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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
1155 .1 .1 Area of TrianglesArea of Triangles
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:
1155 .1 .1 Area of TrianglesArea of Triangles
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
1155 .1 .1 Area of TrianglesArea of Triangles
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
1155 .1 .1 Area of TrianglesArea of Triangles
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
1155 ..22 Sine FormulaSine Formula
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
1155 ..22 Sine FormulaSine Formula
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.)
A.A. Solving a Triangle with Two Angles andSolving a Triangle with Two Angles and Any Side GivenAny Side Given
P. 11
Example 15.5T
1155 ..22 Sine FormulaSine Formula
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
A.A. Solving a Triangle with Two Angles andSolving a Triangle with Two Angles and Any Side GivenAny Side Given
P. 12
Example 15.5T
1155 ..22 Sine FormulaSine Formula
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
A.A. Solving a Triangle with Two Angles andSolving a Triangle with Two Angles and Any Side GivenAny Side Given
P. 13
1155 ..22 Sine FormulaSine Formula
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
1155 ..22 Sine FormulaSine Formula
(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
1155 ..22 Sine FormulaSine Formula
C
c
A
a
sinsin
19
65sin15sin
A
7155.0
7.45A (cor. to 1 d. p.)
B. Solving a Triangle with Two Sides andB. Solving a Triangle with Two Sides and One Non-included Angle GivenOne Non-included Angle Given
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
1155 ..22 Sine FormulaSine Formula
14
48sin16sin
A
8493.0 121.9or 1.58A
B. Solving a Triangle with Two Sides andB. Solving a Triangle with Two Sides and One Non-included Angle GivenOne Non-included Angle Given
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)
1155 ..33 Cosine FormulaCosine Formula
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
1155 ..33 Cosine FormulaCosine Formula
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
1155 ..33 Cosine FormulaCosine Formula
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
1155 ..33 Cosine FormulaCosine Formula
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
1155 ..33 Cosine FormulaCosine Formula
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
1155 ..33 Cosine FormulaCosine Formula
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
1155 ..44 Heron’s FormulaHeron’s Formula
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
1155 ..44 Heron’s FormulaHeron’s Formula
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
1155 ..44 Heron’s FormulaHeron’s Formula
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
Chapter Chapter SummarySummary
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
Chapter Chapter SummarySummary
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
Chapter Chapter SummarySummary
In ABC, area of ABC ,))()(( csbsass
).(2
1 where cbas