Trigonometry

Outcomes

Must: Use label sides of triangles correctly

Should: Be able to do calculations involving trig functions

Could: Use trig ratios to find missing lengths in triangles.

Wednesday 8 August 2012

Using all 3 trig ratios to find missing lengths

Right-angled trianglesA right-angled triangle contains a right angle.

The longest side opposite the right angle is called the hypotenuse.

Wednesday 8 August 2012

The opposite and adjacent sidesThe two shorter sides of a right-angled triangle are named with respect to one of the acute angles.

The side opposite the marked angle is called the opposite side.

The side between the marked angle and the right angle is called the adjacent side.

x

Label the sides

Calculate the following ratiosUse your calculator to find the following to 3 significant figures.

1) sin 79° = 0.982 2) cos 28° = 0.883

3) tan 65° = 2.14 4) cos 11° = 0.982

5) sin 34° = 0.559 6) tan 84° = 9.51

7) tan 49° = 1.15 8) sin 62° = 0.883

9) tan 6° = 0.105 10) cos = 0.55956°

θ

OPPOSITE

HY

PO

TE

NU

SE

A D J A C E N T

The three trigonometric ratios

Sin θ =Opposite

Hypotenuse S O H

Cos θ =Adjacent

Hypotenuse C A H

Tan θ =OppositeAdjacent T O A

Remember: S O H C A H T O A

The sine ratio

The ratio of the length of the opposite sidethe length of the hypotenuse

is the sine ratio.

The value of the sine ratio depends on the size of the angles in the triangle.

θ

OPPOSITE

HY

PO

TE

NU

SE

We say:

sin θ =opposite

hypotenuse

Using sine to find missing lengths

65°

x cm

11 cm

opposite

hyp

S O H C A H TO A Sin = opposite

hypotenuse θ

Sin = opp

hyp θ

Sin 65 = x

11

Sin 65 x 11 = x

9.97 (2dp) = x

32°

x cm

6 cm 72°

x cm

12 cm 54°

x cm

15 cm

(1) (2) (3)

Sin 32 = x 6

Sin 32 x 6 = x

3.18cm (2dp) = x

Sin 72 = x 12

Sin 72 x 12 = x

11.41cm (2dp) = x

Sin 54 = x 15

Sin 54 x 15 = x

12.14cm (2dp) = x

Using sine to find missing lengths

47°

8 cm

x cm

opposite

hyp

S O H C A H TO A Sin = opposite

hypotenuse θ

Sin = opp

hyp θ

Sin 47 = 8

x

x = 8

Sin 47

x = 10.94 cm (2dp)

32°

6 cm

x cm 72°

12 cm

x cm 54°

3 cm

x cm

(1) (2) (3) hyp

opp

Sin 32 = 6 x

x = 6Sin 32

x = 11.32 cm(2dp)

Sin 72 = 12 x

x = 12Sin 72

x = 12.62 cm(2dp)

Sin 54 = 3 x

x = 3Sin 54

x = 3.71 cm(2dp)

Using cosine to find missing lengths

S O H C A H TO A cos = adjacent

hypotenuse θ

53°

x cm

10 cm

hyp

adj

cos = adj

hyp θ

cos 53 = x 10

cos 53 x 10 = x

6.02 (2dp) = x66°

12 cm

x cm

hyp

adj

cos = adj

hyp θ

cos 66 = 12 x

x = 12cos 66

x = 29.50 cm(2dp)

(1)

31°

x cm

9 cm

(2)

49°

22 cm

x cm

hyp

adj

cos = adj hyp θ

cos 31 = x 9

cos 31 x 9 = x

7.71 (2dp) = x

cos = adj hyp θ

cos 49 = 22 x

x = 22cos 49

x = 33.53 cm(2dp)

Using tangent to find missing lengths

S O H C A H TO A tan = oppositeadjacent

θ

71°

x cm

10 cmadj

opp

tan = opp

adj θ

tan 71 = x 10

tan 71 x 10 = x

29.04cm (2dp) = x

318cm xcm

adj

opp

tan = opp adj θ

tan 31 = 8 x

x = 8tan 31

x = 13.31 cm(2dp)

31xcm

7cm

48

xcm

43cm(1) (2)

4.21cm (2dp) = x x = 38.72 cm(2dp)

Using all 3 trig ratios to find missing lengths

S O H C A H TO A

477cm

xcm hyp

opp

sin = opp

hyp θ

sin 47 = 7 x

x = 7sin 47

x = 9.57 cm(2dp)

33°

x cm

opp

adj

7cm

tan = opp adj θ

tan 33 = 7 x

x = 7tan 33

x = 10.78 cm(2dp)

53°

x cm

10 cm

hyp

adj

cos = adj

hyp θ

cos 53 = x 10

cos 53 x 10 = x

6.02cm (2dp) = x

Finding Angles using Trig

S O H C A H TO A

θ

5 cm

8 cm

adj

opp

tan = opp

adj θ

tan = 8 5

θ

= 8 5

θ57.99 (2dp)

tan-1

θ10cm

6cm

hyp

opp

sin = opp

hyp θ

sin = 6 10

θ

= 6 10

θ sin-1

θ = 36.87 (2dp)

Trigonometry 2Objective: Use trig to find missing lengths

and angles in right angled triangles for worded questions. Grade A

Outcomes

Must: Use trig ratios to find missing lengths and angles in triangles.

Should: Use trig to answer worded problems.

Could: Use trig to answer more difficult worded problems.

Wednesday 8 August 2012

Finding side lengthsA 5 m long ladder is resting against a wall. It makes an angle of 70° with the ground.

5 m

70°x

What is the distance between the base of the ladder and the wall?

We are given the hypotenuse and we want to find the length of the side adjacent to the angle, so we use:

cos θ =adjacent

hypotenuse

cos 70° =x5

x = 5 × cos 70°= 1.71 m (to 2d.p.)

Area of a triangle (using ½ ab Sin C)

Outcomes

Must: Understand when you can use this formula for triangle area.

Should: Be able to find the area of a triangle using ½ ab sin C.

Could: Answer exam questions

Wednesday 8 August 2012

Objective Find the area of a triangle using Area = ½ ab sin C

A

B C4cm

5cm

Triangle Area : When to use ½ ab Sin C ?• When you have not been given a perpendicular height (straight height).

• When you have been given two lengths and an angle between them.

35

Area of triangle ABC = ab sin C12

A = ½ x 4 x 5 Sin 35A = ……….cm2

4cm6cm

40

A =½ ab Sin C

A = ½ x 4 x 6 Sin 35A = ……….cm2

6cm

9cm23

A =½ ab Sin CA = ½ x 6 x 9 Sin 35A = ……….cm2

50

4cm

7cm

A = ½ ab Sin CA = ½ x 7 x 4 Sin 50A = ……….cm2

Find the area of the triangles

6cm(1)

35

4cm

7cm

A = ½ ab Sin CA = ½ x 7 x 4 Sin 35A = ……….cm2

6cm(2)

453cm

7cm

A = ½ ab Sin CA = ½ x 5 x 3 Sin 45A = ……….cm2

5cm

(3)

Sine Rule

Outcomes

Must: Use sine rule to find lengths in triangles.

Should: Use sine rule to find angles in triangles.

Could: Answer mixed questions

Wednesday 8 August 2012

Objective Use sine rule to find angles and lengths in triangles. (not right angles triangles)

Sine Rule

• Find the sides and angles of a triangle whether it’s a right angle or not.

C

A B

b

c

a

asin A

=b

sin B=

csin C

6cm y cm4025

Examples: Find y using sine rule

A

a

B

b

aSin A

=b

Sin B

3Sin 25

=y

Sin 40

3Sin 25

x Sin 40 = y

4.56 cm (2dp) = y

C

A B

b

c

a

Sine Rule aSin A

=b

Sin B=

cSin C

Exercise: Find lengths y using sine rule

6cm

y cm34

15 A

a

C

c

ySin 15

=6

Sin 34

y =6

Sin 34 x Sin 15

y = 2.8 cm (1dp)

y

4.2cm 41

22 B

bC

c

4.2Sin 22

=y

Sin 41

4.2Sin 22

x Sin 41 = y

7.4 cm (1dp) = y

(1) (2)

y

8.9cm

6236 B

bC

c

(3)

Y = 5.9 cm

C

A B

b

c

a

asin A

=b

sin B=

csin C

or

sin A sin B sin Ca

=b

=c

Sine Rule

43

x

3.5cm

2.3cm

C

cB

b

Sin Bb

=Sin C

cSin x2.3

=Sin 43

3.5

Sin x =Sin 43

3.5 x 2.3

Sin x = 0.44817….. x = 0.44817….. sin-1

x = 26.6 (1dp)

Use this when finding a length

Use this when finding an angle

C

A B

b

c

a

Sine Rule

Sin Aa

= Sin Bb

= Sin Cc

Exercise: Find angle y using sin rule

4.3cm

2.9 cm63

y A

a

C

c

(1)

Sin Aa

=Sin C

cSin y2.9

=Sin 63

4.3

Sin y =Sin 63

4.3 x 2.9

Sin y = 0.600911….. y = 0.600911….. sin-1

y = 36.9 (1dp)

8.4cm

7.3 cm y

53 B

b

C

c

(2)

Sin Bb

=Sin Y

c

Sin 537.3

=Sin y

8.4Sin 537.3

x 8.4 = Sin y

0.9080…= Sin y

0.9080… sin-1 = y

65.2 (1dp) = y

Sine Rule Practice

Higher GCSE for AQA (Oxford) Book

Page 387 Exercise 4r Q1 – Q19

Cosine Rule

Outcomes

Must: Use cosine rule to find lengths in triangles.

Should: Use cosine rule to find angles in triangles.

Could: Answer mixed questions

Wednesday 8 August 2012

Objective Use cosine rule to find angles and lengths in triangles. (not right angles triangles)

The cosine ruleA

B C

c

a

b

a2 = b2 + c2 – 2bc cos Aa2 = b2 + c2 – 2bc cos A

or

cos A = b2 + c2 – a2

2bc

Use when finding a length of a side.

Use when finding an Angle.

B

C A7 cm

4 cm

48°

a

Example 1Find a

When you are given two lengths and an angle between them use;

a2 = b2 + c2 – 2bc cos Aa2 = 72 + 42 - 2 x 7 x 4 cos 48a2 = 27.52868…..a = 5.25 (2dp)

B

C A7 cm

4 cm

48°

a

Example 1Find a

When you are given two lengths and an angle between them use;

a2 = b2 + c2 – 2bc cos Aa2 = 72 + 42 - 2 x 7 x 4 cos 48a2 = 27.52868…..a = 5.25 (2dp)

Exercise Find the length marked x

2cm

5cm

52x

(1)

98x 9cm

5cm

(2)

a2 = 52 + 22 - 2 x 5 x 2 cos 52a2 = 16.68677…..a = 4.08 (2dp)

a2 = 52 + 92 - 2 x 5 x 9 cos 98a2 = 118.52557…..a = 10.89 (2dp)

The cosine ruleA

B C

c

a

b

a2 = b2 + c2 – 2bc cos Aa2 = b2 + c2 – 2bc cos A

or

cos A = b2 + c2 – a2

2bc

Use when finding a length of a side.

Use when finding an Angle.

(given 2 lengths and an angle)

(given 3 lengths)

Example 1Find angle A

4 cm

8 cm6 cm

A

B

C

You are given 3 sides and asked for an angle.

cos A = b2 + c2 - a2

2bca

b

ccos A = 42 + 62 - 82

2 x 4 x 6cos A = - 0.25

A = - 0.25 cos-1 A = 104. 5 (1dp)

Exercise Find the length marked x

2cm

5cm

x4cm

(1)x

11cm9cm

5cm

(2)

a

b

c

A

cos A = 22 + 52 - 42

2 x 2 x 5cos A = 0.65

A = 0.65 cos-1 A = 49.5 (1dp)

a

A

b

c

cos A = 92 + 52 - 112

2 x 9 x 5cos A = - 0.1666….

A = - 0.166.. cos-1 A = 99. 6 (1dp)

Example 1Find angle A

4 cm

8 cm6 cm

A

B

C

You are given 3 sides and asked for an angle.

cos A = b2 + c2 - a2

2bca

b

c cos A = 42 + 62 - 82

2 x 4 x 6cos A = - 0.25

A = - 0.25 cos-1

A= 104. 5 (1dp)