1515 15.1Area of Triangles 15.2Sine Formula 15.3Cosine Formula Chapter Summary Case Study Trigonometry (2) 15.4Heron’s Formula

15

15.1 Area of Triangles

15.2 Sine Formula

15.3 Cosine Formula

Chapter Summary

Case Study

Trigonometry (2)

15.4 Heron’s Formula

P. 2

Today is the school’sOpen Day. John helps his teacher to hang a banner outside the school building.

In order to calculate the distance between the bottom of the banner and the ground, he should first find out the angles of elevation of the top and bottom of the banner.

Case StudyCase Study

Then he can use the formula

height of the observer’s eyes

to calculate the required distance h.

tantan

tan16

h

I have measured the banner and its length is 16 m. How can I know the distance between the bottom of the banner and the ground? It is not difficult. Let me tell

you how to find it out.

P. 3

1155 .1 .1 Area of TrianglesArea of Triangles

However, if we do not know the height of the triangle, then consider the following two cases:

Substituting h b sin C into (*),

(*)..........2

1heightbase

2

1Area ah

b

hC sin

Cbh sinb

hC )180sin(

b

hC sin

Cbh sin

Cabsin2

1 Area

For a given triangle, if we know the base and the height of the triangle, then:

P. 4


For a triangle with two sides and their included angle given, we can find the area of the triangle using the formula:

Similarly, we can show that:

BcaAbc sin2

1sin

2

1 Area

Remark:

If C is a right angle, the area of ABC becomes ab.

In this case, b is the height of the triangle.

Cabsin2

1 Area

2

1

P. 5

Example 15.1T

Solution:


The figure shows a trapezium with AD // BC. If AD 12 cm,AC 16 cm, AB 9.3 cm, DAC 35 and ABC 80, find the area of the trapezium. (Give the answer correct to 1 decimal place.)

(BAC 35) 80 180 (int. s, AD // BC)

BAC 65 Area of the trapezium

Area of ABC Area of ACD

2cm 35sin(12)(16)2

165sin(9.3)(16)

2

1

2cm 5.122 (cor. to 1 d. p.)

P. 6

Example 15.2T


ABC sin628325

628

325 sin ABC

ABC 39.1494 or 180 39.1494 141or 1.39 (cor. to 3 sig. fig.)

cm 27AB325The area of ABC is cm2. If and , find ABC. (Give the answer correct to 3 significant figures.)

cm 38BC

Solution:

Remarks:From this example, we find two possible values of ABC and we can draw ABC in two different ways.

ABCABC sin)cm 38)(cm 27(2

1 of Area

P. 7

Example 15.3T


AOBOBOAOAB sin2

1 of Area

AOB sin5)(12.5)(12.2

140

512.0sin AOBAOB 30.7971

22 cm 360

7971.30)5.12(πsector theof Area

41.9931cm2 Area of the shaded region (41.9931 40) cm2

2cm 99.1 (cor. to 3 sig. fig.)

The figure shows a sector with centre O and radius 12.5 cm.If the area of OAB is 40 cm2, find the area of the shaded region. (Give the answer correct to 3 significant figures.)

Solution:

P. 8

1155 ..22 Sine FormulaSine Formula

In the last section, we learnt that the area of a triangle, such as ABC in the figure, can be expressed in three different but equivalent forms:

CabBcaAbcABC sin2

1sin

2

1sin

2

1 of Area

If we divide each of the above expressions by , we have: abc2

1

c

C

b

B

a

A sinsinsin

The above results are known as sine formula. In other words,

Sine Formula For any triangle, the lengths of the sides are directly proportional to the sine ratios of the opposite angles.

C

c

B

b

A

a

sinsinsinor

From the sine formula, we havea : b : c sin A : sin B : sin C

P. 9


For any triangle, if we know the values of any two angles, then we can always find the remaining one by using the property of angle sum of triangle.

Moreover, if we know the length of one side, then we can use the sine formula to find the lengths of the other two sides.

A.A. Solving a Triangle with Two Angles andSolving a Triangle with Two Angles and Any Side GivenAny Side Given

P. 10

Example 15.4T


A B C 180 ( sum of )

60C

By sine formula,

C

c

A

a

sinsin

60sin

cm 10

40sin

a

cm60sin

40sin10

a

cm 42.7 (cor. to 3 sig. fig.)

C

c

B

b

sinsin

60sin

cm 10

80sin

b

cm60sin

80sin10

b

cm 4.11

In ABC, A 40, B 80 and c 10 cm. Solve the triangle and correct the answers to 3 significant figures if necessary.

Solution:

‘Solve a triangle’ means to find the measures of all unknown angles and sides of a triangle.

(cor. to 3 sig. fig.)


P. 11

Example 15.5T


40sin

cm 12

105sin

BD

cm40sin

105sin12

BD

18.0326 cm cm 0.18 (cor. to 1 d. p.)

BAD BCD 180 (opp. s, cyclic quad.)

In BCD, BDC 75 60 180 ( sum of )

The figure shows a cyclic quadrilateral ABCD. AB 12 cm, BAD 105, ADB 40 and DBC 60.(a) Find the lengths of BD and BC. (b) Hence find the area of quadrilateral ABCD. (Give the answers correct to 1 decimal place.)

Solution:(a) By sine formula,

105 BCD 180BCD 75

18.0 cm

?

?

?

BDC 45


P. 12

Example 15.5T


cm 2.13

(b) In ABD,ABD 105 40 180 ( sum of )

ABD 35

35sin))([(2

1BDAB

2cm 1.165

The figure shows a cyclic quadrilateral ABCD. AB 12 cm, BAD 105, ADB 40 and DBC 60.(a) Find the lengths of BD and BC. (b) Hence find the area of quadrilateral ABCD. (Give the answers correct to 1 decimal place.)

Solution:

18.0 cm

45

75

By sine formula,

13.2008 cm (cor. to 1 d. p.)

75sin

0326.18

45sin

BC

cm75sin

45sin0326.18

BC

2cm ]60sin))(( BCBD

(cor. to 1 d. p.)

Area of quadrilateral ABCD Area of ABD Area of BCD


P. 13


If two sides and one non-included angle are given, we can also use the sine formula to solve the triangle.

However, there may be more than one set of solution.

Construct a ABC such that AB 3 cm, ABC 30 and (a) AC 4 cm; (b) AC 2 cm; (c) AC 1.5 cm; (d) AC 1 cm. (Use a pair of compasses to construct each of the above lengths of AC.)

B. Solving a Triangle with Two Sides andB. Solving a Triangle with Two Sides and One Non-included Angle GivenOne Non-included Angle Given

Consider the following cases:

For example, in (a), since AC AB, only one triangle can be constructed:

P. 14


(a) AC 4 cm

In the above task, three cases can be concluded: Case 1: Only one triangle can be formed (a) and (c)Case 2: Two triangles are formed (b)Case 3: No triangle can be formed (d)

B. Solving a Triangle with Two Sides andB. Solving a Triangle with Two Sides and One Non-included Angle GivenOne Non-included Angle GivenThe construction:

(b) AC 2 cm

(c) AC 1.5 cm (d) AC 1 cm

C C

C

P. 15

Example 15.6T


C

c

A

a

sinsin

19

65sin15sin

A

7155.0

7.45A (cor. to 1 d. p.)


In ABC, a 15 cm, c 19 cm and C 65, find the possible values of A. (Give the answer correct to 1 decimal place.)Solution:

By sine formula,

Since the angle sum of triangle is 180, (180 45.7) is not a possible value of A.

P. 16

Example 15.7T


14

48sin16sin

A

8493.0 121.9or 1.58A


In ABC, a 16 cm, b 14 cm and B 48. (a) Find the possible values of A.

(Give the answers correct to 1 decimal place.) (b) How many triangles can be formed?

Solution:(a) By sine formula,

B

b

A

a

sinsin

(cor. to 1 d. p.)

(b) Two triangles can be formed.

P. 17

1155 ..33 Cosine FormulaCosine Formula

In the last section, we learnt how to use the sine formula to solve a triangle.

However, in some cases, such as the triangles in the following figures, we cannot solve the triangles using the sine formula.

We are going to learn another formula to solve triangles with two sides and the included angle given, or three sides given.

P. 18

Case 1 : A is an acute angle. Case 2 : A is an obtuse angle.

Let AP x

In PBC, we have h2 a2 (c x)2 (Pyth. theorem)

a2 c2 2cx x2 ....... (2)


Consider the following two cases, we draw a perpendicular line from C to meet AB (or its extension) at P. Let CP h.

In APC, we have h2 b2 x2 (Pyth. theorem) ... (1)

From (1) and (2),a2 c2 2cx b2

a2 b2 c2 2bc cos A

b cos A Let AP x

In PBC, we have h2 a2 (c x)2 (Pyth. theorem)

a2 c2 2cx x2 ....... (2)

In APC, we have h2 b2 x2 (Pyth. theorem) ... (1)

a2 b2 c2 2bc cos A

b cos(180 A) b cos A

a2 b2 c2 2cx

From (1) and (2),a2 c2 2cx b2

a2 b2 c2 2cx

P. 19


In the two cases, we obtain the same result:a2 b2 c2 2bc cos A

Using the same method as shown above, we can also show that b2 a2 c2 – 2ac cos B; c2 a2 b2 – 2ab cos C.

Case 1 : A is an acute angle. Case 2 : A is an obtuse angle.

The above results are known as cosine formulas.

Cosine Formulasa2 b2 c2 2bc cos A b2 a2 c2 2ac cos B c2 a2 b2 2ab cos C

P. 20


Thus, Pythagoras’ theorem is a special case of cosine formula for right-angled triangles.

Notes:When A 90,a2 b2 c2 2bc cos A

b2 c2 ( cos 90∵ 0)

P. 21

Example 15.8T


86BCD

cm 86cos)11)(8(2118 22 BD

cm 7229.172 13.1424 cm

cm 1.13 (cor. to 3 sig. fig.)

In the figure, ABCD is a cyclic quadrilateral withBAD 94, BC 8 cm and CD 11 cm. (a) Find BCD. (b) Find the length of BD.

(Give the answer correct to 3 significant figures.)

Solution:(a) BCD 94 180 (opp. s, cyclic quad.)

(b) By cosine formula,

P. 22


Alternate Forms of Cosine Formula

ab

cbaC

2 cos

222 ac

bcaB

2 cos

222 bc

acbA

2 cos

222

Consider the cosine formula:a2 b2 c2 2bc cos A

bc

acbA

2 cos

222

2bc cos A b2 c2 a2

Similarly, we can express cos B and cos C in terms of a, b and c.

Thus, the alternate forms of the cosine formula are:

These formulas are useful in finding the unknown angles when three sides of a triangle are given.

P. 23

Example 15.9T


bc

acbBAC

2cos

222

)19)(18(2

141918 222

228

163

BAC 44.3640 4.44 (cor. to 1 d. p.)

2cm )3640.44)(sin19)(18(2

1

2cm 6.119 (cor. to 1 d. p.)

In ABC, a 14 cm, b 18 cm and c 19 cm. (a) Find BAC. (b) Hence find the area of ABC. (Give the answers correct to 1 decimal place.)

Solution:(a) By cosine formula, (b) Area of ABC

BACbc sin2

1

P. 24

The formula is credited to the ancient Greek mathematician Heron of Alexandria (about 75 AD).

Another important formula for calculating the area of a triangle is Heron’s formula.

1155 ..44 Heron’s FormulaHeron’s Formula

Heron’s FormulaFor a triangle with known sides as shown in the figure,

area of triangle

where

,))()(( csbsass

).(2

1cbas

The formula is stated as follows.

P. 25


By cosine formula, . ab

cbaC

2cos

222

CC 22 cos1sin )cos1)(cos1( CC

ab

cbaab

ab

cbaab

2

2

2

2 222222

224

))()()((

ba

cbacbabacbac

).(2

1Let cbas

222

4

))()((16sin

ba

csbsassC

22

))()((4

ba

csbsass

22

))()((4

2

1

ba

csbsassab

))()(( csbsass

Consider

Area of ABC

ab

cba

ab

bac

2

)(

2

)( 2222

P. 26

Example 15.10T


cm2

875 xxxs

10x cm

By Heron’s formula, 2cm )810)(710)(510(10area xxxxxxx

2cm )2)(3)(5(10 xxxx24 cm 300x

22 cm 310 x

Find the area of a triangle with sides 5x cm, 7x cm and 8x cm. (Express the answer in surd form.)

Solution:

P. 27

Example 15.11T


cm2

589 s 11 cm

2cm )511)(811)(911(11 2cm 396

2cm 9.19 (cor. to 3 sig. fig.)

In ABC, a (k 5) cm, b 2k cm and c (2k 3) cm. If the perimeter of the triangle is 22 cm, find (a) the value of k; (b) the area of the triangle correct to 3 significant figures.

Solution:(a) Perimeter ∵ 22 cm

∴ (k 5) 2k (2k 3) 22k 4

(b) The sides of ABC are 9 cm, 8 cm and 5 cm respectively.

Area of ABC

P. 28

15.1 Area of Triangles

In ABC,

Chapter Chapter SummarySummary

Abc

Bac

CabABC

sin2

1

sin2

1

sin2

1 of Area

P. 29

15.2 Sine Formula


c

C

b

B

a

A sinsin

sin C

c

B

b

A

a

sinsin

sinor

In ABC, the length of a side is directly proportional to the sine of its opposite angle.

P. 30

15.3 Cosine Formula


ab

cbaC

ac

bcaB

bc

acbA

2cos

2cos

2cosor

222

222

222

In ABC, a2 b2 c2 2bc cos A b2 a2 c2 2ac cos B c2 a2 b2 2ab cos C

P. 31

15.4 Heron’s Formula


In ABC, area of ABC ,))()(( csbsass

).(2

1 where cbas

Documents

1515 15.1Area of Triangles 15.2Sine Formula 15.3Cosine Formula Chapter Summary Case Study Trigonometry (2) 15.4Heron’s Formula