CAPS Mathematics

GRADE 11

Sine, Cosine

and Area Rules

4. Apply the Sine and the Cosine rules to solve

problems in 2-dimensions. Le ss on 4

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 Topic

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

Lesson 1

The Area Rule

CAPS Mathematics

GRADE 11

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

Find the areas of the triangles shown in the diagrams.

Give your answers accurate to 2 decimal digits

40

308 cm

10 cmradius 6 cm

120AOB

1) 2)

Classwork [Area of Triangles]

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

Solution [Area of Triangles]

radius 6 cm120AOB

212) Area sin

2AOB r O

21

6 cm sin1202

215,59 cm

Given:

Solution [Area of Triangles]

Lesson 2

The Sine Rule

CAPS Mathematics

GRADE 11

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

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

1. In , find .

(Correct to two decimal places)

ABC B

Classwork [Sine Rule]

Find .

(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

Solution

1

2

Obtained: 55,02

or 124,98

B

B

Given:

35 acute or obtuseB B

1 (Obtuse)B2 (Acute)B

Solution

2. Find and the area of .QR PQR

67

sin 64 sin80

QR

Given:

80R

67sin 64

sin80QR

61,15 cm

1Area of sin36

2PQR QP QR

167 61,15 sin36

2

21 204,09 cm

Solution

Lesson 3

The Cosine Rule

CAPS Mathematics

GRADE 11

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. )

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

Classwork

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

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

Solution

Lesson 4

Basic Applications:

Problems in 2-D

CAPS Mathematics

GRADE 11

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

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

Activity 1

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

Solution1

2. 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

Activity 2

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

Solution2

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

Solution 2b