CHAPTER 1 REVISIONARY MATHEMATICS

Transcription

1 CHAPTER REVISIONARY MATHEMATICS EXERCISE, Page. Convert the following angles to degrees correct to decimal places (where necessary): (a) 0.6 rad (b) 0.8 rad (c) rad (d).9 rad (a) 0.6 rad 0.6 rad 80 πrad (b) 0.8 rad 0.8 rad 80 πrad (c) rad rad 80 πrad.9 (d).9 rad.9 rad 80 πrad 80. Convert the following angles to radians correct to decimal places: (a) (b) 90 (c) 0 (d) 80 π (a) rad 80 π rad or 0.78 rad π (b) rad 80 rad π rad or.708 rad π (c) 0 0 rad 80 rad π rad or.09 rad π (d) rad π rad or.6 rad 80

2 EXERCISE, Page. Find the cosine, sine and tangent of the following angles, where appropriate each correct to decimal places: (a) 60 (b) 90 (c) 0 (d) 80 (e) 0 (f) 70 (g) 0 (h) - 0 (i) 0 (j) 0 (k) 0 (a) cos sin tan 60.7 (b) cos 90 0 sin 90 tan 90 (c) cos sin tan (d) cos 80 - sin 80 0 tan 80 0 (e) cos sin tan (f) cos 70 0 sin 70 - tan 70 - (g) cos sin tan (h) cos sin tan (i) cos sin tan 0.7 (j) cos 0 0 sin 0 tan 0 (k) cos sin tan

3 EXERCISE, Page 6. If ab. m and bc. m, determine angle θ. It is convenient to use the expression for tan θ, since sides ab and bc are given. Hence, tan θ bc ab. from which, θ tan (0.787 ).º. If ab. m and ac.0 m, determine angle θ. It is convenient to use the expression for cos θ, since sides ab and ac are given. Hence, cos θ ab ac.0 from which, θ cos (0.60) 6.6º. If bc. m and ac 6. m, determine angle θ. It is convenient to use the expression for sin θ, since sides bc and ac are given. Hence, sin θ bc ac 6. from which, θ sin (0.87) 8.97º

4 . If ab.7 cm and bc. cm, determine the length ac From Pythagoras, ac ab + bc from which, ac m. If ab. m and ac 6. m, determine length bc. From Pythagoras, ac ab + bc from which, bc ac - ab from which, ac.6.6 m

5 EXERCISE, Page 7. If b 6 m, c m and B 00, determine angles A and C and length a. Using the sine rule, b c i.e. sin B sin C 6 sin00 sin C from which, sin C and C sin sin (0.66).0 Angle, A Using the sine rule again gives: a b i.e. a sin A sin B bsin A 6 sin 8.96 sin B sin00.8 m. If a m, c m and B 67, determine length b and angles A and C. From the cosine rule, b a + c accos B + cos ()()cos Hence, length, b m

6 Using the sine rule: b c i.e. sin B sin C.0 sin 67 sin C from which,.0 sin C sin 67 and sin C sin and C sin (0.969) 7. Since A + B + C 80, then A 80 - B - C If a m, b 8 m and c 6 m, determine angle A. Applying the cosine rule: a b + c - bc cos A from which, bc cos A b + c - a and cos A b + c a bc (8)(6) 0.87 Hence, A cos If a 0.0 cm, b 8.0 cm and c 7.0 cm, determine angles A, B and C. Applying the cosine rule: a b + c - bc cos A from which, bc cos A b + c - a 6

7 and cos A b + c a bc (8.0)(7.0) Hence, A cos Applying the sine rule: from which, sin 8. sin B sin B 8.0sin Hence, B sin and C PR represents the inclined jib of a crane and is 0.0 m long. PQ is.0 m long. Determine the inclination of the jib to the vertical (i.e. angle P) and the length of tie QR. Applying the sine rule: PR sin0 PQ sin R from which, sin R PQsin0 PR (.0)sin Hence, R sin (or 9.7, which is not possible) P , which is the inclination of the jib to the vertical. Applying the sine rule: 0.0 QR sin0 sin 9.7 from which, length of tie, QR 0.0sin m sin0 7

8 EXERCISE, Page 8. Evaluate A given A ( + + ) A ( + + ) (7) 7. Evaluate A given A [( + ) (6 7] A [( + ) (6 7] [() (- )] [ + ] [8] 8 7. Expand the brackets: (x y + ) (x y + ) (x) (y) + () x y + 6. Expand the brackets: (x y) + (y z) (z x) (x y) + (y z) (z x) x y + y z z + x x + x y + y z z 7x y z. Expand the brackets: x + [y (x + y)] x + [y (x + y)] x + [y x y] x + [- x] x - x 0 6. Expand the brackets: a [{(a b) (a + b)} + b] a [{(a b) (a + b)} + b] a [{a b a b} + b] 8

9 a [{a 7b} + b] a [6a b + b] a [6a b] a 6a + b - a + b or b a 7. Expand the brackets: ab[c + d e(f g + h{i + j})] ab[c + d e(f g + h{i + j})] ab[c + d e(f g + hi + hj)] ab[c + d ef + eg ehi ehj] abc + abd abef + abeg abehi - abehj 9

10 EXERCISE 6, Page 0. Evaluate + A common denominator can be obtained by multiplying the two denominators together, i.e. the common denominator is The two fractions can now be made equivalent, i.e. and so that they can be easily added together, as follows: Evaluate + A common denominator can be obtained by multiplying the two denominators together, i.e. the common denominator is 0 The two fractions can now be made equivalent, i.e. and 0 so that they can be easily added together, as follows: Evaluate Use a calculator to evaluate 8 8 by cancelling

11 . Use a calculator to evaluate Evaluate + as a decimal, correct to decimal places correct to decimal places 7. Evaluate 8 8 as a mixed number Evaluate 7 as a decimal, correct to decimal places correct to decimal places Determine + as a single fraction. x y y and x xy x y xy Hence, + y x y xy + x xy y + x xy or x + y xy

12 EXERCISE 7, Page. Express 0.07 as a percentage %.7%. Express 0.7 as a percentage % 7.%. Express 0% as a decimal number 0% Express 6 as a percentage 00 00% % 68.7% Express as a percentage, correct to decimal places 00 00% % by calculator 8.6% correct to decimal places 6. Place the following in order of size, the smallest first, expressing each as percentages, correct to decimal place: (a) (b) 9 7 (c) 9 (d) 6 00 (a) 00% % 7.% (b) % 900 %.9% 7 7 7

13 (c) 00% 00 %.6% (d) % 600 %.6% Placing them in order of size, the smallest first, gives: (b), (d), (c) and (a) 7. Express 6% as a fraction in its simplest form 6% 6 00 and by dividing the numerator and denominator by gives: 6% Calculate.6% of 0 kg.6% of 0 kg.6 0 kg.8 kg Determine 6% of 7 m 6% of 7 m m 9.7 m 0. Calculate correct to significant figures: (a) 8% of 78 tonnes (b) 7% of 8. grams (c) 7% of. seconds (a) 8% of 78 t 8 78 t 96. t 00 (b) 7% of 8. g g 8.67 g (c) 7% of. s s 0.7 s. Express: (a) 0 kg as a percentage of t (b) 7 s as a percentage of min (c). cm as a percentage of. m It is essential when expressing one quantity as a percentage of another that both quantities are in the same units.

14 (a) tonne 000 kg, hence 0 kg as a percentage of t 0 00% 000 % (b) minutes s, hence 7 s as a percentage of minutes 7 00% % (c). m cm, hence. cm as a percentage of. m. 00% 0.6 %. A computer is advertised on the internet at 0, exclusive of VAT. If VAT is payable at 0%, what is the total cost of the computer? VAT 0% of Total cost of computer Express mm as a percentage of 867 mm, correct to decimal places. mm as a percentage of 867 mm 00% %. When signing a new contract, a Premiership footballer s pay increases from,00 to,00 per week. Calculate the percentage pay increase, correct significant figures. Percentage change is given by: new value original value 00% original value i.e. % increase % 00% 8.7% A metal rod.80 m long is heated and its length expands by 8.6 mm. Calculate the percentage increase in length. % increase % 00%.7%

15 EXERCISE 8, Page. Evaluate ++ 7 by law of indices 8. Evaluate in index form by law of indices. Evaluate by law of indices 6. Evaluate by law of indices by law of indices 9. Evaluate by law of indices 6. Evaluate 7 6

16 by laws and of indices Evaluate by laws and of indices 8. Evaluate by law of indices 0 9. Evaluate by law of indices Evaluate by laws and of indices 7. Evaluate (7 ) in index form (7 ) by law of indices. Evaluate ( ) ( )

17 . Evaluate 7 in index form by laws and of indices. Evaluate (9 ) ( 7) in index form ( ) ( ) ( ) ( ) (9 ) 8 ( 7) 8 by laws, and of indices. Evaluate (6 ) ( 8) ( ) ( ) ( ) ( ) 6 (6 ) ( 8) by laws, and of indices 6. Evaluate ( ) + by law of indices 7. Evaluate + by laws, and of indices 8. Evaluate

18 by laws and of indices 9. Simplify, giving the answer as a power: z z z z z + 8 z by law of indices 0. Simplify, giving the answer as a power: a a a a a a a a by law of indices. Simplify, giving the answer as a power: n 8 n 8 8 n n n n by law of indices. Simplify, giving the answer as a power: b b b b b + b by law of indices. Simplify, giving the answer as a power: b b b b b b b b b or by laws and of indices. Simplify, giving the answer as a power: c c c + 8 c c c c 8 c c c c c c c c by laws and of indices. Simplify, giving the answer as a power: m m m m 6 8

19 6 + 6 m m m m m + 7 m m m m 7 m by laws and of indices 6. Simplify, giving the answer as a power: (x )(x) x 6 + (x )(x) x x x x x x x or by laws, and of indices 7. Simplify, giving the answer as a power: ( x ) ( ) x x x by law of indices 8. Simplify, giving the answer as a power: ( ) y ( y ) y y 6 y or 6 by laws and of indices 9. Simplify, giving the answer as a power: ( ) t t + ( t t ) ( t ) ( t ) 8 t by laws and of indices 7 0. Simplify, giving the answer as a power: ( ) c ( c ) c c by law of indices 7 7. Simplify, giving the answer as a power: a a 9

20 a ( a ) ( a ) a a 9 a or 9 by laws and of indices. Simplify, giving the answer as a power: b b ( b ) b b or by laws and of indices. Simplify, giving the answer as a power: b b 7 b b 7 7 ( b ) ( b ) 0 b by laws and of indices. Simplify, giving the answer as a power: ( s ) 9 s ( s ) s or 9 s by laws and of indices. Simplify, giving the answer as a power: p qr p q r pqr p qr p q r pqr p q r p q r 6 7 pqr 6. Simplify, giving the answer as a power: xyz x yz xyz x yz x y z y x z x yz or 0

21 EXERCISE 9, Page. If apples and bananas cost. and apples and 6 bananas cost., determine how much an apple and a banana each cost. Let an apple A and a banana B, then: From equation (), A + B () A + 6B () A B and A B 9 0.6B () From equation (), A 6B and A 6B 60..B () Equating () and () gives: 9 0.6B 60..B i.e..b 0.6B and 0.9B. and B. 0.9 Substituting in () gives: A 9 0.6() 9 8 Hence, an apple costs 8p and a banana costs p. If 7 apples and oranges cost.6 and apples and oranges cost., determine how much an apple and an orange each cost. Let an apple A and an orange R, then: 7A + R 6 () A + R () Multiplying equation () by gives: A + R 79 ()

22 Multiplying equation () by gives: A + R 0 () Equation () equations () gives: 9A from which, A 9 8 Substituting in () gives: (8) + R 79 i.e. R 79 (8) i.e. R 0 and B 0 7 Hence, an apple costs 8p and an orange costs 7p. Three new cars and four new vans supplied to a dealer together cost 9000, and five new cars and two new vans of the same models cost Find the respective costs of a car and a van. Let a car C and a van V, then working in 000 s: C + V 9 () C + V 99 () Multiplying equation () by gives: 0C + V 98 () Equation () equations () gives: 7 C 0 from which, C 0 7 Substituting in () gives: () + V 9 i.e. V 9 () i.e. V 8 and V 8 Hence, a car costs 000 and a van costs 000

23 . In a system of forces, the relationship between two forces F and F is given by: F + F - 6 F + F - 8 Solve for F and F F + F - 6 () F + F - 8 () Multiplying equation () by gives: F + F - 0 () Multiplying equation () by gives: 9F + F - () Equation () equation () gives: 6F from which, F 6. Substituting in () gives: (.) + F - 6 i.e. F (.) i.e. F -. and F. -. Hence, F. and F -.. Solve the simultaneous equations: a + b 7 a b Adding equations () and () gives: a 0 a + b 7 () a b ()

24 from which, a 0 Substituting in () gives: + b 7 i.e. b 7 Hence, a and b 6. Solve the simultaneous equations: 8a - b a + b 8a - b () a + b () Multiplying equation () by gives: a b 0 () Multiplying equation () by gives: 9a + b () Equation () + equations () gives: a 6 from which, a 6 6 Substituting in () gives: 8 b i.e. i.e. 8 - b b and b - Hence, a 6 and b - EXERCISE 0, Page. (b). (d). (a). (d). (a) 6. (b) 7. (c) 8. (a) 9. (b) 0. (c). (b). (d). (a). (c). (a) 6. (c) 7. (d) 8. (b) 9. (a) 0. (c)