TrigonometryTrigonometry

Cosine Rule Finding a Length

Sine Rule Finding a length

Mixed Problems

Sine Rule Finding an Angle

Cosine Rule Finding an Angle

Area of ANY Triangle

Learning IntentionLearning Intention Success CriteriaSuccess Criteria

1.1. Know how to use the sine Know how to use the sine rule to solve REAL LIFE rule to solve REAL LIFE problems involving problems involving lengths.lengths.

1. To show how to use the sine rule to solve REAL LIFE problems involving finding the length of a side of a triangle .

Sine RuleSine Rule

C

B

A

Sine RuleSine Rule

a

b

c

The Sine Rule can be used with ANY triangle as long as we have been given enough information.

Works for any Triangle

a b c= =

SinA SinB SinC

Deriving the rule

B

C

A

b

c

a

Consider a general triangle ABC.

The Sine Rule

Draw CP perpendicular to BA

P

CPSinB CP aSinB

a

CP

also SinA CP bSinAb

aSinB bSinA

aSinBb

SinA

a bSinA SinB

This can be extended to

a b cSinA SinB SinC

or equivalentlySinA SinB SinCa b c

Calculating Sides Calculating Sides Using The Sine RuleUsing The Sine Rule

10m

34o

41o

a

Match up corresponding sides and angles:

sin 41oa

10

sin 34o

Rearrange and solve for a. 10sin 41

sin34

o

oa 10 0.656

11.740.559

a m

Example 1 : Find the length of a in this triangle.

A

B

C

sin sin sino

a b c

A B C

Calculating Sides Calculating Sides Using The Sine Using The Sine

RuleRule

10m133o

37o

d

sin133od

10

sin 37o

10sin133

sin 37

o

od

10 0.731

0.602d

=

12.14m

Match up corresponding sides and angles:

Rearrange and solve for d.

Example 2 : Find the length of d in this triangle.

C

D

E

sin sin sino

c d e

C D E

What goes in the Box What goes in the Box ??

Find the unknown side in each of the triangles below:

(1)12cm

72o

32oa

(2)

93o

b47o

16mm

A = 6.7cm

B = 21.8mm

Learning IntentionLearning Intention Success CriteriaSuccess Criteria

1.1. Know how to use the sine Know how to use the sine rule to solve problems rule to solve problems involving angles.involving angles.

1. To show how to use the sine rule to solve problems involving finding an angle of a triangle .

Sine RuleSine Rule

Calculating Angles Calculating Angles

Using The Sine Using The Sine RuleRule

Example 1 :

Find the angle Ao

A

45m

23o

38m

Match up corresponding sides and angles:

45

sin oA 38

sin 23o

Rearrange and solve for sin Ao

45sin 23sin

38

ooA = 0.463 Use sin-1 0.463 to find Ao

1sin 0.463 27.6o oA

sin sin sin

a b c

A B C

B

C

Calculating Angles Calculating Angles

Using The Sine Using The Sine RuleRule

143o

75m

38m

X

38

sin oX

75

sin143o

38sin143sin

75

ooX = 0.305

1sin 0.305 17.8o oX

Example 2 :

Find the angle Xo

Match up corresponding sides and angles:

Rearrange and solve for sin Xo

Use sin-1 0.305 to find Xo

Y

Z

sin sin sin

x y z

X Y Z

What Goes In The Box What Goes In The Box ??

Calculate the unknown angle in the following:

(1)

14.5m

8.9m

Ao

100o(2)

14.7cm

Bo

14o

12.9cm

Ao = 37.2o

Bo = 16o

Learning IntentionLearning Intention Success CriteriaSuccess Criteria

1.1. Know when to use the Know when to use the cosine rule to solve cosine rule to solve problems.problems.

1. To show when to use the cosine rule to solve problems involving finding the length of a side of a triangle .

Cosine RuleCosine Rule

2. 2. Solve problems that Solve problems that involve finding the length involve finding the length of a side.of a side.

C

B

A

Cosine RuleCosine Rule

a

b

c

The Cosine Rule can be used with ANY triangle as long as we have been given enough information.

Works for any Triangle

cos2 2 2a =b +c - 2bc A

Deriving the rule

A

B

C

a

b

c

Consider a general triangle ABC. We require a in terms of b, c and A.

Draw BP perpendicular to AC

b

Px b - x

BP2 = a2 – (b – x)2

Also: BP2 = c2 – x2

a2 – (b – x)2 = c2 – x2

a2 – (b2 – 2bx + x2) = c2 – x2

a2 – b2 + 2bx – x2 = c2 – x2

a2 = b2 + c2 – 2bx*

a2 = b2 + c2 – 2bcCosA*Since Cos A = x/c x = cCosA

When A = 90o, CosA = 0 and reduces to a2 = b2 + c2

1

When A > 90o, CosA is negative, a2 > b2 + c2 2

When A < 90o, CosA is positive, a2 > b2 + c2 3

The Cosine Rule

The Cosine Rule generalises Pythagoras’ Theorem and takes care of the 3 possible cases for Angle A.

a2 > b2 + c2

a2 < b2 + c2

a2 = b2 + c2

A

A

A

1

2

3

Pythagoras + a bitPythagoras - a bit

Pythagoras

a2 = b2 + c2 – 2bcCosA

Applying the same method as earlier to the other sides produce similar formulae

for b and c. namely:b2 = a2 + c2 – 2acCosB

c2 = a2 + b2 – 2abCosC

A

B

C

a

b

c

The Cosine Rule

The Cosine rule can be used to find:

1. An unknown side when two sides of the triangle and the included angle are given (SAS).

2. An unknown angle when 3 sides are given (SSS).

Finding an unknown side.

Cosine RuleCosine Rule

How to determine when to use the Cosine Rule.

Works for any Triangle

1. Do you know ALL the lengths.

2. Do you know 2 sides and the angle in between.

SASOR

If YES to any of the questions then Cosine Rule

Otherwise use the Sine Rule

Two questions

Using The Cosine Using The Cosine RuleRule

Example 1 : Find the unknown side in the triangle below: L5m

12m

43o

Identify sides a,b,c and angle Ao

a =

L b =

5 c =

12 Ao = 43o

Write down the Cosine Rule.

Substitute values to find a2.a2 =

52 + 122 - 2 x 5 x 12 cos 43o

a2 =

25 + 144

- (120 x

0.731 )

a2 =

81.28 Square root to find “a”.

a = L = 9.02m

Works for any Triangle

Example 2 :

Find the length of side M.

137o17.5 m

12.2 m

MIdentify the sides and angle.

a = M

b = 12.2 C = 17.5 Ao = 137o

Write down Cosine Rule

a2 = 12.22 + 17.52 – ( 2 x 12.2 x 17.5 x cos 137o )

a2 = 148.84 + 306.25 – ( 427 x – 0.731 )Notice the two negative

signs.a2 = 455.09 + 312.137

a2 = 767.227

a = M = 27.7m

Using The Cosine Using The Cosine RuleRuleWorks for any Triangle

What Goes In The What Goes In The Box ?Box ?

Find the length of the unknown side in the triangles:

(1)78o

43cm

31cmL

(2)

8m

5.2m

38o

M

L = 47.5cm

M =5.05m

Learning IntentionLearning Intention Success CriteriaSuccess Criteria

1.1. Know when to use the Know when to use the cosine rule to solve cosine rule to solve REAL LIFE problems.problems.

1. To show when to use the cosine rule to solve REAL LIFE problems involving finding an angle of a triangle .

Cosine RuleCosine Rule

2. 2. Solve Solve REAL LIFE problems problems that involve finding an that involve finding an angle of a triangle.angle of a triangle.

C

B

A

Cosine RuleCosine Rule

a

b

c

The Cosine Rule can be used with ANY triangle as long as we have been given enough information.

Works for any Triangle

cos2 2 2a =b +c - 2bc A

Finding Angles Finding Angles Using The Cosine RuleUsing The Cosine Rule

Consider the Cosine Rule again:We are going to change the subject of the formula to cos Ao

Turn the formula around:b2 + c2 – 2bc cos Ao = a2

Take b2 and c2 across.-2bc cos Ao = a2 – b2 – c2

Divide by – 2 bc.2 2 2

cos2

o a b cA

bc

Divide top and bottom by -12 2 2

cos2

o b c aA

bc

You now have a formula for finding an angle if you know all three sides of the triangle.

Works for any Triangle

Write down the formula for cos Ao

2 2 2

cos2

o b c aA

bc

Label and identify Ao and a , b and c.

Ao = ? a = 11b = 9 c = 16

Substitute values into the formula.

2 2 29 16 11cos

2 9 16oA

Calculate cos Ao .Cos Ao =0.75

Use cos-1 0.75 to find Ao

Ao = 41.4o

Example 1 : Calculate the

unknown angle Ao .

Finding Angles Finding Angles Using The Cosine RuleUsing The Cosine Rule

Works for any Triangle

Example 2: Find the unknown

Angle yo in the triangle:

Write down the formula.

2 2 2

cos2

o b c aA

bc

Identify the sides and angle.

Ao = yo a = 26 b = 15 c = 13

2 2 215 13 26cos

2 15 13oA

Find the value of cosAo

cosAo = - 0.723The negative tells you the angle is obtuse.

Ao = yo = 136.3o

Finding Angles Finding Angles Using The Cosine RuleUsing The Cosine Rule

Works for any Triangle

What Goes In The Box ?What Goes In The Box ?

Calculate the unknown angles in the triangles below:

(1)

10m

7m5m Ao

Bo

(2) 12.7c

m

7.9cm

8.3cm

Ao =111.8o

Bo = 37.3o

Learning IntentionLearning Intention Success CriteriaSuccess Criteria

1.1. Know the formula for the Know the formula for the area of any triangle.area of any triangle.

1. To explain how to use the Area formula for ANY triangle.

Area of ANY TriangleArea of ANY Triangle

2.2. Use formula to find area of Use formula to find area of any triangle given two any triangle given two length and angle in length and angle in between.between.

Labelling TrianglesLabelling Triangles

A

B

C

A

aB

b

Cc

Small letters a, b, c refer to distancesCapital letters A, B, C refer to angles

In Mathematics we have a convention for labelling triangles.

F

E

D

F

E

D

Labelling TrianglesLabelling Triangles

d

e

f

Have a go at labelling the following triangle.

General Formula forGeneral Formula forArea of ANY TriangleArea of ANY Triangle

Consider the triangle below:

Ao Bo

Co

ab

c

h

Area = ½ x base x height 1

2A c h

What does the sine of Ao equal

sin o hA

b

Change the subject to h. h = b

sinAoSubstitute into the area formula

1sin

2oA c b A

1sin

2oA bc A

Area of ANY TriangleArea of ANY Triangle

A

B

C

A

aB

b

Cc

The area of ANY triangle can be found by the following formula.

sin1

Area = ab C2

sin1

Area = ac B2

sin1

Area = bc A2

Another version

Another version

Key feature

To find the areayou need to knowing

2 sides and the angle in between (SAS)

Area of ANY TriangleArea of ANY Triangle

A

B

C

A

20cmB

25cm

Cc

Example : Find the area of the triangle.

sinC1

Area = ab2

The version we use is

30o

120 25 sin30

2oArea

210 25 0.5 125Area cm

Area of ANY TriangleArea of ANY Triangle

D

E

F

10cm

8cm

Example : Find the area of the triangle.

sin1

Area= df E2

The version we use is

60o

18 10 sin 60

2oArea

240 0.866 34.64Area cm

What Goes In The Box What Goes In The Box ??

Calculate the areas of the triangles below:

(1)

23o

15cm

12.6cm

(2)

71o

5.7m

6.2m

A = 36.9cm2

A = 16.7m2

Key feature

Remember (SAS)

Learning IntentionLearning Intention Success CriteriaSuccess Criteria

1.1. Be able to recognise the Be able to recognise the correct trigonometric correct trigonometric formula to use to solve a formula to use to solve a problem involving problem involving triangles.triangles.

1. To use our knowledge gained so far to solve various trigonometry problems.

Mixed problemsMixed problems

SOH CAH TOA

25o

15 mAD

The angle of elevation of the top of a building

measured from point A is 25o. At point D which is

15m closer to the building, the angle of elevation is

35o Calculate the height of the building.

T

B

Angle TDA =

145o

Angle DTA =

10o

o o

1525 10

TDSin Sin

o15 2536.5

10Sin

TD mSin

35o

36.5

o3536.5TB

Sin

o36.5 25 0. 93TB Sin m

180 – 35 = 145o

180 – 170 = 10o

sin sin sin

t d a

T D A

Exam Type Questions

A fishing boat leaves a harbour (H) and travels due East for 40 miles to a marker buoy (B). At B the boat turns left and sails for 24 miles to a lighthouse (L). It then returns to harbour, a distance of 57 miles.

(a) Make a sketch of the journey.

(b) Find the bearing of the lighthouse from the harbour. (nearest degree)

H40 miles

24 miles

B

L

57 miles

A

2 2 257 40 242 57 40

CosAx x

A 20.4o

90 0 020.4 7 oBearing

Exam Type Questions

A

The angle of elevation of the top of a column measured from point A, is 20o. The angle of elevation of the top of the statue is 25o. Find the height of the statue when the measurements are taken 50 m from its base

50 m

Angle BCA =

70o

Angle ACT = Angle ATC =

110o

65o

o 5020Cos

AC o

5020

ACCos

53.21 m

o o

53.215 65

TCSin Sin

o

53.21 5 (1 )

655.1

SinTC m dp

Sin

B

T

C

180 – 110 = 70o 180 – 70 = 110o 180 – 115 = 65o

20o

25o

5o

SOH CAH TOA

53.21 (2 )m dp

Exam Type Questions

sin sin sin

t d a

T D A

2 2 2

2b c a

CosAbc

An AWACS aircraft takes off from RAF Waddington (W) on a navigation exercise. It flies 530 miles North to a point (P) as shown, It then turns left and flies to a point (Q), 670 miles away. Finally it flies back to base, a distance of 520 miles.

Find the bearing of Q from point P.

2 2 2530 670 5202 530 670

CosPx x

48.7oP

180 22948.7 oBearing

P

670 miles

W

530 miles

Not to Scale

Q

520 miles

Exam Type Questions