CHAPTER 10- SOLUTION OF TRIANGLES

Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 10-Solution Of Triangles

CHAPTER 10- SOLUTION OF TRIANGLES

10.1 SINE RULE 10.1.1 Verifying the sine rule

A (1)

c

b

h

B

C

D

a A

h

= sin B c h = c sin B

c h

B

1

D

A h

= sin C b h = b sin C

b h

D

2

C

Compare

1

and

2

,

sin C = c sin B b sin sin B b

=

sin C c

or

b sin sin B

=

c sin C

Page | 132



(2)

A c

E b

a

B

C

A t

= sin A c t = c sin A

c E

1

t

B E t

t

B

C

a

Compare

1

and

2

= sin C a t = c sin C

2

,

c sin A = a sin C

sin A

=

sin C

a

c

or

a

c

=

sin A

sin C

From the first solution we know that

sin B

sin C

=

b

c

or

b

c

=

sin B

sin C

From the second solution we know that

sin A

=

sin C

a

c

or

a

sin A

c

=

sin C

Hence,

sin A a

=

sin B b

or

=

sin C c

or

a sin A

=

b sin B

=

c sin C

Page | 133



10.1.2 Using the sine rule Example 1:

C

60 °

A

40 ° 5cm

B

The diagram above shows a triangle ABC. Calculate (a) the length of BC (b) the length of AC Solution:

From the given information, we know that ∠ ACB = 180 ° − 60° − 40 ° =

80°

(a) Using the sine rule,

BC

=

5

sin 60° sin 80° 5 sin 60° BC = sin 80° =

4.3969cm

(b) Using the sine rule,

AC

=

5

sin 40° sin 80° 5 sin 40° BC = sin 80° =

3.2635cm

Page | 134



Example 2:

C 135 °

8cm

A

B

12 cm

The diagram above shows a triangle ABC. Calculate (a) ∠BAC (b) the length of AC (a) Using the sine rule,

sin ∠BAC

=

8 sin ∠BAC = = ∠BAC = =

sin 135° 12 8 sin 135° 12 0.4714 −1

sin (0.4714) 28°8'

(b) At first, calculate the angle ∠ ABC ∠ ABC = 180 ° − 135 ° −

28 °8'

= 16°52'

Using the sine rule,

AC

=

12

sin 16°52' sin 135° 12 sin 16°52' AC = sin 135° =

4.9239cm

Page | 135



EXERCISE 10.1 1. ABC is a triangle where AB = 12cm , AC = 8cm and ∠ ABC = 30° . Find two possible values of ∠CAB

2. In diagram below, KLM is a straight line.

20 cm 12 cm

K

8cm

L

M

Calculate (a) ∠ JLK (b) ∠LJM

3. In diagram below, ABC and BED are straight lines, E is the mid-point of BD.

C

6.6cm

B E

D

9.8cm

A Given that sin ∠CBD = 0.7 , calculate (a) the length of BC (b) ∠BEA 4. Find the value of θ in each of the following triangles. (a)

(b)

C

P 4.4cm 35 °

θ θ

6cm

6.7 cm

40 °

A

Q

9cm

B

R Page | 136



10.2 AN AMBIGUOUS CASE

C

b

A

a

a

B1

B2

An ambiguous case occurs when ∠ A , length of AC are fixed. While a < b. There are two possible triangles that can be constructed.

C C b

b

a A

B

A

a

B

Example: ABC is a triangle with ∠ A = 28° . AB= 14 cm and BC = 9cm. Solve the triangle.

B

14 cm

9cm

28 °

A

C

Solution:

B 14 cm

9cm 9cm

28 °

A

C 1

C 2

To solve the triangle, we have to find ∠ ABC , ∠ ACB and the length of AC. There are two possible triangles that can be constructed.

Page | 137



B B ?

14 cm

14 cm

?

9cm

9cm 28 ° ?

A

sin ∠ ACB =

C

?

sin 28°

=

14

?

A

C

?

sin ∠ ACB

28 °

9 14 sin 28° 9

=

0.7303

∠ ACB =

46°55'

For another one triangle, ∠ ACB = 180° − 46°55' = 133°5'

B B ?

14 cm

14 cm

9cm 28 ° 133 °5'

A

C

?

∠ ABC = 180° −

?

9cm 46 °55'

28 °

A

C

?

46°55'−28° , 180° − 105°5'−28°

= 105°5' , = 18°55'

AC

=

sin 18°55' AC =

9 sin 28° 9 sin 18°55'

sin 28° = 16.2149cm

,

AC

=

sin 105°5' AC =

9 sin 28°

9 sin 105°5'

sin 28° = 18.51cm

Page | 138



EXERCISE 10.2 1. Diagram below shows triangle PQR.

P

6.2cm 130 ° Q

R

4.8cm

Calculate: (a) the length of PQ (b) The new length of PR if the lengths PQ, QR and ∠QPR are maintained. 2. Diagram below shows two triangles ABC and CDE . The two triangles are joined at C such that AE and BD are straight lines. The ∠CED is an obtuse angle.

A

7cm 4cm

B

9cm 5cm

D

C 6.5cm

E (a) Calculate (i) ∠ ACB (ii) ∠DEC (b) The straight line CE is extended to F such that DE = DF. Find the area of triangle CDF .

Page | 139



10.3 COSINE RULE 10.3.1 Verifying the cosine rule

A

c

b

h

a − x

x D

B

C

a A x

= cos B c x = c cos B

c h

2

2

c = x + h B

x

1 2

2

D

A 2

2

b = h + (a − x ) b h

a − x

D

Substitute

b

2

=

a

2

+

=

h 2 + a 2 + x 2 − 2ax

=

a + h + x − 2ax

2

2

2

3

C and

1

c

2

2

−

2

into

3

,

2a(c cos B)

Hence, 2

2

2

2

2

2

a = b + c − 2 bc cos A b = a + c − 2ac cos B c

2

=

2 2 a + b − 2ab cos C

Page | 140



EXERCISE 10.3 1. Given a triangle ABC , AB = 7.3 cm, AC = 9.3 cm and ∠CAB = 65° . Calculate the length of BC . 2. Given a triangle PQR, PQ = 7 cm, QR = 9 cm and PR = 15 cm. Calculate the length of ∠PQR . 3. Diagram below shows a triangle PQR.

Q

10cm 12cm R 13cm

P

Calculate ∠PQR . 4. In diagram below, KMN is an equilateral triangle. H is the midpoint of KN and KL = 8 cm.

K

8cm H L M

N

12cm

Caclulate (a) the length of LH (b) ∠KLH 5.

P 12cm 8cm S

x Q

10.7cm

R

In diagram above, calculate (a) the length of PR (b) the value of x.

Page | 141



10.4 AREA OF TRIANGLE A

The formula for area of triangle that is

1 c

2

b

× base ×

height can only be used in

situation where there is right angle triangle.

h

In the situations that the triangle is not a B

a

right-angled triangle, we cannot use the

C

formula.

SinC =

h

b h = b sin C SinA =

1

h

AB h = C sin B Area =

1 2

This formula can be used to find the area of

2

×a×h

3

all types of triangle as long as there is enough information given. The sine of an angle is multiplied by the length of line that

Substitute

Area =

1 2

1

× a × b sin

1

=

2

1 2

=

2

2

joining to form the angle. For example, sine A is multiply by AB and AC t hat are the lines

C

into

× a × c sin

1

,

that joining to form the angle A.

ab sin C

Substitute

Area =

into

3

3

,

B

ac sin B

Hence,

Area = Area = Area =

1 2 1 2 1 2

ab sin C ac sin B bc sin A

Page | 142



EXERCISE 10.4 1. PQR is a triangle where PQ = 7.3 cm, QR = 9.6 cm and PR = 14.7 cm. Calculate (a) the area of ∆PQR (b) the height of P from QR 2.

A

10 .9cm 8.2cm

B

C 6.4cm

In diagram above, calculate the area of triangle ABC . 3. In diagram below, BCD is a straight line.

A

10 .6cm

6.5cm 73 °

B

C

Calculate

5.7cm

D

(a) ∠ ACD (b) the length of AB (c) the area of ∆ ABC CHAPTER REVIEW EXERCISE 1. A

The diagram shows a triangle ABC. (a) If the length of PQ and PR and the size of ∠ ACB are

9.2cm

maintained, sketch and label another triangle different from ∆ ABC in the figure.

6.5cm

(b) Calculate the two possible values of BC .

33 °

B

C

Page | 143



R

2.

40 ° 10 cm

S 6cm

P

15 cm

In diagram above, sin ∠PSR =

5 6

Q

where ∠PSR is an obtuse angle. Calculate

(a) the length of PR, correct to two decimal places (b) ∠PQR (c) the area of the whole diagram 3. Diagram below shows triangle ABC and triangle AED. AEC is a straight line.

A 5cm

8.5cm

E

D

B

15 .6cm 8cm

C

Given that ∠BAC = 60 ° , AB = 5 cm. BC = 8 c,. AE = 8.5 cm an ED = 15.6 cm. Calculate (a) the length of EC 2

(b) ∠ AED , if the area of triangle AED is 54 cm . 4. Diagram below shows a right prism with an isosceles triangular base where DE =DF = 10 cm. FE = 8 cm and AD = 7 cm.

Calculate

A

C

(a) the angle between the line AE and the base FED (b) ∠FAE

7cm

D F 10 cm

8cm

E

Page | 144

CHAPTER 10- SOLUTION OF TRIANGLES

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