Find the value of ain each case so the gradient of the line joining the points is equal to the given value. a (2, 5) and (4,a gradient 3 b(6, 1) and (a(, 3) gradient 5 Solution a Gradient

(1)

SkillSHEET

answers

SkillSHEET 1.1

Using gradient to find the value of a

parameter

The gradient of the line joining two points (x₁, y₁) and (x₂, y₂) is given by .

Try these

Find the value of a in each case so the gradient of the line joining the points is equal to the given value. a (3, 5) and (6, a) gradient 4 b (6, 2) and (4, a) gradient 1 c (3, a) and (1, 8) gradient −1 d (4, a) and (6, 5) gradient −7 e (−1, 2) and (a, 6) gradient −2 f (4, 3) and (a, 15) gradient 6 g (a, −3) and (−2, 1) gradient 4 h (a, 9) and (3, 3) gradient −3 i (2, 1) and (5, a) gradient −6 j (−6, 5) and (a, −2) gradient 2

y₂–y₁ x₂–x₁

---Find the value of a in each case so the gradient of the line joining the points is equal to the given value.

a (2, 5) and (4, a) gradient 3 b (6, 1) and (a, 3) gradient 5

Solution Solution

a Gradient =

=

∴ = 3

a− 5 = 6 (Multiply both sides by 2) a= 11 (Add 5 to both sides)

b Gradient =

=

∴ = 5

2 = 5(a − 6) (Multiply both sides by (a − 6)) 2 = 5a – 30 (Expand the brackets)

32 = 5a (Add 30 to both sides) 6.4 =a (Divide both sides by 5) y₂–y₁

x₂–x₁

---a–5 4–2

---a–5 2

---y₂–y₁ x₂–x₁

---3–1 a–6

---2 a–6

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Domain and range

A relation can be described by a listed set of ordered pairs, a graph or an algebraic rule. For example, a relation with the set of ordered pairs {(−1, −2), (0, 0), (1, 2), (2, 4)} can be described by this graph:

The relation can also be summarised by the rule {(x,y): y= 2x; x∈ {−1, 0, 1, 2}} Domain= {−1, 0, 1, 2}, Range = {−2, 0, 2, 4}

Domain is the set of all first elements of the set of ordered pairs. Range is the set of all second elements of the set of ordered pairs. The range can be found by applying the domain values in the given rule.

Try these

For each of the following relations, list the set of possible ordered pairs and state the domain and range. 1 {(x,y): y=x + 2; x∈ {0, 1, 2, 3, 4}} 2 {(x,y): y=−x; x∈ {−5, −1, 0, 3, 5}} 3 {(x,y): y= 1 − 3x; x∈ {−3, −1, 0, 1, 3}} 4 {(x,y): y= 2x2; x∈ {−2, −1, 0, 1, 2}} 5 {(x,y): y=x2 − 4; x∈ {−3, −1, 0, 1, 2}} 6 { (x,y): y=−x2; x∈ {−1, 0, 1, 2, 3, 4}} 7 {(x,y): y =x2; x∈ R} 8 {(x,y): y= 5x− 1; x∈ {−3, −1, 0, 1, 3}} 9 {(x,y): y=−3x; x∈ {−2, −1, 0, 1, 2}} 10 {(x,y): y= 4x; x∈ {−1.5, −1, 0, 1, 1.5}} 11 {(x,y): y =−3x2; x∈ {−3, −1, 0, 1, 3}} 12 {(x,y): y=x2 + 9; x∈ {−2, −1, 0, 1, 2}} 13 {(x,y): y= 2x2; x∈ {−3, −1, 0, 1, 3}} 14 {(x,y): y= 3x2 + 7; x∈ R+}

y

x 0

–1 –2

2 4

1 2 3

For the relation {(x, y): y =x2; xŒ {-3, -2, -1, 0, 1, 2, 3}}, list the set of ordered pairs and state the

domain and range.

Solution

Substitute each x-value into the rule y=x2 to find the corresponding y-value. When x=−3, y= (−3)2= 9

When x=−2, y= (−2)2= 4 When x=−1, y= (−1)2= 1 When x= 0, y= (0)2= 0 When x= 1, y= (1)2= 1 When x= 2, y= (2)2= 4 When x= 3, y= (3)2= 9

The set of ordered pairs is: {(−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9)} Domain = {−3, −2, −1, 0, 1, 2, 3}

Range = {0, 1, 4, 9}

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SkillSHEET

answers

SkillSHEET 2.2

Substitution

For a given function f(x), finding f(2) means evaluating the function when x is 2. This means substituting the value of 2 for x in the rule for f(x). (It is similar to finding y for a particular x-value by substituting into the rule for y.)

Try these

1 For the function f(x) = 3x− 1 find

a f(−1) b f(2) c f(0)

d f(−4) e f(5) f f(k)

2 For the equation g(a) =a2− 3a− 7 find

a g(−3) b g(−1) c g(0)

d g(1) e g(3) f g(k)

3 For the polynomial h(z) = 8 −z3 find

a h(−2) b h(−1) c h(0)

d h(1) e h(2) f h(k)

4 Given the function f(x) = (x− 4)2+ 5, find

a f(−4) b f(0) c f(4)

5 The volume, V(t) m3, of water in a lake at time t hours is given by V(t) = 5 − 3 sin . Find the volume of water at

a t= 0 b t = 1 c t= 3 d t=

(where necessary, give your answer correct to 3 decimal places)

6 During a physics experiment, the kinetic energy, E, (in kJ) of a moving trolley at time t (in seconds) is modelled by the equation E(t) = 15e−2t. Find the energy dissipated at

a t= 0 b t= 1 c t= 3 d t= 6

Give your answers correct to 3 decimal places.

If f(x) =x3+ 4x−2, find f(−2). If t(n) =n2+ 4n, find t(5).

Solution Solution

f(−2) = (−2)3+ 4(−2) − 2 f(−2)=−8 − 8 − 2 f(−2)=−18

t(5) = (5)2+ 4(5) t(5)= 25 + 20 t(5)= 45

1

WORKED

E

xample

WORKED

E

xample

2

πt 2 ---   

6

π

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---Transposition of equations

An equation can be rearranged or transposed so that a particular pronumeral becomes the subject of the equation. Transposition means to transfer a quantity or term from one side of an equation to the other.

Try these

1 Complete each of the following transpositions to make y the subject.

a a+y=x b y−b=z c y2=c

d d= 4x+y e 2y−x= 5e f + 3 = 7

g c=y −mx h A=xy i V=xzy

2 Transpose each of the following to make y the subject.

a by+a=c b mx=y−c c +z=x

d k− 2p= e 15 = 3(x+y) f w(y−a) = 4w

g x2+y2= 25 h =x i x =y2

j xy= 1 k y2+ 4 =x l y2 − = 0 m y2= (x + 4)2 n y2+x2− 16 = 0 o 4x2+ 4y2= 64

Rearrange the equation K=pt2 to make t the Rearrange the equation P= 2x+ 2y to make y the subject. subject.

pt2 =K (reverse the equation) t2= (divide both sides by p)

t= ± (take the square root of both sides)

2x+ 2y=P (reverse the equation)

2y=P− 2x (subtract 2x from both sides) y= (divide both sides by 2) K

p

----K p

----p–2x 2

---1

WORKED

E

xample

WORKED

E

xample

2

y x

--y 8

---y x

--y+1

(5)

---SkillSHEET

answers

SkillSHEET 3.1

Expanding perfect squares

If (A + B)2 is expanded the following rule is obtained:

(A + B)2 = (A + B)(A + B) =A2 + AB + BA + B2 =A2 + 2AB + B2

or (First term + Second term)2 = (First term)2 + Twice the product of terms + (Second term)2

Try these

Expand each of the following:

a (3x + 4)2 b (5x− 1)2 c (7x + 2)2

d (x− 7)2 e (2x + 7)2 f (3 − 5x)2

g (5x + 2y)2 h (4x− 11)2 i (8 + x)2

Expand the following.

a (2x + 3)2 b (3x− 5)2

a (2x + 3)2 First term = 2x Second term = 3

(First term)2= (2x)2= 4x2 Product of terms = 2x× 3 = 6x Twice the product of terms = 2 × 6x

Twice the product of terms= 12x (Second term)2= (3)2= 9

(2x + 3)2= 4x2+ 12x + 9

b (3x− 5)2 First term = 3x Second term =−5

(First term)2= (3x)2= 9x2

Product of terms = 3x×−5 =−15x Twice the product of terms = 2 ×−15x

Twice the product of terms=−30x (Second term)2= (−5)2= 25

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Expanding difference of squares

If (A+B)(A−B) is expanded the following rule is obtained:

(A+B)(A−B) =A2−AB+BA−B2 =A2−B2

or (First term + Second term)(First term − Second term) = (First term)2− (Second term)2

Try these

1 Which of the following can be expanded using the rule for difference of squares?

a (3x+ 4)(3x− 4) b (5x− 1)(5x+ 2) c (7x− 2)(7x + 2) d (1 − 2x)(1 + 2x) e (2x− 5)(2x− 5) f (7x− 3)(5x+ 2) g (3x− 2y)(3x+ 2y) h (3x+ 5)(4x− 5) i (2x− 5)2 2 Expand each of the following.

a (3x+ 4)(3x− 4) b (5x− 1)(5x+ 1) c (7x− 2)(7x+ 2) d (1 − 2x)(1 + 2x) e (2x+ 5)(2x− 5) f (9x− 2)(9x+ 2) g (3x− 2y)(3x+ 2y) h (4x + 5)(4x− 5) i (5x+ 7)(5x− 7)

Expand the following:

a (3x+ 7)(3x− 7) b (2x− 9)(2x+ 9)

a (3x+ 7)(3x− 7) First term = 3x Second term = 7

(First term)2= (3x)2= 9x2 (Second term)2= (7)2= 49 So (3x+ 7)(3x− 7) = 9x2− 49

b (2x− 9)(2x+ 9) First term = 2x Second term = 9

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SkillSHEET

answers

SkillSHEET 3.3

Gradient of a straight line

The gradient, m, of a straight line which passes through the points (x₁, y₁) and (x₂, y₂), is given by the formula: m=

The gradient of a straight line remains constant.

Try these

1 Calculate the gradient of each of the following, and hence state whether the gradient is positive or negative. a rise = 3, run = 1 b rise = 8, run = 4 c rise = 15, run =−2

d run = 7, rise =−21 e run =−5, rise = 75 f run = 35, rise = 18

g h i

j k l

2 Find the gradient of the line joining each of the following pairs of points:

a (5, 2) and (7, −3) b (−2, 2) and (4, 10) c (−1, −3) and (−5, 1) d (1, 1) and (5, 5) e (2.5, 3) and (7.5, 6) f (−3, −2) and (−8, −12).

rise run

--- y2–y1 x₂–x₁ ---=

State whether the gradient Find the gradient of the line joining the points of the line joining the points (−1, 4) and (5, −7).

(1, 1) and (5, 4) in the graph at right is positive or negative. Calculate the gradient.

The line has a positive gradient as it rises from left to right.

Rise = 4 − 1 = 3 Run = 5 − 1

= 4

m= =

So the gradient is .

m=

m =

m =−

So the gradient is − . y x 0 (5, 4) (1, 1) rise run ---3 4 ---3 4

---y₂–y₁ x₂–x₁

---−7–4 5–−1

---11 – 5+1 ---11 6 ---11 6

---1

WORKED

E

xample

WORKED

E

xample

2

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Rounding

Decimal numbers can be rounded to a number of decimal places or to a number of significant figures.

When rounding to a number of significant figures we count from the first non-zero digit in the number. If necessary place values are filled out with zeros.

Try these

1 Round each of the following correct to the number of decimal places indicated in the brackets.

a 0.256 822 54 [2] b 1.151 616 51 [3] c 234.143 343 [1]

d 0.001 242 88 [3] e 235.555 46 [4] f 3.281 572 3 [0]

2 Round each of the following correct to the number of significant figures shown in the brackets.

a 0.256 822 54 [2] b 1.151 616 51 [3] c 234.143 343 [5]

d 0.001 242 88 [3] e 235.555 461 [4] f 3.281 572 3 [3]

3 Round each of the following correct to the number of significant figures shown in the brackets.

a 453 [2] b 45 673 059 [4] c 24 589.356 41 [2]

d 41 284 225 [4] e 12.36 [1] f 98 724 557 553 [2]

Round each of the following calculator displays, correct to 2 decimal places.

a 0.483 245 848 7 b 2.728 128 162 c 4.875

THINK WRITE

a We are rounding to 2 decimal places so consider the third decimal place. This digit (3) is less than 5 so all decimal places from the third place on are ignored.

a 0.48

b We are rounding to 2 decimal places so consider the third decimal place. This digit (8) is greater than 5 so the second decimal place is increased by 1.

b 2.73

c We are rounding to 2 decimal places so consider the third decimal place. This digit is 5 so the second decimal place is increased by 1.

c 4.88

1

WORKED

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xample

Round each of the following calculator displays to three significant figures.

a 0.000 582 325 b 2365.487 36

THINK WRITE

a Count the first three significant figures beginning with the first non-zero.

a

Consider the fourth significant figure (3). As this is less than 5 it, and all subsequent decimal places, are ignored.

0.000 582

b Count the first three significant figures beginning with the first non-zero.

b

The fourth significant figure is a 5 so the third significant figure is increased by 1.

All places after the third significant figure up to the decimal point must be filled with zeros.

2370 1

2

1

2

3

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SkillSHEET

answers

SkillSHEET 4.2

Calculating trigonometric ratios

The sine, cosine or tangent of an angle can be obtained simply from a scientific or graphics calculator. For example, to calculate sin 30°:

1. with a scientific calculator: Enter 30 then press .

2. with a graphics calculator: Press then enter 30 and press . Caution: Before entering the angle, check that the calculator is in DEGree mode.

Try these

1 Find the value of the following, giving answers correct to 4 decimal places.

a sin 15° b sin 34° c sin 79° d sin 20°46′ e sin 37°25′ f sin 83°17′ g cos 15° h cos 34° i cos 79° j cos 20°46′ k cos 37°25′ l cos 83°17′ m tan 15° n tan 34° o tan 79° p tan 20°46′ q tan 37°25′ r tan 83°17′ 2 Find the value of the following, giving answers correct to 3 decimal places.

a sin 90° b cos 90° c tan 180° d sin 220°50′ e cos 139°25′ f tan 283°30′ g sin 115° h cos 234° i tan 370° j sin 180° k cos 180° l tan 90°

SIN

SIN ENTER

Find: a sin 35∞ (to 4 decimal places) b sin 35∞30¢ (to 4 decimal places).

a sin 35°= 0.5736 b 30′ means 30 minutes or 0.5° so sin 35°30′ is the same as sin 35.5°.

sin 35°30′= 0.5807

1

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Find: a cos 45∞ (to 3 decimal places) b cos 45∞20¢ (to 3 decimal places).

a cos 45°= 0.707 b cos 45°20′= 0.703

(Note: cos 45°20′ is the same as cos 45.333°)

2

WORKED

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xample

Find: a tan 60∞ (to 2 decimal places) b tan 60∞52¢ (to 2 decimal places).

a tan 60°= 1.73 b tan 60°52′= 1.79

(Note: tan 60°52′ is the same as tan 60.867°)

3

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Try these

3 Find the value of θin each of the following. Where appropriate, give your answer in degrees and minutes. a sin θ°= 0.5 b sin θ°= 0.3145 c sin θ°= 0.7974 d sin θ°= 0.866 e cos θ°= 0.5 f cos θ°= 0.3145 g cos θ°= 0.7974 h cos θ°= 0.866 i tan θ°= 0.5 j tan θ°= 0.3145 k tan θ°= 0.7974 l tan θ°= 0.866 m tan θ°= 0.95 n tan θ°= 3.145 o tan θ°= 1.79 p tan θ°= 2.567

Find the value of angle θ, if sin θ∞ = 0.55, giving your answer in degrees and minutes.

Solution

θ° = sin–1_0.55

θ° = 33.3670° (degrees only) or 33°22′ (degrees and minutes)

Find the value of angle θ, ifcos θ∞ = 0.55, giving your answer in degrees and minutes.

Solution

θ° = cos−1_0.55

θ° = 56.632 99° or 56°38′

5

WORKED

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xample

Find the value of angle θ, iftan θ∞ = 55, giving your answer in degrees and minutes.

Solution

θ° = tan−1_0.55

θ° = 28.8108° or 28°49′

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SkillSHEET

answers

SkillSHEET 4.3

Labelling right-angled triangles

A trigonometric function is a function of any angle (θ) inside a right-angled triangle and can be defined as the ratio of two sides of a right-angled triangle.

Sine (abbreviated as sin), cosine (abbreviated as cos) and tangent (abbreviated as tan) are used to define the ratios of the sides.

Try these

1 For each of the following right-angled triangles, identify the side labelled x in respect to the given angle,

θ. Where appropriate, also identify the side labelled y.

a b c

d e f

g h i

2 For each of the triangles in question 1, identify which trigonometric function (sine, cosine or tangent) could be used to write an equation to find x.

sin θ =

or

cos θ =

or

tan θ =

or opposite

hypotenuse

---Hypotenuse Opposite

side

θ

Hypotenuse

Opposite side θ

adjacent hypotenuse

---Hypotenuse

Adjacent sideθ

Hypotenuse Adjacent

side θ

opposite adjacent

---Opposite side

Adjacent sideθ

Opposite side Adjacent

side θ

θ

10 cm

x

θ

10 cm x

θ 10 cm

x

θ 10 cm

8 cm x

θ 10 cm x

θ 6 cm

x

θ 15 mm

x y

θ

6 cm

x y

θ 8 cm x

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Degrees and minutes

An angle θ may be given in degrees and minutes or you may need to express your answer for the size of an angle in degrees and minutes.

Remember: 60 minutes = 1 degree or 60′= 1° (Also 60 seconds = 1 minute) Ensure that your calculator is set to degree mode.

Convert 4∞30¢ into degrees. Convert 23.45∞ into degrees and minutes.

With a TI graphics calculator:

Press 4, [ANGLE] 1:° then 30, [ANGLE] 2: ¢. Pressing will show the angle in degrees as a decimal.

So 4°30′= 4.5° Alternatively:

30′= °

30′= 0.5°

So 4°30′= 4°+ 0.5°

So 4°30′= 4.5°

With a TI graphics calculator:

Press 23.45, [ANGLE] 4: DMS. Pressing will show the angle in degrees and minutes (and seconds).

So 23.45°= 23°27′ Alternatively: 0.45°= 0.45 × 60′

0.45°= 27′

So 23.45°= 23°+ 0.45°

So 23.45°= 23°+ 27′

So 23.45°= 23°27′

2nd 2nd

ENTER

30 60

---2nd _▼

ENTER

1

WORKED

E

xample

WORKED

E

xample

2

Find a sin 4∞30¢, b cos 4∞30¢ and c tan 4∞30¢. Find the angle, q∞, in degrees and minutes given Express your answers correct to 3 decimal a sin q∞= 0.5, b cos q∞= 0.65 and c tan q∞= 1.02. places.

Using a TI graphics calculator:

a Press and enter the angle using [ANGLE]. Pressing will display the required value.

So sin 4°30′= 0.078

b Similarly, first press then enter the angle. So cos 4°30′= 0.997

c Again first press the required trigonometric function before entering the angle.

So tan 4°30′= 0.079

Using a TI graphics calculator: a Press [SIN–1_]_{and enter 0.5.}

For sin θ°= 0.5, θ° = 30°

b Press [COS] and enter 0.65. Change the angle from degrees to degrees and minutes by pressing [ANGLE] 4: DMS and

.

For cos θ°= 0.5, θ° = 49.458 39…° For cos θ°= 0.5, θ° = 49°28′

(Round up to 28′ as the number of seconds is over 30.)

c Press [TAN] and enter 1.02. Convert to degrees and minutes as before.

For tan θ°= 1.02, θ° = 45.5672…° = 45° 34′

SIN 2nd

ENTER

COS

2nd

2nd _▼

ENTER

2nd

3

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Try these

1 Convert each of the following angles from degrees and minutes into degrees. Give your answers correct to 3 decimal places.

a 50°30′ b 89°20′ c 90°15′

d 120°5′ e 157°33′ f 184°39′

g 206°7′ h 268°56′ i 300°9′

j 305°3′ k 328°19′ l 409°45′ 2 Convert each of the following angles in degrees into degrees and minutes.

a 50.5° b 89.333° c 90.25°

d 12.083 33° e 17.15° f 4.94°

g 26.784° h 268.5666° i 30.79°

j 35.35° k 28.12° l 140.65°

3 Find the value of each of the following, giving your answers correct to 4 decimal places.

a sin 50°30′ b cos 89°20′ c tan 90°15′

d sin 120°5′ e cos 157°33′ f tan 184°39′

g sin 206°7′ h cos 268°56′ i tan 300°9′

j sin 305°3′ k cos 328°19′ l tan 409°45′

4 Find each angle, θ°, expressing your answer in degrees and minutes.

a sin θ°= 0.866 b cos θ°= 0.866 c tan θ°= 0.866

d sin θ°= 0.75 e cos θ°= 0.75 f tan θ°= 0.75

g sin θ°= 0.238 h cos θ°= 0.238 i tan θ°= 0.238

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Angles of elevation and depression

When you look up at an object in the sky your line of sight will make an angle with the ground. Assuming that the ground is horizontal this angle is called an angle of elevation.

The angle of elevation is the angle starting from the horizontal and then turning upwards.

When you look down at an object on the ground your line of sight will make an angle with a line drawn horizontally from your eyes. This angle is called an angle of depression.

The angle of depression is the angle starting from the horizontal and then turning downwards.

The angle of elevation

The horizontal

The angle of depression The horizontal

Which of the following are angles of elevation? Which of the following are angles of depression?

a b a b

c c

a This is an angle of elevation as the angle is shown upwards from a horizontal line.

b This is not an angle of elevation but shows an angle of depression.

c There is no horizontal line so it does not show an angle of elevation.

a There is no horizontal line so it does not show an angle of depression.

b This shows an angle of elevation not an angle of depression.

c This is an angle of depression as the angle is shown downwards from a horizontal line.

1

(15)

Try these

State whether the following angles are angles of elevation, angles of depression or neither.

1 2 3 4

5 6 7 8

(16)

Using trigonometric ratios

Trigonometric ratios deal with right-angled triangles. The sides of the triangle are named according to their position with respect to a specific angle. The hypotenuse is the longest side length of the right-angled tri-angle.

Trigonometric ratios: SOH CAH TOA means

sin θ= , cos θ= and tan θ= Hypotenuse

Opposite side with respect to the angleθ θ

Adjacent side with respect to the angleθ

opposite

hypotenuse

--- adjacent

hypotenuse

--- opposite

adjacent

---Find the value of x. Express your answer Find the value correct to 3 decimal places. of x.

Express your answer correct to 2 decimal places.

Given information is:

Angle is 47°, hypotenuse length is 18 cm. The side marked x is the opposite side. Use the sine ratio:

sin θ =

sin 47°=

= sin 47°

x= 18 × sin 47° x= 13.164 cm

Angle = 30°, hence the adjacent side = 45 mm. The side marked x is the hypotenuse.

Use the cosine ratio:

cos θ=

cos 30°=

x cos 30°= 45 x=

x = 51.96 mm

45 mm

30°

x

18 cm

x

47°

opposite hypotenuse

---x 18

---adjacent hypotenuse

---45 x

---45 cos 30°

---1

(17)

Try these

1 For the following right-angled triangles state which trigonometric ratio would be appropriate in finding the side length marked x.

a b c

d e

2 Find the side length marked x for each of the triangles in question 1. Express your answers correct to 3 decimal places.

Find the value of the angle θ, to the nearest degree.

Solution

Opposite side is 24 mm and adjacent side is 37 mm. Use tangent ratio:

tan θ =

tanθ= 33°

24 mm

37 mm θ

opposite adjacent

---24 37

---3

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6 cm

x

45°

7 cm

x

15°

4 cm

x

20°

13 cm

x

73°

10 cm

(18)

c d

e

22 cm

15 cm θ

39 mm

10 cm 12 cm

θ

0.3 cm 5.9 cm

θ

0.52 m 1.34 m

(19)

SkillSHEET

answers

SkillSHEET 4.7

Bearings

Bearings can be given in two ways, as a conventional bearing or as a true bearing.

A conventional bearing expresses the direction firstly in terms of north or south and secondly in terms of east or west.

For example:

In the diagram at right the arrow makes an angle of 35° with the north. Because it is to the east of north the bearing is written as

North 35° East or N35°E.

In the second diagram at right the arrow makes an angle of 62° with the south. Because it is to the west of south the bearing is written as South 62° West or S62°W.

A true bearing expresses the direction in terms of the angle made with North, the angle always being drawn clockwise.

For example:

In the diagram at right the angle drawn clockwise from North is 35°. Thus its true bearing is 35°T. (The T signifies a true bearing.)

In the second diagram at right the angle drawn clockwise from North is 180°+ 62° = 242°. Thus its bearing is 242°T.

35° N

S

E W

62° N

S

E W

35° N

S

E W

62° N

S

(20)

3 4

5 6

7 8

26°

S

E W

51°

S

E W

72° N

S

E W

81° N

S

E W

40° N

S

E W

30° N

S

E W

43°

S

E W

46°

S

(21)

SkillSHEET

answers

SkillSHEET 5.1

Converting degrees to radians

It is often desirable to give answers in exact form rather than an approximate decimal form. For example is an exact value but 1.4142 is an approximate value correct to 4 decimal places.

In trigonometry, when changing degrees to radians, exact answers in terms of π or an approximate answer to a given number of decimal places can be given.

Remember: To convert degrees to radians, multiply by .

Try these

1 Change the following to radians, leaving your answer in π form.

a 5° b 10° c 15° d 20° e 25° f 30°

g 45° h 50° i 75° j 80° k 85° l 90°

2 Change the following to radians. Give i the exact answer ii an approximate answer to 2 decimal places. a 125° b 150° c 180° d 220° e 215° f 270° g 300° h 310° i 330° j 345° k 350° l 360° 3 Change the following to radians. Give exact answers. (Hint: the same method is used for negative angles.)

a −30° b −60° c −90° d −150° e −180° f −270° 2

π

180

---Change 60∞ to radians. Give answer in π form. Change 225∞ to radians. Give:

a the exact answer

b an approximate answer to 2 decimal places.

60°= ×

60°=

1

×

3

(Cancel by dividing top and bottom by 60)

60°=

225°= ×

225°=

5

×

4

(Cancel by dividing top and bottom by 45)

225°=

a 225°=

b 225°= (Substitute 3.141 59 …

for π or use the π value on your calculator)

225°≈ 3.93c (Round to 2 decimal places) 60

1 --- π

180

---60 1 --- π

180

---π

3

---225 1 --- π

180

---225 1 --- π

180

---5π 4

---5×3.141 59 . . . 4

---1

(22)

Converting radians to degrees

When using polar coordinates, it is necessary to know how to use both degrees and radians. However, pro-tractors are usually in degrees and so, if the angle is given in radians, it must be changed into degrees if it is to be plotted using a protractor.

The relationship that needs to be remembered is:

π radians = 180 degrees or πc= 180° To convert radians to degrees: × .

Note that if an angle does not have a degree symbol, then it may be in radians. Angles in radians can be

written with or without the radian symbol. For example, is the same as .

Try these

1 Convert each of the following to degrees.

a b c d e f g h i j 3π

2 Convert each of the following to degrees and minutes.

a 3c b 1c c 0.75c d 1.5c e 1.25c f 2.5c g 0.5c h 0.25c i 0.6c j 0.1c 180 π ---πc 2 --- π 2

---Convert each of the following to degrees: Convert each of the following to degrees and minutes:

a b c . a 2c b 0.65c c 4.3c.

a = ×

= = 30°

b = ×

= = 240°

c = ×

= = 315°

a 2c= 2 ×

=

= 114.591 56° (Divide by π on your calculator) = 114°35′ (Convert to degrees and minutes)

b 0.65c= 0.65 ×

=

= 37.242 257° = 37°15′

c 4.3c= 4.3×

=

= 246.371 85° = 246°22′ p

6

--- 4p 3

--- 7p 4 ---π 6 --- π 6 --- 180 π ---180 6 ---4π 3 --- 4π

3 --- 180 π ---720 3 ---7π 4 --- 7π

4 --- 180 π ---1260 4 ---180 π ---360 π ---180 π ---117 π ---180 π ---774 π

---1

WORKED

E

xample

WORKED

E

xample

2

π

4

--- 4π

3

--- 5π 6

--- 3π 4

--- 5π 3

--- π 6

--- 3π

8

--- 2π 15

(23)

---SkillSHEET

answers

SkillSHEET 5.3

Period and amplitude of sine and cosine

graphs

In trigonometry, the period is defined as the smallest interval of x in which a trigonometric function f(x) takes on all values before repeating itself.

The amplitude is the halfway position between the maximum and minimum values of the periodic function f(x).

The graph of y= sin x, at right, has a period of 2π and an amplitude of 1.

The graph of y = cos x, at right, has a period of 2π and an amplitude of 1.

In general:

Note: |A| is the magnitude of A or A written as a positive number.

You can verify your results by viewing the graph of each function on a graphics calculator.

Press and ensure that Degree is selected. Enter the function in the Y= editor, press and choose 7: ZTrig in order to display the graph. Press and hold down the right arrow key to update the graph as it traces. Alternatively, you can download and use the program for Trigonometric graphs on the Maths Quest CD-ROM accessed via the icon on page 223.

Function y=f(x) Period Amplitude

y = A sin a(x+b) + B |A|

y = A cos a(x +b) + B |A| y

x

0 π _π π π

π _—3 2_π

2 5 — 2 –

2 – 2

1

Amplitude

Period –1

–π –

y

x 0 _π 3 _— ₂_π

2 5 — 2 –

2 – 2

1 Amplitude

Period –1

–π –π π π π

2π a

---2π a

---State the period and amplitude of the graph of State the period and amplitude of the graph of

y= 4 sin 2x. y=-cos 0.5x.

Period =

Period=π

Amplitude =|4|

Amplitude= 4

Period =

Period= 4 Amplitude =|−1|

Amplitude= 1 2π

2

--- 2π

0.5

---1

WORKED

E

xample

WORKED

E

xample

2

MODE ZOOM

(24)

1 Write down the period and amplitude of each of the following.

a y= 4 sin x b y=−sin 3x c y=−5 sin 4x

d y= 2 cos 3x e y=−3 cos x f y= 4 cos 2x

g y= 5 cos x h y=−2 cos x i y= sin x

2 A tuning fork is vibrating so that its position, x (mm), at time t (s) is given by the formula: x= 2 sin .

State:

a the amplitude b the period of the vibration.

3 The water level, h (metres), in a lake at time t (hours) is found to have the rule: h= 4 − 2 cos 3(t+1)

State the period and amplitude of the water level.

4 A rubber duck in a water tank is oscillating up and down in the water. Its height, h (cm), above the normal (calm) water level at time t (s) is given by:

h= 7 cos 2(t+π) + 10.

a State the period and amplitude of this relationship for the height. b Calculate the maximum and minimum height of the water.

π

2

--- π

4

--- 3

8 --- π

4

(25)

---SkillSHEET

answers

SkillSHEET 6.1

Finding trigonometric values and angles

The sine, cosine and tangent of any angle can be obtained from a scientific or graphics calculator by simply pressing the button for the appropriate function.

Note that the calculator must be in the same mode as the given angle. That is, if the angle is in degrees, the calculator must also be in the DEG (i.e. degree) mode.

If the value of sin or cos or tan of a certain angle is known, the size of that angle can be found by using an appropriate inverse function of your calculator.

Use a calculator to find the value of each of the following, giving answers correct to 4 decimal places.

a sin 45° b cos 8° c tan 53°

Solution

a Ensure that your calculator is in the degree mode. To find the sine of a given angle, press the following sequence of buttons on your scientific calculator: (or press instead of , if using a graphics calculator).

Record the number shown on the display and round off to 4 decimal places. sin 45° = 0.707 106 781

sin 45°≈ 0.7071

Note that the sign ≈ is read as ‘approximately equal to’.)

b To find the cosine of the given angle, press the following sequence of buttons on your calculator: . Copy the number from the display and round off to 4 decimal places.

cos 8° = 0.990 268 068

cos 8°≈ 0.9903

c To find the tangent of the given angle, press the following sequence of buttons on your calculator: . Record the number shown on the display and round off to 4 decimal places. tan 53° = 1.327 044 822

tan 53°≈ 1.3270

SIN 4 5 = ENTER =

COS 8 =

TAN 5 3 =

1

WORKED

E

xample

For each of the following, find the value of θ, giving answers correct to 1 decimal place.

a sin θ° = 0.5505 b cos θ° = 0.5505 c tan θ° = 0.5505

Continued over page

Solution

a To find the value of an angle when given the value of its sine, press [SIN–1], enter 0.5505 and press (or press in place of , if you are using a graphics calculator).

Record the number shown on the display and round it off to 1 decimal place. (Your answer is in degrees, provided that the calculator was in the degree mode prior to your making the calculations.)

sin θ° = 0.5505

sin°θ= sin−1 0.5505

sin° θ= 33.401 321 82

sin° θ≈ 33.4°

2nd

= ENTER =

(26)

Try these

1 Find the value of each of the following, giving answers correct to 4 decimal places.

a sin 15° b sin 34° c sin 79° d sin 20° e sin 37° f sin 83°

g cos 15° h cos 34° i cos 79° j cos 20° k cos 37° l cos 83° mtan 15° n tan 34° o tan 79° p tan 20° q tan 37° r tan 83°

2 For each of the following, find the value of θ, giving answers correct to 1 decimal place.

a sin θ° = 0.5 b sin θ° = 0.3145 c sin θ° = 0.7974 d sin θ° = 0.866

e cos θ° = 0.5 f cos θ° = 0.3145 g cos θ° = 0.7974 h cos θ° = 0.866

i tan θ° = 0.5 j tan θ° = 0.3145 k tan θ° = 0.7974 l tan θ° = 0.866 mtan θ° = 0.95 n tan θ° = 3.145 o tan θ° = 1.79 p tan θ° = 2.567 Record the number shown on the display and round it off to 1 decimal place.

cos θ° = 0.5505

cos°θ= cos−1 0.5505

cos° θ= 56.598 678 18

cos° θ≈ 56.6°

c To find the value of an angle when given the value of its tangent, press , enter 0.5505 and press .

Record the number shown on the display and round it off to 1 decimal place. tan θ° = 0.5505

tan°θ= tan−1 0.5505

tan° θ= 28.832 783 64

tan° θ≈ 28.8°

2nd [TAN-1]

(27)

SkillSHEET

answers

SkillSHEET 6.2

Exact values of trigonometric ratios

The exact value of the three trigonometric ratios for angles of 30°, 45° and 60° can be found using the right-angled triangles below.

Try these

Find the exact value of each of the following.

1 sin 30° 2 sin 45° 3 sin 60°

4 cos 30° 5 cos 45° 6 cos 60°

7 tan 30° 8 tan 45° 9 tan 60°

1

1 45°

√2

2

1 30°

60°

√3

Find the exact value of cos 60°.

THINK WRITE

Find the 60° angle in one of the triangles above.

Write the formula for cos θ. cos θ =

Substitute the values for θ, adj and hyp. cos 60° = 1

2 _hyp---adj

3 1₂

(28)

Negative and rational powers

A negative power can be expressed as a positive power using the following rule.

a−n= For example, 3−5= , x−2=

A surd can be expressed as a fractional power using the following rule.

= For example, = , =

This rule can be extended to powers of surds and surds of powers.

=

( )

m= For example, = ,

( )

7=

Try these

1 Express each of the following with positive index numbers and hence evaluate them.

a 7−2 b 8−3 c 10−5 d e f

g 9−3 h i 6−4 j k 5−4 l

2 Express each of the following with positive index numbers.

a b

(

)

2 c d

(

)

3 e f

1 an --- 1 35 --- 1 x2 ---a n 1 an

--- 3 7 7

1 3 ---x 5 _x 1 5 ---am n a n _a m n ----x2 3 x 2 3 ---x 5 _x 7 5

---Express each of the following with positive Express each of the following with positive index index numbers and hence evaluate them. numbers.

a 3-4 b 5-3 c a b

(

)

5 c

a 3−4=

3−4=

a =

=

b 5−3=

5−3=

b ( )5=

(

)

5

=

c = 6−(−2) = 62 = 36

c =

=

1 6–2

--- 5 _x4 3 ₂₇ 3 _x–5

1 34 ---1 81 ---x4 5 x4 ( ) 1 5 ---x 4 5 ---1 53 ---1 125 ---27 3 ₂₇ 1 3 ---27 5 3 ---1

6–2

--- 3 x–5 (x–5)

1 3 ---1 x5 ---    1 3 ---1 x 5 3 ---

----1

WORKED

E

xample

WORKED

E

xample

2

1

3–4

--- 1

5–2

--- 1

6–3

---1

7–3

--- 1

2–8

--- 1

3–5

---x5 3

243

5 7 _b3

256

4 5 _x–4

(29)

SkillSHEET

answers

SkillSHEET 7.2

Substitution in exponential functions

One way to sketch a graph of an exponential function is to construct a table of values. This involves substi-tuting various x-values to find corresponding y-values and to generate coordinate points for plotting. To construct a table of values:

1. consider a set of x-values such as x=−3, −2, −1, 0, 1, 2, 3

2. substitute each x-value into the rule for the exponential function to find the corresponding y-value.

Try these

Construct a table of values for each of the following exponential functions.

1 y= 2x 2 y= 7x 3 y= 3−x 4 y= 5−x

Construct a table of values for the function y= 4x.

The table of values for y= 4x is

Solution

Consider the x-values −3, −2, −1, 0, 1, 2, 3. When x=−3, y= 4−3

=

= = 0.0156

When x=−2, y= 4−2 =

= = 0.0625 When x=−1, y= 4−1

= = 0.25

When x= 0, y= 40 = 1

When x= 1, y= 41 = 4

When x= 2, y= 42 = 16 When x= 3, y= 43

= 64

x −3 −2 −1 0 1 2 3

y 0.0156 0.0625 0.25 1 4 16 64

1

43

---1 64

---1

42

---1 16

---1 4

(30)

Solving indicial equations

Indicial equations are those where the variable is part of the power (or index). The following are examples

of indicial equations: 3n= 7 and ( )3n–1 = 0.7.

To solve equations of the type shown above, we use the logarithm law: log_aAn=nlog_a. This law enables us to remove the variable from the power. Once this is achieved, an equation takes linear form and can then be solved normally.

Try these

Solve each of the following equations for n, expressing, where necessary, your answers correct to 2 decimal places.

1 5 × (3)n–1 = 405 2 12 × (5)n–1 = 23 437 500 3 100 × ( )n–1 = 3.125

4 6561 × ( )n–1 = 256 5 1000 × (0.9)n–1 = 100 6 2000 × (0.95)n–1 = 1

7 1000 × (1.05)n= 10 000 8 2 × (1.1)n= 25 9 10 × (2)n–1 = 640

4 9

---Solve each of the following equations for n.

a 3 ¥ (2)n–1= 6144 b 100 ¥ (0.95)n–1= 50

Solution

a 3 × (2)n–1=₆₁₄₄

(2)n–1=_{6144 ÷ 3 (Divide both sides by 3.)}

(2)n–1=₂₀₄₈

To remove n from the power, take the logarithm to any base of both sides of the equation. It is always convenient to choose either base 10 or base e, as these can readily be evaluated using a calculator. So, select either of these two bases, say base 10.

log₁₀ (2)n–1=_log 10 2048

(n −1)log ₁₀2= log₁₀2048 (Apply logarithmic law logaA n₌

nlogaA.)

n− 1 = (Divide both sides by log₁₀2.) n− 1 = 11 (Evaluate RHS using calculator.)

n = 12 (Add 1 to both sides to find the value of n.)

b 100 × (0.95)n–1= 50

(0.95)n–1= 50 ÷ 100 _{(Divide both sides by 100.)}

(0.95)n–1= 0.5

log₁₀ (0.95)n–1= log

10 0.5 (Take logarithm to the base 10 of both sides of the equation.)

(n− 1)log₁₀ 0.95= log₁₀ 0.5 (Apply logarithmic law logaA n₌

nlogaA.)

n− 1 = (Divide both sides by log₁₀0.95.)

n− 1 = 13.51 (Evaluate RHS using calculator and round off to 2 decimal places.) n= 14.51 (Add 1 to both sides to find the value of n.)

log₁₀ log₁₀2

---log₁₀0.5 log₁₀0.95

---WORKED

E

xample

1 2

(31)

---SkillSHEET

answers

SkillSHEET 8.2

Transposition of formulas

Equations (or formulas) can be rearranged so that a nominated pronumeral is the subject of the equation. This means that the nominated pronumeral is by itself on one side of the equation.

The process of rearranging an equation or formula is known as transposition. Transposition, in fact, means to transfer a quantity or term from one side of an equation to the other.

Try these

1 Transpose each of the following to make y the subject.

a a+y=x b y−b=z c y2=c

d d= 4x+y e 2y −x= 5e f + 3 = 7

g c=y−mx h A=xy i V=xzy

2 Transpose to make y the subject in each of the following formulas.

a 9y+ 27x= 3 b 5x=y− 12 c + 9 =x d 3y− 6x− 15 = 0 e 15 = 3(x+y) f w(y+a) = 4w g 7x− 4y− 35 = 0 h =x i x =y2

j xy= 1 k mx=y−c l y2 − = 0 m y2− (x + 4)2= 0 n y+x2− 16 = 0 o 4x2+ 4y2= 64

Transpose P= 2x+ 2y to make y the subject. Transpose y=pt2 to make t the subject.

P= 2x+ 2y

2x+ 2y=P (Reverse the equation) 2y=P− 2x (Subtract 2x from both sides)

y= (Divide both sides by 2)

y=pt2

pt2 =y (Reverse the equation) t2= (Divide both sides by p)

t= ± (Take the square root of both sides) P–2x

2

---y p

---1

WORKED

E

xample

WORKED

E

xample

2

y x

--y 8

---y+1

(32)

---Substitution into the simple interest

formula

Simple interest can be calculated using the formula I= , where I= interest ($)

P= principal invested at the start ($) r= rate of interest (per period) T= time (the number of periods)

In this formula, the units of rate (r) and time (T) must correspond to each other. For example, if the rate is given per annum (per year), time must also be in years. If this is not so, the value of time

needs to be converted to the units of the rate prior to substituting into the formula. Note that the rate per year (or per annum) is abbreviated as pa.

PrT 100

---Change the value of time T according to the Find the simple interest for each of the following: period given in the rate r. a a loan of $10 000 at a rate of 2% pa for a time

a r= 2% pa; T= 15 months period of 15 months

b r= 6% per month; T= 3 years b an investment of $500 invested at a rate of 6% per month for the term of 3 years.

a Since the rate is given as per year (pa), convert 15 months to years by dividing the number of months by 12.

T= 15 months

T = years (Divide by 12 to convert to years.)

T = 1 years (Change into a mixed number.)

T = 1 years (Simplify.)

b Since the rate is given as per month, convert 3 years to months by multiplying the number of years by 12.

T= 3 years T = 3 × 12 months T = 36 months

a Since the rate is per year and the time is in months, convert 15 months to years first.

T= 15 months T = years

T = 1 years

Write the formula for the simple interest and identify the values of the pronumerals.

I = , P= 10 000, r= 2, T= 1

Substitute the values of the pronumerals into the formula and evaluate.

I= = $250

b Since the rate is given per month and the time is in years, convert 3 years to months first.

T= 3 years

T = 3 × 12 months T = 36 months

Write the formula for the simple interest and identify the values of the pronumerals.

I = , P= 500, r= 6, T= 36

Substitute the values of the pronumerals into the formula and evaluate.

I= = $1080

15 12 ---3 12 ---1 4 ---15 12 ---1 4 ---PrT 100 --- 1 4

---10 000×2×11₄

---100

---PrT 100

---500×6×36 100

---1

(33)

Try these

1 Change the value of time T according to the period given in the rate r:

a r= 3% pa T= 24 months b r= 0.02% per day T= 31 days c r= 5 % pa T= 9 months d r= 0.1% per week T= 1 years e r= 7% pa T= 18 months f r= 4.75% per month T= 52 weeks g r= 2.5% pa T= 13 weeks h r= 0.15% per week T= 21 days i r= 7% pa T= 150 days j r= 0.009% per day T= 4 years

2 Find the simple interest on an investment of $5000 at the rate and time period as given in questions 1 a to j.

3 4

--- 1

(34)

---Simple interest and arithmetic

progressions

An arithmetic progression is a sequence of numbers for which the difference between consecutive terms is constant.

For example

and 2, 4, 6, 8, … and 8, 6 , 5, 3 , 2, …

are both arithmetic progressions.

The Simple Interest earned over a number of years forms an arithmetic progression.

Try these

1 For each of the following sequences find: ii the 20th term

ii an expression for the nth term

a 1, 3, 5, 7, … b 20, 18, 16, 14, … c 4, 8, 12, 16, … d 0, , 1 , 2, … e 4.2, 5.4, 6.6, 7.8, …

2 By considering the simple interest earned in year 1, year 2 and year 3, show that the interest earned forms an arithmetic progression.

a P= $2500, r= 8% b P= $120 000, r= 3% c P= $10 000, r= 2.5%

1 2

--- 1

2

---For the sequence 1, 4, 7, 10, …, find:

a the 20th term b an expression for the nth term

Solution

a The sequence is an arithmetic progression with the difference between terms equal to 3. The 20th term will be the first term plus 19 times the difference.

T₂₀= 1 + 19 × 3

T₂₀= 58

b The nth term, T_n, is: T_n= 1 + (n− 1)3.

1

WORKED

E

xample

Show that the Simple Interest earned by investing $4000 at 7% p.a. forms an arithmetic progression.

Solution

The interest after 1 years is I= = = $280 The interest after 2 years is I= = = $560

The interest after 3 years is I= = = $840

The sequence $280, $560, $840, … forms an arithmetic progression with the difference between consecutive terms being $280.

PrT 100

--- $4000×7×1 100

---PrT 100

--- $4000×7×2 100

---PrT 100

--- $4000×7×3 100

---2

WORKED

E

xample

2 3 --- 1

(35)

---SkillSHEET

answers

SkillSHEET 9.1

Percentage skills

There are some percentages and their equivalent fractions that are worth memorising. This knowledge can speed up calculations, particularly if a calculator is not available.

Try these

1 Find the value of:

a 10% of 69 b 12 % of 800 c 30% of 170

d 33 % of 99 e 66 % of 240 f 100% of 84

g 2000% of 12 h 50% of 450 i 33 % of 720

j 40% of 9.50 k 12 % of 16.16 l 70% of 1.50

m 25% of 500 n 75% of 352 o 40% of 53.50

2 In a survey at Cooper Middle School it was found that: 66 % of students owned a computer

25% of students rode a bike to school

10% of students have not travelled out of the state of Victoria and 75% of the boys played a club sport of some type.

If there are 550 boys and 422 girls enrolled at Cooper Middle School, find the number of students in each category surveyed.

3 A jacket costing $200 is to be discounted by 12 %. Find the discount and, hence, the new price. 4 During the 1980s the average cost of living increased by 33 %. If at the beginning of 1980 the rent of a

flat was $108 per week, what would be the estimated increase in rent and hence the cost of the same flat at the end of 1989?

10% = 12 % = 25% = 33 % =

50% = 66 % = 100% = 1 200% = 2 etc.

1 10 --- 1 2 --- 1 8 --- 1 4 --- 1 3 --- 1 3 ---1 2 --- 2 3 --- 2 3

---Find 25% of 60. A book costing $56 is to be discounted by 20%. Find the discount and, hence, the new price.

25% =

So 25% of 60 is of 60 which is 15. Therefore, 25% of 60 = 15.

20% is the same as 2 × 10%.

Since 10% = and of $56 = $5.60 then

2 × $5.60 = $11.20. Discount is $11.20.

New price = $56.00 − $11.20

New price= $44.80

1 4 ---1 4 --- 1 10 --- 1 10

---1

WORKED

E

xample

WORKED

E

xample

2

(36)

---Finding the median

The median of a set of scores is defined as the middle score when the data are arranged in order of size. The position of the median can be found using the formula

Median position = score where n is the number of scores.

For example, if there are 15 scores the median position is the score, that is the 8th score.

If there are 14 scores, the median position is the score, that is the 7 th score.

The 7 th score is the average of the 7th score and 8th score.

Try these

Find the median of each of the following sets of data.

1 2, 5, 8, 9, 2, 4, 6, 1, 8, 3, 5 2 10, 8, 18, 13, 7, 12, 13, 15, 9

3 22, 35, 28, 41, 27, 14, 16 4 102, 109, 98, 103, 99, 96, 108, 92, 110 5 2, 5, 8, 9, 2, 4, 6, 1, 8, 3, 7, 2 6 22, 35, 28, 32, 41, 27, 14, 16

7 12, 8, 18, 13, 7, 12, 13, 15, 9, 16 8 12, 8, 18, 13, 7, 12

9 34, 76, 12, 78, 39 10 203, 87, 156, 54, 100, 234

n+1 ( )th

2

---15+1 ( )th

2

---14+1 ( )th

2

--- 1

2 ---1

2

---Find the median of the following data: Find the median of the following data: 2, 3, 7, 4, 9, 1, 5, 8, 6, 11, 12, 3, 4. 8, 13, 7, 9, 9, 9, 12, 8, 14, 11, 12, 13.

First write the data in order.

1, 2, 3, 3, 4, 4, , 6, 7, 8, 9, 11, 12. There are 13 scores, so n= 13.

Median score = score

Median score= the 7th score

Median score= 5

First write the data in order.

7, 8, 8, 9, 9, , 12, 12, 13, 13, 14. There are 12 scores, so n= 12.

Median score = score

Median score= the 6 th score

Median score= average of 6th and 7th scores

Median score= Median score= 10 5

13+1 ( )th

2

---9, 11

12+1 ( )th

2

---1 2

---9+11 2

---1

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SkillSHEET

answers

SkillSHEET 11.1

Sampling without replacement

Two markers are drawn out from a bag containing 12 markers, 7 of which are green, 3 red and 2 blue. A third marker is now drawn out. The chances of it being blue will obviously depend on what colours were removed on the first two draws.

If the two drawn were both green, there would be 5 green, 3 red and 2 blue left in the bag. As there are

10 markers left in total, Pr(blue on third draw) = (2 out of 10 are blue).

If, of the two drawn, one was green and one was blue, there would be 6 green, 3 red and 1 blue left in the

bag. So Pr(blue on third draw) = (1 out of 10 are blue).

Try these

1 A bag contains 3 red, 6 purple and 5 white markers. If two markers are removed from the bag, what is left in the bag if:

a one red and one white are removed? b both removed are purple?

c one purple and one white are removed?

2 For each part of question 1, calculate the probability that, if a third marker is removed, it will be a purple one. 3 A bag contains 4 pink, 8 orange and 6 black cubes. If two cubes are removed from the bag, what is left

in the bag if:

a one black cube and one orange cube are removed? b both removed cubes are pink?

c one pink cube and one black cube are removed?

4 For each part of question 3, calculate the probability that if a third cube is removed it will be a black one. 5 A bag contains 10 green, 9 blue and 2 orange cubes. If two cubes are removed from the bag, what is left

in the bag if:

a one blue cube and one green cube are removed? b both removed cubes are orange?

c one blue cube and one orange cube are removed?

6 For each part of question 5, calculate the probability that if a third cube is removed it will be an orange one.

2 10

---1 10

---A bag contains 6 red, 4 blue and 3 yellow For each part of worked example 1, calculate cubes. If two cubes are removed from the the probability that if a third cube is removed, it bag, what is left in the bag if: will be a red one.

a one red and one blue are removed?

b both cubes removed are yellow?

c one red and one yellow are removed?

d both cubes removed are red?

a 5 red, 3 blue and 3 yellow a 5 out of 11 are red

So Pr(red on third draw) =

b 6 red, 4 blue and 1 yellow b 6 out of 11 are red

c 5 red, 4 blue and 2 yellow c 5 out of 11 are red

d 4 red, 4 blue and 3 yellow d 4 out of 11 are red

5 11

---6 11

---5 11

---4 11

---1

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Converting a fraction or a decimal to a

percentage

A relative frequency is given usually as a fraction or as a decimal. Converting this to a percentage frequency can be done as follows:

To change a fraction to a percentage: multiply by (This can be done on your calculator.)

To change a decimal to a percentage: multiply by 100 (This can be done by moving the decimal point two places to the right.)

Try these

1 Convert each of the following fractions to a percentage. Where appropriate, express your answers correct to 2 decimal places.

a b c d e

f g h i j

k l m n o

2 Convert each of the following decimals to a percentage.

a 0.73 b 0.56 c 0.06 d 0.86 e 0.2

f 0.673 g 0.9134 h 0.0895 i 0.735 j 0.034

k 0.12 l 0.0043 m 0.342 n 0.02 o 0.0425

100 1

---Convert each of the following to a percentage. Convert each of the following to a percentage.

a b a 0.23 b 0.1875

a = × %

= %

= 46%

b = × %

= %

= 38.57%

Note that this answer is expressed correct to 2 decimal places.

a 0.23 = 0.23 × 100%

= 23% (Move the decimal point 2 places to the right)

b 0.1875 = 0.1875 × 100%

= 18.75% (Move the decimal point 2 places to the right)

23 50 --- 27 70 ---23 50 --- 23 50 --- 100 1 ---2300 50 ---27 70 --- 27 70 --- 100 1 ---2700 70

---1

WORKED

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xample

WORKED

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---SkillSHEET

answers

SkillSHEET 12.1

Gradient of a straight line

The gradient, m, of a straight line which passes through the points (x₁, y₁) and (x₂, y₂), is given by the formula: m=

The gradient of a straight line remains constant.

Try these

1 Calculate the gradient of each of the following, and hence state whether the gradient is positive or negative. a rise = 3, run = 1 b rise = 8, run = 4 c rise = 15, run =−2

d run = 7, rise =−21 e run =−5, rise = 75 f run = 35, rise = 18

g h i

j k l

2 Find the gradient of the line joining each of the following pairs of points:

a (5, 2) and (7, −3) b (−2, 2) and (4, 10) c (−1, −3) and (−5, 1) d (1, 1) and (5, 5) e (2.5, 3) and (7.5, 6) f (−3, −2) and (−8, −12).

rise run

--- y2–y1 x₂–x₁ ---=

State whether the gradient Find the gradient of the line joining the points of the line joining the points (-1, 4) and (5, -7).

(1, 1) and (5, 4) in the graph at right is positive or negative. Calculate the gradient.

The line has a positive gradient as it rises from left to right.

Rise = 4 − 1 = 3 Run = 5 − 1

= 4

m= =

So the gradient is .

m=

m =

m =−

So the gradient is − . y x 0 (5, 4) (1, 1) rise run ---3 4 ---3 4

---y₂–y₁ x₂–x₁

---−7–4 5–−1

---11 – 5+1 ---11 6 ---11 6

---1

WORKED

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WORKED

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Displacement and distance

Consider an object moving in a straight line only. Two quantities which can be used when describing the motion of the object are displacement of an object and distance travelled by the object.

Displacement describes the change in an object’s position, relative to a reference point. (That is, final position minus initial position.)

Distance describes how far an object has travelled. (That is, the total length travelled from the starting point to the finishing point.)

Try these

Every weekday Peter walks to one of two local parks from his home in Ivanhoe. On each of his return trips he stops at a different coffee shop for some refreshment. Calculate Peter‘s displacement and total distance travelled by the time he stops for his coffee for each of the five weekdays.

1 Monday a Displacement =

b Total distance =

2 Tuesday a Displacement =

b Total distance =

3 Wednesday a Displacement =

b Total distance =

4 Thursday a Displacement =

b Total distance =

5 Friday a Displacement =

b Total distance =

Find: a the displacement, and b the distance travelled by a person following the straight course below.

Solution

a The displacement is +5 m from the starting point.

b The distance travelled is 15 m from start to finish.

0 m 5 m 10 m start

finish

WORKED

E

xample

0 km 3 km 6 km Home

Coffee shop #1

Park #1

0 km 3 km 6 km