, find the B length of the timer em [B] m 16 m 08. m 1 n open right-circulr cone hs verticl ngle mesuring 40 o nd se rdius of 30 cm. Find the cpcity of the cone in litres refrigertor is tipped ginst verticl wll so it P cn e serviced. It mkes n ngle of 70 o with the 1 m horizontl floor. How high is point ove the floor? m 70 Q N 14 From n oserver O, the ngles of elevtion to the ottom nd the top of flgpole re 36 o nd 38 o respectively. Find the height of the flgpole. C B 36 O m 15 The ngle of depression from the top of 150 m high cliff to ot t se is 7 o. How much closer to the cliff must the ot move for the ngle of depression to ecome 19 o? 16 helicopter flies horizontlly t 100 km h 1. n oserver notices tht it tkes 0 seconds for the helicopter to fly from directly overhed to eing t n ngle of elevtion of 60 o. Find the height of the helicopter ove the ground. 17 [C] is stright shore line nd B is ot out t se. Find the shortest distnce from the ot to the shore if nd C re 5 km prt. 5 km B C

46 46 BCKGROUND KNOWLEDGE 18 regulr pentgonl grden plot is to e constructed with sides of length 0 m. d m Find the width of lnd d m required for the plot. 0 m

47 NSWERS 47 EXERCISE 1 p 15 3 c 4 d 1 e 4 p p p f 6 g h 6 5 p p c p 5 d 8 p 5 e p 5 f 9 p 3 g 3 p 6 h 3 p 3 p p 3 c p 5 d 4 p e 3 p 3 f 3 p 5 g 4 p 3 h 3 p 6 i 5 p j 4 p 5 k 4 p 6 l 6 p 3 4 p 3 8 p c 5 p 6 d 10 p 3 e 3 p 3 f p p 5 p 3 c 7 p f 3 p 6 g 4 p 3 h 5p 7 7 EXERCISE B d p 5 e 5 p i p 7 j p 6 1 :59 10 : c : d : e : f 4: g 4: h 4:07 10 i 4: j 4: k 4: : m mm c mm d 1: o C e c 100 d e f 86 g h :004 0:05 c 0:001 d 0: e 0: f 0:86 g 0: h 0: : m c light yers d 0: mm 6 1: : c 5: d 1:36 10 e : f 1: : km 9: km c 9: km 8 1: m : m c 9: m EXERCISE C 1 The set of ll rel x such tht x is greter thn 5. The set of ll integers x such tht x is less thn or equl to 3. c The set of ll y such tht y lies etween 0 nd 6. d The set of ll integers x such tht x is greter thn or equl to, ut less thn or equl to 4. x is, 3 or 4. e The set of ll t such tht t lies etween 1 nd 5. f The set of ll n such tht n is less thn or greter thn or equl to 6. fx j x>g fx j 1 <x6 5g c fx j x 6 or x > 3g d fx j 16 x 6 3, x Z g e fx j 0 6 x 6 5, x Z g f fx j x<0g 3 c x x -5 4 x d e EXERCISE D 1 10x 10 9x c 5x +5y d 8 8x e 1 f cnnot e simplified x x c 4 3 d 3x 3 16x +11x x 3 3 EXERCISE E c 4x d x =10 x>6 c x = 4 5 d x =51 e x< 10 f x =14 g x 6 9 h x =18 i x = 3 x =5, y = x = 3, y = 8 c x =, y =5 3 d x = 45 11, y = 18 e no solution f x =66, y = EXERCISE F c 16 d 18 e f 3 c 6 d 6 e 5 f 1 g 1 h 5 i 4 j 4 k l 3 x = 3 no solution c x =0 d x =4or e x = 1 or 7 f no solution g x =1or 1 3 h x =0or 3 i x = or 14 5 EXERCISE G x 1 x +5x +3 3x +10x +8 c 10x + x d 3x + x 10 e 6x +17x +14 f 6x 13x +5 g 15x +11x 1 h 15x 11x + i x 17x +1 j 4x 16x +15 k x 3x l 4x x +6 x 36 x 64 c 4x 1 d 9x 4 e 16x 5 f 5x 9 g 9 x h 49 x i 49 4x j x k x 5 l 4x 3 3 x +10x +5 x +14x +49 c x 4x +4 d x 1x +36 e x +6x +9 f x +10x +5 g x x + 11 h x 0x i 4x +8x +49 j 9x +1x +4 k 4x 0x +5 l 9x 4x y =x +10x +1 y =3x 6x +7 c y = x +6x +7 d y = x 4x 15 e y =4x 4x +0 f y = 1 x 4x 14 g y = 5x +35x 30 h y = 1 x +x 4 i y = 5 x +0x 40 8 x 5 x +1x +19 3x +3x 16 c x +6x 6 d x x+5 e x 16x+33 f 3x+4 g 7x+8 h 7x +18x +1 i x +19x 3 j 7x 16x +

48 48 NSWERS EXERCISE H 1 3x(x+3) x(x+7) c x(x 5) d 3x(x 5) e (3x 5)(3x+5) f (4x+1)(4x 1) g (x )(x+) h 3(x+ p 3)(x p 3) i 4(x+ p 5)(x p 5) j (x 4) k (x 5) l (x ) m (4x +5) n (3x +) o (x 11) (x + 8)(x +1) (x + 4)(x +3) c (x 9)(x +) d (x + 7)(x 3) e (x 6)(x 3) f (x + 3)(x ) g (x )(x +1) h 3(x 11)(x 3) i (x +1) j (x+5)(x ) k (x 8)(x+3) l (x 6)(x 1) m 3(x 1) n (x +1) o 5(x 4)(x +) 3 (x 3)(x +4) (3x + 1)(x ) c (7x )(x 1) d (3x )(x+1) e (x 3)(x+1) f (5x 3)(x+1) g (x+1)(x 6) h (3x+7)(x 4) i (4x+3)(x 1) j (5x+3)(x 3) k (3x 1)(x+8) l (3x+)(x+1) m (x + 3)(x 1) n (6x + 1)(x 3) o 3(x + 7)(x ) p (3x )(3x 5) q (4x 9)(x+3) r (4x+3)(3x+1) s (6x+1)(x+3) t (5x 4)(3x ) u (7x + 5)(x 3) 4 (x + 4)(x +1) ( x)(5 3x) c 3(x + )(x +7) d 4(x + 5)(x + 11) e x(x +3) f 5(x +3) g (x )(3x + 6) h (x 1)(x 1) 5 (x + 7)(x 1) (x + 1)(3 x) c 1(x +1) d 4x(x +4) e (3x + )(x +4) f h(x + h) g 1(x +1) h 5(3x 4)(x 4) i 3(x + 9)(5x +9) 10 v = v = EXERCISE J x 5 d 14 x 4 x +5 x + 5x + d x +1 3 c µ sc 1 m 0 m = c p m m0 m p 8 3 c c : ms 1 7x 6x 5 (x 5)(x ) 8x +6x +8 x 16 x +3 x e 8x x x 4 x +3 e x +1 EXERCISE K.1 1 Hint: Consider s EC, BDC x +4 c 4x +17 f 1 1 3x c x 1 x +3 f 1 x 1 (x )(x 3) d 3x +3x +13 (x 3)(x +1) Hint: Let M e on [B] so tht [PM]? [B]. Let N e on [C] so tht [PN]? [C]. Join [PM], [PN] nd consider the two tringles formed. EXERCISE K. 1 x =:4 x =:8 c x = d x =6 3 e x =7 f x =7: 1:35 m tll EXERCISE I 1 x = x = d x = t c g x = d 3y 7 0 y e x = 5 h x = c y c x = d 1 3y f x = i x = y c m EXERCISE L.1 1 x =0:663 x =4:34 c x =:3 4:54 m 4:17 m 3 37 m 4 87:1 km N 74 km 46 km z = c 3 = F m q 4 r = ¼ d x = 3 q n D z = d r = C ¼ c z = d 3 x = 5p N c d = V lh d z = p d K = c r = 3 q 3V 4¼ 5 = d n l =5T c = p + c d l = gt 4¼ e = P f h = ¼r ¼r g r = E I R 6 = d k q S 8 t = 9 v = uf u f h q = p B 1:9 7 r = 3 q 3V 4¼ 15:81 s i 9:5 cm ii 10:9 cm :1 cm EXERCISE L. 1 50:3 m 17:3 cm 3 8:60 m 4 53:4 m 5 44:4 km EXERCISE M 1 p 13 units p 34 units c p 0 units d p 10 units (, 3) (, 1) c (3 1, 1 1 ) d (, 1 ) c 1 d 0 e 4 f undefined 4 3 c undefined d 0 e 4 f 5 prllel, grdients 5 prllel, grdients 1 7 c perpendiculr, grdients 1, d neither, grdients 4, 1 3 e neither, grdients 7, 1 5 f perpendiculr, grdients, c 1 4 d 3 e 1 5 f undefined 7 y =x 7 y = x +4 c y =3x 15 d y = 3x +4 e y = 4x +9 f y = x x y =4 3x +y =11 c x 3y =4 d 4x + y =6 e 3x y =0 f 4x +9y = 1 4

49 NSWERS 49 9 x 3y = 3 5x y =1 c x y =3 d 4x 5y =10 e x y = 1 f x +3y = 5 10 y = x =6 c x = 3 11 Eqution of line Grdient x-int. y-int. x 3y = x +5y = c y = x +5 5 d x =8 undef. 8 no y-int. e y =5 0 no x-int. 5 f x + y = g 4x + y = h x 3y = yes no c yes 13 (4, ) (, 3) c ( 3, 6) d (4, 0) e prllel lines do not meet f coincident lines 14 y = 4 x =5 c x = 1 d y = e y =0 f x =0 15 3x 4y = 19 x y =7 c y =3x d x +3y =19 e x y =5 f x +3y =13 16 x 8y = 83 8x + y =41 c 9x y =3 for 5 6 x 6 7 d (8, ) 17 x 3y = 16 x y = 3 c x =5 18 (4 3 4, 8) 19 ( 1, 7) 8:03 km c yes (6:16 km) 3 0 (13:41, 8) or ( 5:41, 8) 1 Hint: Use the distnce formul to find the distnce from the centre of the circle to point P. i (x 4) +(y 3) =5 ii (x+1) +(y 5) =4 iii x + y = 100 iv (x 1) +(y 3) =8 centre (1, 3), rdius units centre (0, ), rdius 4 units c centre (0, 0), rdius p 7 units 3 centre (, 3), rdius p 0 units Hint: Sustitute (4, 1) into eqution of circle. c x +y =6 4 (3, 3) 6 BN =60 o,b N =30 o BN =1,N= p 3 c i sin 60 o p = 3, cos 60o = 1, tn 60o = p 3 ii sin 30 o = 1 p, cos 30o = 3, tn 30o = p 1 3 EXERCISE N. 1 x =17: x = 57 c x =15:1 d x =7:10 e x = 554 f x = 457 µ =69:5 o µ =76: o c µ =60:0 o d µ =73:4 o e µ =19:5 o f µ =77:9 o g µ =9:06 o h µ =34:7 o 3 µ =56:4 =4:76 c =48: 4 C =6:40 m, =38:7 o, C =51:3 o R =39 o, PQ =8:10 m, PR =1:9 m 5 x =:65, µ =37:1 x =6:16, µ =50:3, y =13:0 EXERCISE N.3 1 x =4:13 =75:5 c =41:0 µ = 36:9 r = 11:3 c = 61:9 3 7:99 cm 4 89: o 5 47: o EXERCISE N.4 1 :4 o 11:8 cm o 4 36:5 cm 5 x =3:44 =51: o EXERCISE N :3 m 371 m 1:6 km m 4 1:58 o 5 ngle of elevn. =6:4 o, ngle of depn. =6:4 o 6 418:5 m m 8 7:0 m 9 16: ms 1 11:5 o 10 µ =1:6 11 9:56 m 1 77:7 litres 13 : m 14 10:95 m m m 17 3:17 km 18 30:8 m EXERCISE N.1 1 3, sin µ = 3 5, cos µ = 4 5, tn µ = 3 4 1, sin µ = 5 1, cos µ = 13 13, tn µ = 5 1 c p 11, sin µ = 5 p 6, cos µ = 11 6, tn µ = p 5 11 d p 5, sin µ = p, cos µ = p 1, tn µ = 5 5 e p 5, sin µ = p 4, cos µ = p 6, tn µ = p p f 15, sin µ = 7 8, cos µ = 15 8, tn µ = 7 sin µ = p 3, tn µ = p 3 cos = c sin = 4 5, cos = 3 5 p 15 p 5 3, tn = p 5 3 sin µ = c, cos µ = c, tn µ = c c 4 i ii iii iv i complement ii complement 5 p sin 45 o = 1 p, cos 45 o = 1 p, tn 45 o =1