NCS Mathematics

DVD Series

Sine, Cosine

and Area Rules

4. Apply the Sine and the Cosine rules to solve

problems in 2-dimensions. Le ss on 4

In this DVD we will:

1. Calculate the area of a triangle given an angle and the two

adjacent sides. Lesson 1

2. Apply the Sine Rule for triangles to calculate an unknown

side or an unknown angle of a given triangle. Le ss on 2

Outcomes for this DVD

3. Apply the Cosine Rule for triangles to calculate an unknown

side or an unknown angle of a given triangle. L e ss on 3

NCS Mathematics

DVD Series

Lesson 1

The Area Rule

sin opposite

hypotenuse

opposite hypotenuse

Trigonometric Ratios

In a , the 3 trigonometric ratios for an angle

are defined as follows:

right angled triangle

cos adjacent

hypotenuse

adjacent

tan opposite

adjacent

Some basic definitions – a reminder

Consider a non-right angled triangle ABC.

, and are the sides opposite angles , and respectively.

( This is the conventional way of labelling a triangle ).

a b c A B C

A B

C

b a

c N

h

1Area of base height

2

12

Area c h --- (1)

In ,ACN sin Ah

bsin b A h

Substituting for h in (1)

12

Area c sin b A12

sinArea bc A

The area formula of a triangle

Draw the perpendicular, , from to . h C BA

Any angle can be used as such in

area formula, so

12

sin Bca12

sinab C 12

sinbc AArea = = =

90

A similar argument gives the same formula for the area

if is obtuse i.e. B B

The formula always uses

2 sides and the

angle formed by those sides (Included )

A B

C

b

a

c N

h

90

Different forms of the area formula

c

b a a

C

B A

c

b a a

C

B A

c

b a a

C

B A

Three possible approaches to find

the area of a triangle

12

Area sin ab C

Any angle can be used in the formula, so

12

Area sin bc A

12

Area sin Bca

Find the area of PQR.

We know PQ and RQ so use the included angle Q

The area of a triangle – Example 1

Solution: We must use the angle formed by the

2 sides with the given lengths.

64

1Area of sin

2PQR QP QR Q

218 7 sin 64 cm

2

225,2 cm

r

B

A

C

r

21sin

2 Area r

Find the area of .ABCA useful application of the area formula occurs when we

hav 2 radiie a triangle formed by and of a a cho cirrd cle.

The area of a triangle – Example 2

1

Area sin2

CA CB C

But CA CB r

Tutorial 1: Area of a TriangleFind the areas of the triangles shown in the diagrams.

Give your answers accurate to 2 decimal digits

PAUSE

• Do Tutorial 1

• Then View Solutions

40

308 cm

10 cmradius 6 cm

120AOB

1) 2)

Tutorial 1: Problem 1: Area of a Triangle: Solution

40

308 cm

10 cm

11) Area sin

2XYZ XY YZ Y

180 40 30 110Y

1sin

2z x Y

218 10 sin 110 cm

2

237,59 cm

Tutorial 1: Problem 2: Area of a Triangle: Solution

radius 6 cm

120AOB

212) Area sin

2AOB r O

21

6 cm sin1202

215,59 cm

Given:

NCS Mathematics

DVD Series

Lesson 2

The Sine Rule

One way to find unknown sides and angles in

is by using the :

non - right angled

triang Sine Rs ulele

The Sine Rule for Triangles

In ACN, sinh

Ab

sinh b A

Suppose is a scalene triangleABC

In , sinh

BCN Ba

sinh a B

sin sinb A a B

sin sinor

A B

a b

ab

c N

Drop CN AB

h

C

A B

b a

c

A

B C

b

a

c

h

Now sin sin

sin sin

h c B b C

B C

b c

can be turned so that is the base.

We then get

ABC BC

The Complete Sine Rule for Triangles

sin sin sinSo

A B C

a b c

When do we use the Sine Rule?The sine rule can be used in a triangle when:

Two angles and a side are given

Two sides and the non-included angle are given

To calculate second side

To calculate second angle

sin sinSolution: Use

A B

a b

sinsin

a BA

b10 sin 62

sin12

A

180 62 47,4 70,6 CThus

Application of the Sine Rule - Example 1

ABC A CIn , find the size of angles and .

47,4 A is opposite the shorter of the 2 given sides.

62 must be an acute angle.

(Only one possibility as can be seen from sketch)

A

A A

sin 5 sin 48sin

4

p QP

q

2

1

68,3

180 68,3 111,7

or

P

P

Application of the Sine Rule - Example 2

is opposite the longer of the 2 given sides.

48 can be an acute or obtuse angle.

( Two possibilities as can be seen from sketch below)

P

P P

1P

2Psin sin Use Solution :

Q P

q p

5, 4 48 .

.

PQR

QR PR Q

P

In it is given that:

and

Determine

13sin 55

sin 29

z

22,0 z

sin sin

z y

Z Y

In , find the length .XYZ XY

Application of the Sine Rule - Example 3

As the unknown is a side, we use the sine rule in

its reciprocal form. The unknown side is then at the top.

Solution :

sin

sin

y Zz

Y

Tutorial 2: Sine Rule

2. In , find and the area of PQR QR PQR

1. In , find .

(Correct to two decimal places)

ABC B

PAUSE DVD

• Do Tutorial 2

• Then View Solutions

Tutorial 2: Problem 1: Sine rule: SolutionFind .

(2 decimal places)

B

10sin35sin

7B

1 55,02B

2or 180 55,02 124,98B

Given:

35 acute or obtuseB B

sin sin sin35 sin

7 10

A B B

a b

Tutorial 2: Problem 1: Why two solutions?

1

2

Obtained: 55,02

or 124,98

B

B

Given:

35 acute or obtuseB B

1 (Obtuse)B 2 (Acute)B

2. Find and the area of .QR PQR

67

sin 64 sin80

QR

Tutorial 2: Problem 2: Sine rule: Solution

Given:

80R

67sin 64

sin80QR

61,15 cm

1Area of sin36

2PQR QP QR

167 61,15 sin36

2

21 204,09 cm

NCS Mathematics

DVD Series

Lesson 3

The Cosine Rule

2 2 2

2 2 2

2 cos

or

2 sin

b a c ac B

c a b ab C

c

b a a

C

B A

The Cosine Rule for is given by:ABC

The Cosine Rule for Triangles

2 2 2 2 cosa b c bc A

We use this form to find the third side when

two sides and included angle are given.

Symmetry also implies that:

Proof of the Cosine Rule

b a

In :CAD

2 2 2cos and x

A b x hb

In :BCD

22 2 2 2 22a h c x h c cx x

x c x

h

2 22 2 2cos2a c xb x c b A

2 2 2 cosb c bc A

Proofs for symmetrical results are similar.

A second form of the Cosine Rule 2 2 2 2 cos a b c bc AKnow:

2 2 22 cos bc A b c a 2 2 2

cos2

b c aA

bc

We use this form to find any angle of

a triangle when we know all 3 sides.

7 P R

Q

6

p

120

Find in the p PQR

Applications of the Cosine Rule - Example 1

Apply the Cosine Rule

2 2 2 2 cosp q r qr P

2 2 27 6 2 7 6 cos120p 127 11,3 1 decimal accuracyp

6

Y

Z 8

4

X

2 2 28 6 4cos

2(8)(6)

X

Find in the X XYZ

Solution: Use the Cosine Rule

Applications of the Cosine Rule - Example 2

29,0 ( 1 dec )X

2 2 2

cos2

y z xX

yz

2 2 2 2 cos c b a ba C

A

B C

c

30

Sine rule: sin sin

B C

b c

Find side and in the given . c B ABC

15b

19a

Cosine rule:

Applications of the Cosine Rule Example 3

2 2 215 19 2(15)(19)cos30 c

9,61 c ( 2 decimal places )

15 sin 30sin

9,61

B

51,3 B ( 1 dec. )

Tutorial 3: Cosine Rule

2. Find all the angles in , giving your

answers to one decimal place accuracy.

XYZ

1. Given with 6 cm; 4 cm and

60 . Find correct to 2 decimal digits.

ABC AB BC

ABC AC

PAUSE DVD

• Do Tutorial 3

• Then View Solutions

Tutorial 3: Problem 1: Cosine Rule: Solution

2 2 2 2 cosAC BC AB BC AB ABC

Given:Find (2 dec accuracy):AC

2 24 6 2 4 6 cos60

28

28 5,29 cmAC

Tutorial 3: Problem 2: Cosine Rule: Solution

sin sin sin 4sin 48.2Now sin

7

X Y y XY

x y x

Given:

Determine all angle measures of XYZ.

2 2 2

cos2

y z xX

yz

2 2 24 9 7Hence cos

2 4 9X

48,2X

25,2Y

Then 180 106,6Z X Y

NCS Mathematics

DVD Series

Lesson 4

Basic Applications:

Problems in 2-D

42 65

Problems in 2 dimensions: Example 1 1. Points and are in the same horizontal plane as ,

the foot of a vertical tower . 42 ; 65

and 25 . Calculate .

A B C

PC B PAC

AB m PC

P

CAB25 m

65 42 23BPA

25

sin 42 sin 23

AP

25sin 4242,81 m

sin 23AP

sin 65PC

AP

sin65 42,81sin65PC AP 38,8 m

23

Sine rule:

2. In the figure represents a proposed tunnel.

and are visible from a point .

The three points are in the same plane.

QR

Q R P

Q R

P

Given:

100 m; 60 m

and 110

QP PR

QPR

Calculate the length of tunnel.

100 60110

2 2 2100 60 2 100 60 cos110QR

133 mQR

Problems in 2 dimensions: Example 2

Tutorial 4: Part 1: Problems in 2 D

1. From the ends of a bridge , 101 metres long,

the angles of depression of a point on the ground

directly under the bridge is 42,2 and 70,1 .

Find the height, , of the bridge und

AB

P

h

er this point.

42,2 70,1A B101 m

P

h PAUSE DVD

• Do Tutorial 4 Part 1

• Then View Solutions

Tutorial 4: Part 1: Suggested Solution

42,2 70,1A B101 m

P

h

Question: Find h

180 42,2 70,1 67,7APB

101

sin 70,1 sin 67,7

AP

101 m sin 70,1

sin 67,7AP

102,65 mAP But sin 42,2102,65

h

102,65 sin 42,2 68,95 mh

Tutorial 4: Part 2: Problems in 2-D2. is a wall of a room, being the

line of the ceiling. is a picture rail,

with being directly below .

= 2 metres; and

2cos(a) Prove that

sin( )

(b)

ABCD AD

EF

E A

AE ACB x ECB y

xEC

x y

Find the length and height

of the wall if 33 and 20 .x y

2 m

xy

PAUSE DVD

• Do Tutorial 4 Part 2

• Then View Solutions

Tutorial 4: Part 2(a) Solution

2 m

xy

Now and ACE x y CAD x

x

Hence, 90CAE x

From :AEC

2

sin 90 sin

EC

x x y

2sin 90

sin

xEC

x y

2cos

sin

x

x y

2(a) Prove that

2cos

sin( )

xEC

x y

2 m

xy

Know:

2cos

sin

33 and 20

xEC

x y

x y

2cos337,46 m

sin13EC

cos cosBC

y BC EC yEC

Length of room 7,46 m cos20BC

7,01 m

Tutorial 4: Part 2: Length of room

End of the DVD on

Sine, Cosine and Area Rules

REMEMBER!

•Consult text-books for additional examples.

•Attempt as many as possible other similar examples

on your own.

•Compare your methods with those that were

discussed in the DVD.

•Repeat this procedure until you are confident.

•Do not forget:

Practice makes perfect!