PYTHAGORAS’THEOREM,TRIGONOMETRY ,BEARINGS AND THREE–DIMENSIONAL PROBLEMS 1. The Pythagoras’ Theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides in a right - angled triangle. 2. For a right angled Triangle ABC, A h p Where h is the hypotenuse. B b C The hypotenuse is the side opposite the right angle and is also the longest side of a triangle. EXAMPLE1: In the right-angled triangle ABC, D is a point on BC. Given that AD = 15 cm, BD = 5 cm and CD = 6 cm 1.1 PYTHAGORAS’ THEOREM: h 2 = p 2 + b 2
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PYTHAGORAS’THEOREM,TRIGONOMETRY
,BEARINGS AND THREE–DIMENSIONAL
PROBLEMS
1. The Pythagoras’ Theorem states that the square of the hypotenuse is
equal to the sum of the squares of the other two sides in a right - angled
triangle.
2. For a right angled Triangle ABC, A
h p
Where h is the hypotenuse. B b C
The hypotenuse is the side opposite the right angle and is also the longest side of a
triangle.
EXAMPLE1:
In the right-angled triangle ABC, D is a point on BC.
Given that AD = 15 cm, BD = 5 cm and CD = 6 cm
1.1 PYTHAGORAS’ THEOREM:
h2 = p2 + b2
Find: A
a) AC
b) AB
B 5 D 6 C
SOLUTION:
A
B 5 D 6 C
a) Using Pythagoras’ Theorem on ΔADC,
AC2 + CD2 = AD2
AC2 + 62 = 152
= 152 - 62
= 225 - 36
15
15
= 189
AC = √���
= 13.75 cm
a) Using Pythagoras’ Theorem on ΔADC,
AC2 + CD2 = AD2
AC2 + 62 = 152
= 152 - 62
= 225 - 36
= 189
AC = √���
= 13.75 cm
b) Using Pythagoras’ Theorem on ΔABC,
AB2 + BC2 = AC2
AB2 = 112 + ( 13.75 )2
= 121 + 189
= 310
AB = √���
= 17.60 cm
EXAMPLE2:
PQR is an isosceles triangle. Given that PQ = PR = 19 cm and QR = 32 cm , find the
area of triangle PQR .
Take positive square root on both sides
since length cannot be negative.
Take positive square root on both sides
since length cannot be negative .
BC = 5 + 6 = 11 cm
AC = 13.75 cm
SOLUTION:
P
19 19
Q 16 T 16 R
QT = 32 ÷ 2 = 16 cm
Using Pythagoras’ Theorem on ΔPQT,
PT2 + TQ2 = AD2
PT2 + 162 = 192
PT2 = 192 - 162
= 361 - 256
= 105
PT = √���
= 10.2 cm
Take positive square root on both sides
since length cannot be negative .
PT bisects QR
Area of ΔPQR
= �
� X QR X PT
= �
� X 32 X 10.2
= 163.2 cm2
3. The Converse of Pythagoras’ Theorem is also true.
In a triangle, if the square of the longest side is equal to the squares of the other
two sides, then the angle opposite the longest side is a right angle .
EXAMPLE3:
Area of Δ = �
� X Base X Height
In triangle ABC, if A
c2 = a2 + b2 , then
c b
ABC is a right - angle triangle with the right angle
Opposite side c.
B a C
Triangle XYZ has sides XY = 9 cm, YZ = √��cm and XZ = 7 cm. Show that triangle XYZ is
a right - angle triangle.
SOLUTION:
YZ2 + XZ2 = (√�� )2 72
= 81
= 92
= XY2
ΔPQR right - angle. (Converse of the Pythagoras’ Theorem)
For a right - angle triangle ABC,
A
(Hypotenuse) (Opposite)
z y
B x C
(Adjacent)
1.2 TRIGONOMETRIC RATIOS FOR ACUTE ANGLES:
sin = ��������
���������� =
�
�
cos = ��������
���������� =
�
�
tan = ��������
�������� =
�
�
TIPS FOR STUDENTS:
sin, cos, tan are abbreviation for sine, cosine and tangent. Hypotenuse – side opposite the right angle, the longest side.
Opposite – side opposite the angle .
Adjacent – side adjacent to angle .
To remember the ratio, use TOA CAH SOH .
T = �
� C =
�
� S =
�
�
( sin = ��������
���������� , cos =
��������
���������� , tan =
��������
�������� )
APPLICATIONS OF TRIGONOMETRIC RATIOS TO
RIGHT – ANGLE TRIANGLES:
We use trigonometric ratios to find the unknown sides or angles in a right - angled
triangle.
EXAMPLE:
In the diagram, BD = 6 cm, CD = 12 cm, BAD = 450 and BD is perpendicular to AC
Calculate D
a) BDC, 12
b) AD .
A B C
SOLUTION:
a) In ΔDBC, D
cos BDC = �
��
BDC = cos-1
12
6 6 12
= cos-1
2
1 = 600
B C
b) In ΔDAB, D
sin 450 = �
��
AD =
045sin
6 6
=
2
1
6 = 6 X √�
A B
6
450
cos = ��������
����������
sin = ��������
����������
450
= 8.46 cm (Correct to 3 sig .fig.)
TIPS FOR STUDENTS:
If the degree of accuracy is not stated in the question and if the answer is not exact, the answer should be given correct to 3 significant figures . This means that working should be performed correct to 4 significant figures or more .
Answer in degree should be given correct to 1 decimal place.
Check that your calculator is in ‘degree’ mode.
The trigonometric identities for an obtuse angle are:
1.3 TRIGONOMETRIC RATIOS FOR OBTUSE ANGLES:
sin (1800 - ) = sin
cos (1800 - ) = - cos
TIPS FOR STUDENTS:
When is acute, it lies in the first quadrant; sin and cos both are positive. When is obtuse, it lies the second quadrant ; sin is positive while cos is negative. y
sin ( 1800 - ) = �
� = sin
cos ( 1800 - ) = −�
� = - cos
r y 1800 - y x
-x O x
EXAMPLE1:
In the diagram, ABC is a right – angled triangle. BA is produced to P and BC is
produced to Q .
Given that AB = 15 cm, BC = 8 cm and AC = 17 cm, express as a fraction
a) tan CAB , P
b) sin PAC , A
c) cos ACQ .
15 17
B C Q
SOLUTION:
a) tan CAB = �
��
b) sin PAC = sin ( 1800 - CAB )
= sin CAB
= �
��
c) cos ACQ = cos ( 1800 - ACB )
tan = ��������
��������
= -cos ACB
= - �
��
EXAMPLE:
In is an obtuse angle and sin = ��
�� , find without the use of a calculator, the value
of cos .
SOLUTION:
sin = ��
�� ( Given )
Using Pythagoras’ Theorem, y
A
OB2 + 242 = 252 25
OB2 = 252 - 242 24
= 49 B (- 7) O x
OB = √��
= 7 units
cos = ��
��
= −�
��
= - �
��
In any triangle,
a) A + B + C = 1800
b) The shortest side is opposite the smallest angle and the longest side is opposite
the largest angle,
c) The sum of the length of the two shorter sides is always greater than the length of
the longest side .
A
c b
B a C
THE SINE RULE:
1.4 SOLUTIONS OF TRIANGLES: THE SINE RULE,THE COSINE
RULE AND AREA OF A TRIANGLE:
1. In any triangle ABC, the Sine Rule states that:
A
c b
B a C
2. The alternative form of the Sine Rule is also useful :
3. We use the Sine Rule when we have:
a) Any 2 angles and 1 side or
b) 2 side and 1 angle opposite one of those sides .
THE COSINE RULE:
1. In any triangle ABC, the Cosine Rule states that:
�
��� � =
�
��� � =
�
��� �
��� �
� =
��� �
� =
��� �
�
A
c b
B a C
2. The alternative from of Cosine Rule is shown below.
TIPS FOR STUDENTS:
We use trigonometric ratios ( sine, cosine and tangent ) of acute angles to solve right – angled triangles . We use the Sine Rule and the Cosine Rule to solve triangles that are not right – angled .
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
Use this to find the remaining side
when given 2 sides and an included
angle .
cos A = bc
ccb
2
222
cos B = ac
bca
2
222
cos C = ab
cba
2
222
Use this ti find an angle if all
three saides are known .
AREA OF A TRIANGLE:
To find the area of a triangle given 2 sides and the included angle, use
A
c b
B a C
EXAMPLE1:
In the diagram, BCD is a straight line. BAC = 1000.
ACB = 520 , AC = 6 cm and CD = 8 cm .
Calculate A
a) AB , 1000
6
b) AD , 520
B C 8 D
c) The area of triangle ACD.
Area of Δ ABC = �
� ab sin C
= �
� bc sin A
= �
� ac sin B
SOLUTION:
A
1000
6
280 520 1280
B C 8 D
a) ABC = 1800 – 1000 – 520 ( sum of Δ )
= 280
Using the Sine Rule on Δ ABC ,
052sin
AB
=
028sin
6
AB = 028sin
6
X sin 520
≈ 10.1 (correct to 3 sig . fig.)
b) ACD = 1800 – 520 (adj . s on a str . line)
= 1280
Using the Cosine Rule on Δ ACD ,
AD2 = 82 + 62 – 2 ( 8 ) ( 6 ) cos 1280
= 159.1
Sine Rule A
��� �
� =
��� �
� c b
B a C
Cosine Rule A
a2 = b2 + c2 – 2bc cos A c b
B a C
AD = √���. �
≈ 12.6 cm (correct to 3 sig . fig.)
c) Area of Δ ACD
= �
� X 8 X 6 X sin 1280
≈ 18.9 cm2 (correct to 3 sig.fig.)
EXAMPLE2:
In the diagram, BAD = 480, AD = 11.2 cm, BD = 8.6 cm, BC = 9.8 cm and CD = 5.1 cm .
a) ABD , A 11.2 D
480
b) CBD , 8.6 5.1
c) The area of triangle BCD,
B 9.8 C
d) The shortest distance from B to CD .
SOLUTION:
Area of Triangle A
= �
� bc sin A c b
B a C
A 11.2 D
480
8.6 5.1
31.330
B 9.8 C
a) Using the Sine Rule on Δ ABD ,
2.11
sin ABD
=
6.8
48sin 0
sin ABD = 6.8
48sin 0
X 11.2
= 0.9678
ABD = 1800 – 75.420
≈ 104.60 (correct to 1 d.p.)
b) Using the Cosine Rule on Δ BCD,
cos CBD = )8.9)(6.8(2
1.58.96.8 222
= 0.8542
CBD = 31.330
Sine Rule A
��� �
� =
��� �
� c b
B a C
ABD is given as an obtuse angle in
the diagram.
Cosine Rule A
cos A = bc
ccb
2
222 c d
B a C
≈ 31.30 (correct to 1 d.p.)
c) Area of Δ BCD
= �
� X 8.6 X 9.8 X sin 31.330
= 21.91
≈ 21.9 cm2 ( correct to 3 sig . fig. )
d) Let h cm be the shortest distance from B to CD.
D
�
� X CD X h = Area of Δ BCD 8.6 5.1
�
� X 5.1 X h = 21.91 h
h = ��.�� � �
�.� B 9.8 C
≈ 8.59 cm ( correct to 3 sig . fig. )
The shortest distance from B to CD is 8.59 cm .
TIP FOR STUDENTS:
The shortest distance from B to CD is the perpendicular distance from B to CD.
Area Of Triangle A
= �
� bc sin A c b
B a C
1. A bearing is an angle that tells the direction of one place from another.
2. Bearings are always
a) Measured from the North.
b) Measured in a Clockwise direction and
c) Written as a Three – Digit number (0000 to 3600)
e.g.-
N N N
B F
0350
A 1150
3350
D
The bearing of B The bearing of D The bearing of F
From A is 0350. From C is 1150. From E is 3350.
EXAMPLE1:
In the diagram, ABC represents a horizontal triangular field. B is 10 m from A on a
bearing of 0480. C is 18 m from B and 20 m from A.
1.5 BEARINGS:
Calculate N
a) ABC,
B
b) The bearing of B from C,
480 10 18
c) The area of triangular field ABC. A
20
C
Given that a water hydrant, H, is due north of A. The bearing of B from H is 0720.
d) Find the distance AH.
SOLUTION:
N
B
480 86.180
10 18
A
20
C
a) Using the Cosine Rule on Δ ABC,
cos ABC = )10)(18(2
201018 222
= 0.06667
CBD = 86.180
≈ 86.20 (correct to 1 d.p.)
b) ABP = 480 ( alt. s, MA || QC ) N
ABP = 86. 180 – 480 N
= 38.180
B 38.180
BCQ = 38.180 ( alt. s, BP || QC ) 10 480
480 P
Bearing of B from C A 18
= 3600 – 38.180 ( s, at a point ) N
≈ 321.80 ( correct to 1 d.p. ) 20 38.180
Q
C
c) Area of Δ ABC
= �
� X 18 X 10 X sin 86.180
Cosine Rule A
cos A = bc
ccb
2
222 c d
B a C
Area of Triangle A
= �
� bc sin A c b
B a C
= 89.8 m2
≈ 89.8 m2 (correct to 3 sig . fig.)
The area of the triangular field is 89.8 m2
d) AHB = 1800 - 720 ( adi. s, on a str. line )
= 1080
ABH = 1800 - 1080 - 480 ( adi. Δ sum of)
= 240
Using the Sine Rule on Δ ABH ,
024sin
AH
=
0108sin
10
AH = 0108sin
10
X sin 240
≈ 4.28 m2 (correct to 3 sig.fig.)
EXAMPLE 2:
A, B and C are three points on horizontal ground. C is 32 m from A on a bearing of
1280. BAC = 1100 and AB = 43 m.
Sine Rule A
��� �
� =
��� �
� c b
B a C
a) Calculate N
i) BC,
A 1280
ii) The area of triangle ABC, 1100
43 32
iii) The bearing of B from C,
B C
b) A boy walks from point B to point C until he reaches a point X, where AX is
minimum.
Find the length of AX.
SOLUTION:
a) i) Using the Cosine Rule on Δ ABC ,
BC2 = 322 + 432 – 2 ( 32 ) ( 43 ) cos 1100
= 3814
AD = √����
= 61.76
≈ 61.8 m (correct to 3 sig. fig.)
ii) Area of Δ ABC
= �
� X 32 X 43 X sin 1100
= 646.5
≈ 647 m2 (correct to 3 sig . fig.)
Cosine Rule A
a2 = b2 + c2 – 2bc cos A c b
B a C
Area of Triangle A
= �
� bc sin A c b
B a C
iii) PAB = 1100 + 1280 - 1800 = 580
= 580
QBA = PAB (alt. s, QB || AP)
= 580 N
Using the Sine Rule on Δ ABC,
A 1280
32
sin ABC
=
76.61
110sin 0
Q 43 P 32
sinQBA = 76.61
110sin 0
= 32 580 29.140
= 0.4869 B 61.76 C
ABC = 29.140
QBC = 580 + 29.140
≈ 87.10 ( correct to 1 d.p. )
The bearing of C from B is 087.10
b) The boy closest to A when AX is to BC .
In Δ ABX ,
sin 29.140 = ��
��
AX = 43 X sin29.140
≈ 20.8 m (correct to 3 sig . fig.)
Alternative method:
�
� X BC X AX = Area of Δ ABC
�
� X 61.76 X AX = 646.5
AX = ���.� � �
��.��
≈ 20.9 m ( correct to 3 sig . fig. )
LINE OF SIGHT
OBSERVER
ANGLE OF ELEVATION
HORIZONATAL LINE
ANGLE OF DEPRESSION
LINE OF SIGHT
1.6 ANGLE OF ELEVATION AND ANGLE OF DEPRESSION:
When you look at the airplane, the angle between the line of sight and the horizontal
line is called the Angle of Elevation .
When you look down at the car, the angle between the horizontal line and the line of
sight is called the Angle of Depression .
EXAMPLE1:
a) The angle of elevation of the foot of a lighthouse from a boat is 360. The
lighthouse is located on the edge of a cliff which is 80 m high .
ANGLE OF ELEVATION AND ANGLE OF DEPRESSION:
Horizontal B
b
a
A HORIZONTAL O
a = angle of elevation of B from A .
b = angle of depression of A from B .
a = b ( alt. s )
i) Find the distance from the boat to the base of the cliff .
ii) Given that the lighthouse is 45 m high, find the angle of elevation of the top of
the lighthouse from the boat .
b) An owl flies from the top of a 10 m tall tree at an angle of depression of 280 to
catch a rat on the ground .
Find the distance
i) The owl flew to catch the rat,
ii) Of the rat from the base of the tree.
SOLUTION:
a) i) In ΔABC, D
tan 360 = ��
�� Lighthouse
AB = 036tan
80 45
= 110.1 Cliff
≈ 110 m ( correct to 3 sig . fig. )
The distance from the best boat to the base
of the cliff is 110 m . 80
360
A (110.1) B
ii) In ΔABD,
tanDAB = ��
��
= 1.110
)4580(
DAB ≈ 48.60 (correct to 1 d.p.)
The angle of elevation of the top of the lighthouse from the boat is 48.60.
b) i) sin 280 = ��
��
R
PR = 028sin
10
280
≈ 21.3 m ( correct to 3 sig . fig. ) 10
The owl flew 21.3 m to catch the rat .
280
P Q
ii) tan 280 = ��
��
PQ = 028sin
10
≈ 18.8 m ( correct to 3 sig . fig. )
The rat is 18.8 m from the base of the tree .
EXAMPLE2:
In the diagram, A, B and C are three points on horizontal ground . The bearing of C
from A is 0630 while the bearing of B from A is 0860 . AC = 15 m and AB = 21 m .
a) Find BC. T
b) A vertical mast TC stands on C. N C
The angle of elevation of T from A is 320.
Find 15
i) TC, 860
ii) The angle of depression of B from T . 630
A B
21
SOLUTION:
BAC = 860 - 630 T
= 230
N C
15
860
630
A 230 B
21
b) i) Using the Cosine Rule on Δ ABC ,
BC2 = 152 + 212 – 2 ( 15 ) ( 21 ) cos 230
= 86.08
BC = √��. ��
= 9.278
≈ 9.28 m (correct to 3 sig . fig.)
b) i) tan 320 = ��
�� T
TC = 15 X tan 320
= 9.373
≈ 9.37 m ( correct to 3 sig . fig. )
A 15 C
ii) Angle of depression = Angle of elevation = TBC
tan TBC = �.���
�.���
TBC ≈ 45.30 ( correct to 1 d.p. )
The angle of depression of B from T is 45.30.
Cosine Rule A
a2 = b2 + c2 – 2bc cos A c b
B a C
320
To solve a given Three – Dimensional problem:
a) Reduce in to a problem in a plane, i.e. look for right – angled triangles in the
relevant planes, and then
b) Apply the Pythagoras’ Theorem or trigonometric ratios (sin , cos or tan)
to these triangles to find the unknowns.
EXAMPLE1:
The diagram shows a wedge. ABEF is a rectangle and CE and DF are vertical lines.
Given that AB = DC = FE = 24 cm , BE = 18 cm and CE =12 cm .
Find
a) AE, D C
b) AEB, 12
E
c) AC, 18
d) CAE . A 24 B
1.7 THREE – DIMENSIONAL PROBLEMS:
F
SOLUTION:
a) Using Pythagoras’ Theorem on ΔADC, E
AE2 = 242 + 182
= 900 18
AE = √���
= 30 cm A 24 B
b) In ΔABE,
tan AEB = ��
��
AEB ≈ 53.10 ( correct to 1 d.p. )
c) Using Pythagoras’ Theorem on ΔAEC, C
AC2 = 302 + 122
= 1044
AC = √����
≈ 32.3 cm ( correct to 3 sig . fig. ) A 30 E
d) In ΔAEC,
tan CAE = ��
��
CAE ≈ 21.80 (correct to 1 d.p.)
EXAMPLE:2
VABCD is a pyramid with a rectangular base . X is the point of intersection of the
diagonals of the base and V is vertically above X . Given that AB = 12 cm, AC = 20 cm
and VX = 16 cm, calculate
a) BC, V
b) AC,
c) AVX . 16
D C
20
X
A 12 B
a) Using Pythagoras’ Theorem on ΔABC, C
BC2 = 122 + 162
BC2 = 256 20
BC = √���
= 16 cm A 12 B
b) Using Pythagoras’ Theorem on ΔVAX, V
AC2 = 302 + 122
= 1044 16
AC = √����
≈ 32.3 cm (correct to 3 sig. fig.) A 10 X
c) In Δ VAX,
tan AVX = ��
��
AVX ≈ 32.00 (correct to 1 d.p.)
RELATION RETWEEN SIDES AND ANGLES OF A TRIANGLE:
1. A triangle consists of three sides and three angles called elements
of the triangle.
In any triangle ABC,
A, B, C denotes the angles of the triangle at the vertices.
A + B + C = 1800
A
2. The sides of the triangle are denoted by a, b, c opposite b c
to the angles A, B and C respectively .
B C
a
Fig (1)
1.1
3. a + b + c = 2s = The perimeter of the triangle .
THE SINE RULE:
In a triangle ABC, prove that
�
��� � =
�
��� � =
�
��� � = 2R
Where, R is the circum radius of the triangle.
PROOF: Let S be the circumcentre of the triangle ABC. First prove that �
��� � = 2R
CASE (I) : Let S be an acute angle. Let P be any point on the circle. Join BP,
Which pass through S . Join CP, so that BCP = 90O. BAC = a =
BPC
(angles in the same segment) . A
P
FROM Δ BPC, sin B��C = ��
��
B
C
sin �� = �
�� ,
�
��� � = 2R Fig ( 2 )
1.2
S
CASE (II) : Let A be right angle, ie., �� = 900 ( Fig 3 ), Then BC is the diameter.
BC = a = 2R A
sin �� = ��
�� =
�
��
B C
�
��� � = 2R Fig ( 3)
CASE (III) : Let A be an obtuse angle ( Fig 4 ) . join BP, passing through S .
Join CP, so that B��P = 900 .
Now B��C = 1800 – ( B��C ) = 1800 – A A
( Since ABPC is a cyclic quadrilateral )
B C
From Δ BPC, sin ( B��C )
= ��
�� Fig ( 4)
i.e.- sin (1800 – A ) = �
��
sin A = �
��
�
��� � = 2R
S
a
S
a
�
��� � = 2R is true for all values of A .
Similarly, we can prove, �
��� � =
�
��� � = 2R ,
Thus, or a = 2R sinA, b = 2R sinB, c = 2R
sinC .
THE COSINE RULE:
In any triangle ABC, prove that
a2 = b2 + c2 - 2bc cos A
b2 = c2 + a2 - 2ca cos B
c2 = a2 + b2 - 2ab cos C
A C C
c b b a b a
B a D C A c B D A C B
( Fig 5 ) ( Fig 6 ) ( Fig 7 )
�
��� � =
�
��� � =
�
��� � = 2R .
1.3
PROOF: Case ( I ) Let A be an acute angle ( Fig 5 )
Draw CD AB . From Δ BDC
BC2 = BD2 + DC2 = ( AB – AD )2 + DC2
= AB2 - 2AB . AD + AD2 + DC2
BC2 = AB2 - 2AB.AD + AC2 ( Since AD2 + DC2 = AC2 )
a2 = c2 – 2c.AD + b2
But from Δ ADC,
Cos A = ��
��
AD = A cos A = b cos A
a2 = c2 – 2c.b.cos A + b2 Or
a2 = b2 + c2 – 2bc cos A
Case ( II ) Let A be right angled, ie., �� = 900 ( Fig 6 )
BC2 = BA2 + AC2 ie., a2 = b2 + c2
But a2 = b2 + c2 – 2bc cos A
a2 = b2 + c2 – 2bc cos 900
a2 = b2 + c2 , which is true for a right angled triangle .
Case ( III ) Let A be obtuse angle, ie., A > 900 ( Fig 7 ) .
Draw CD BA produced .
From Δ BDC BC2 = BD2 + DC2 = ( BA + AD )2 + DC2
BC2 = BA2 + 2BA . AD + AD2 + DC2
BC2 = BA2 + 2AB . AD + AC2 ( Since, AD2 + DC2 = AC2 )
a2 = c2 – 2.C . AD + b2
But from Δ ADC,
Cos ( D�� C ) = ��
�� , ( D�� C = 1800 – A )
Cos ( 1800 – A ) = ��
�� =
��
�
- cos A = ��
� , AD = b cos A
a2 = b2 + 2bc ( – b cos A ) + c2
a2 = b2 + c2 – 2bc cos A
Similarly , we can prove, b2 = c2 + a2 - 2ca cos B
c2 = a2 + b2 - 2ab cos C
Tips for Students:
The above formulae can be written as:
Cos A = bc
acb
2
222 , Cos B =
ac
bac
2
222 , Cos C =
ab
aba
2
222 ,
These results are useful in finding the cosines of the angle when numerical
values of the sides are given. Logarithmic computation is not applicable since
the formulae involve sum and difference of terms. However, logarithmic
method can be applied at the end of simplification to find angle
THE PROJECTION RULE:
In this rule, we show how, one side of a triangle can be expressed in terms of other
two sides . It is called projections rule.
a = b cos C + c cos B ,
b = c cos A + a cos C , c = a cos B + b cos A.
PROOF: Let C be an acute angle
Draw AD BC produced.
In Fig ( i ) BC = BD + DC
[ NOTE : BD is called projection of AB on BC and DC is the projection of AC ]
a = BD + DC …… (1) A
From Δ BDA , cos B = ��
�� BD = AB cos B = c cos B
From Δ CDA , cos C = ��
�� CD = AC cos C = b cos C c b
From (1) c cos B + b cos C = b cos C + c cos B
B D C
a
(Fig 8)
CaseII: When C is a right angle, ie., �� = 900 ( Fig 9 ) .
1.4
cos B = ��
�� =
�
� , a = c cos B ……(2) A
Since �� = 900 ,
cos B = cos 900 = 0, c a
We get, a = b cos 900 + c cos B a = c cos B ……(3)
B a C
( Fig 9 )
Case III : When c is obtuse angle ( Fig (iii) )
From Δ ABD, BC = BD - CD ……(4) A
From Δ ABD , cos B = ��
�� =
��
� , BD = c cos B From Δ ACD
Cos (1800 – C ) = ��
�� =
��
� c b
cos A = ��
� , CD = - b cos C
1800 – C
From (4) BC = c cos B – ( - b cos C ) B a C D
( Fig 10 )
ie., a = c cos B + b cos C = b cos C + c cos B
Similarly, b = c cos A + a cos C , c = a cos B + b cos A .
THE LAW OF TANGENTS:
1.5
In any Δ ABC , Prove that :
1. � − �
� � � =
2tan
2tan
BA
BA
,
2. � − �
� � � =
2tan
2tan
CB
CB
,
3. � − �
� � � =
2tan
2tan
AC
AC
PROOF: Using sine rule,
� − �
� � � =
�� ��� � − �� ��� �
�� ��� � � �� ��� � =
��� � − ��� �
��� � � ��� �
=
2cos2.
2cos2
2sin2.
2cos2
BABA
BABA
, = cos
2
BA.tan
2
BA
� − �
� � � =
2tan
2tan
BA
BA
tan
1cotsin gu
Similarly, other two results can be proved by changing sides and angles in cycle
order.
EXPRESSIONS FOR HALF ANGLES IN TERMS OF a,b,c:
In any triangle ABC, prove that
1. sin �
� =
bc
CSbS ,
2. cos �
� =
bc
aSS ,
3. tan �
� =
aSS
cSbS
PROOF: (1) We know that 2 sin2 A = 1 – cos A
2 sin2 �
� = 1 -
bc
acb
2
222 ( Using cosine rule for A )
2 sin2 �
� =
bc
acbbc
2
2 222
=
bc
bccba
2
2222 =
bc
cba
2
22
=
bc
cbacba
2
=
bc
cbacba
2
1.6
=
bc
cSbS
2
2222 Since a + b + c = 2s
a + b = 2s - c
a + c = 2s - c
2 sin2 �
� =
bc
csbs
2
22 (Divide by 2)
sin �
� = ±
bc
cSbS
If A is acute, then sin �
� is always positive.
2. 2 sin2 �
� = 1 + cos A = 1 +
bc
acb
2
222 (Using cosine rule for A )
2 sin2 �
� =
bc
acbbc
2
2 222
=
bc
acb
2
22
2 sin2 �
� =
bc
acbacb
2
=
bc
ass
2
222 Using a + b + c = 2s
a + b = 2s – c
sin �
� =
bc
cSbS
Dividing by 2, we get
cos2 �
� =
bc
ass = ±
bc
aSS
Since
�
� is acute, cos
�
� is always positive and therefore,
cos �
� =
bc
aSS
3. tan
�
� =
2cos
2sin
A
A
=
bc
assbc
csbs
= ass
csbs
Similarly, we can show that
sin �
� =
ac
csas , cos
�
� =
ac
bss , tan
�
� =
bss
csas
sin �
� =
ac
bsas , cos
�
� =
ac
css , tan
�
� =
css
bsas
WORKED EXAMPLES
1. If a = 3, b = 4, c = 5, in a triangle ABC , find the value of
a.) sin
2
C b.) sin 4C + cos 4C
SOLUTION:
Since, c2 = b2 + a2 is satisfied by the given sides, they form right angled triangle.
52 = 42 + 32
C = 900, sin
2
C = sin
2
900
= sin 450 = 2
1
And, sin 4C + cos 4C = sin( 4 X 900) + cos( 4 X 900 )
= sin 3600 + cos 3600
= 0 + 1 = 1
2. Prove that a sin ( B – C ) + b sin ( C – A ) + C sin ( A – B ) = 0 .
SOLUTION:
Now, a sin( B – C ) = 2R sin A . sin( B – C ) ( Since, a = 2R sin A )
= 2R sin A .sin( B – C )
Since A + B + C = 1800, B + C = 1800 – A
sin( B + C ) = sin A
= 2R sin ( B + C ) sin ( B – C ) = 2R sin [ sin2 B – sin2 C ]
Similarly, b sin ( C – A ) = 2R [ sin2 C – sin2 A ]
C sin ( A – B ) = 2R [ sin2 A – sin2 B ]
L. H. S. = a sin ( B – C ) + b sin ( C – A ) + C sin ( A – B )
= 2R [ sin2 B – sin2 C ] + 2R [ sin2 C – sin2 A ] + 2R [ sin2 A – sin2 B ]
= 2R [ sin2 B – sin2 C + sin2 C – sin2 A + sin2 A – sin2 B ]
3. Prove that, in a ΔABC , CB
CB
sin
sin
=
2
22
a
cb
SOLUTION:
L. H. S. = CB
CB
sin
sin X
CB
CB
sin
sin
= cB
CB
sin
sinsin 22
Since sin ( B + C ) sin ( B - C )
= sin2 B – sin2 C
= A
CB2
22
sin
sinsin
sin( A + B ) = sin C
in ΔABC
=
2
2
2
2
2
2
4
44
R
aR
C
R
b
( using sine rule )
=
2
2
2
22
4
4
R
aR
cb
= a
cb 22 =
R. H. S.
4. Prove that a(b cos C – c cos B) = b2 - c2
SOLUTION:
L. H. S. = a( b cos C – c cos B ) = ab cos C – ac cos B
= ab bc
aba
2
222
- ac
bc
bac
2
222 (Using cosine rule)
= bc
aba
2
222
-
bc
bac
2
222
= 2
22 22 cb
= b2 - c2 = R. H. S.
5. Prove that 2
22
a
cb
sin2A +
2
22
b
ac sin2B +
2
22
c
ba sin2C = 0
SOLUTION:
Now 2
22
a
cb
sin2A =
2
22
a
cb
X
2sin A cos A (Since, sin 2A = 2 sin A cos
A)
= 2
22
a
cb
X
R
a
2 X
bc
acb
2
222
(Using sine rule and cosine rule)
=
abcR
acbcb
.4
22222
Similarly, 2
22
b
ac sin2B =
abcR
bacac
.4
22222
2
22
c
ba sin2C =
abcR
cbaba
.4
22222
L. H. S =
abcR
acbcb
.4
22222
+
abcR
bacac
.4
22222
+
abcR
cbaba
.4
22222
= abcR.4
1 [ b4 – c4 – ( b2 – c2 ) a2 + c4 – a4 – ( c2 – a2 ) b2 + a4 – b4 – ( a2 – b2 ) c2]
= abcR.4
1[ 0 ] = 0 = R. H. S
6. Find the greatest side of the triangle, whose sides are x2 + x + 1 , 2x + 1 , x2 – 1.
SOLUTION:
Let a = x2 + x + 1, b = 2x + 1, c = x2 – 1
Then, a is the greatest side. Therefore �� is the greatest angle.
cos A = bc
acb
2
222
=
1122
11122
22222
xx
xxxx
= 1222
22211214423
2324242
xxx
xxxxxxxxx
cos A = 1222
12223
23
xxx
xxx
= 2
1
= - cos600 = cos( 1800 – 600 )
= 2
1
= cos1200 A = 1200
Therefore, the greatest angle is 1200
7. If sin 2A + sin 2B = sin 2C in a Δ ABC, Prove that either �� = 900 or �� = 900.
SOLUTION:
sin 2A + sin 2B = sin 2C
2 sin
2
22 BA.cos
2
22 BA = sin 2C Using sin C + sin D
= 2 sin
2
DC . cos
2
DC
2 sin ( A + B ) . cos ( A – B ) = 2sin C cos C ∵ sin ( A + B ) =
Sin( 1800 – C ) = sin C
2sin C cos ( A – B ) = 2sin C cos C
Dividing by 2 sin C both sides , we get ,
Cos( A – B ) = cos C
Also, Cos( A – B ) = cos -C [ Since, cos( -C ) = cos C ]
A – B = ± C
When A – B = C , A = B + C
But, A + B + C = 1800, gives
A - B = 1800, i.e., 2A = 1800 , �� = 900
When A – B = - C , B = A + C
A + B + C = 1800, gives
B + B = 1800, i.e., 2B = 1800 , �� = 900
Therefore, triangle is right angled triangle.
8. Prove that 2
2cos
a
A
-
2
2cos
b
B =
2
1
a -
2
1
b
SOLUTION:
L. H. S = 2
2cos
a
A
-
2
2cos
b
B (Using cos 2A = 1 – 2 sin2 A )
= 2
2sin21
a
A
-
2
2sin21
b
B
= 2
1
a -
2
2sin2
a
A
-
2
1
b +
2
2sin2
b
B
= 2
1
a -
2
1
b - 2
2sin
a
A + 2
2sin
a
B
= 2
1
a -
2
1
b - 2
2
2
1
R + 2
2
2
1
R Since,
a
Asin =
R2
1
b
Bsin =
R2
1
= 2
1
a -
2
1
b = R. H. S
SUMMARY AND KEY POINTS
1.The Pythagoras’ Theorem states that the square of the hypotenuse is
equal to the sum of the squares of the other two sides in a right - angled
triangle.
2. For a right angled Triangle ABC, A
h p
h2 = p2 + b2
Where h is the hypotenuse. B b C
The hypotenuse is the side opposite the right angle and is also the longest side of a
triangle.
3. The Converse of Pythagoras’ Theorem is also true.
In a triangle, if the square of the longest side is equal to the squares of the other
two sides, then the angle opposite the longest side is a right angle.
In triangle ABC, if A
c2 = a2 + b2 , then
c b
ABC is a right - angle triangle with the right angle
Opposite side c.
B a C
4. TRIGONOMETRIC RATIOS FOR ACUTE ANGLES: For a right - angle triangle ABC, A
sin = ��������
���������� =
�
�
cos = ��������
���������� =
�
� (Hypotenuse) z y (Opposite)
tan = ��������
�������� =
�
�
B (Adjacent) C
sin, cos, tan are abbreviation for sine, cosine and tangent.
Hypotenuse – side opposite the right angle, the longest side.
Opposite – side opposite the angle .
Adjacent – side adjacent to angle .
To remember the ratio, use TOA CAH SOH.
T = �
� C =
�
� S =
�
�
(sin = ��������
���������� , cos =
��������
���������� , tan =
��������
�������� )
We use trigonometric ratios to find the unknown sides or angles in a right - angled
triangle.
5. Important Note:
If the degree of accuracy is not stated in the question and if the answer is not
exact, the answer should be given correct to 3 significant figures . This means
that working should be performed correct to 4 significant figures or more .
Answer in degree should be given correct to 1 decimal place.
Check that your calculator is in ‘degree’ mode.
6. TRIGONOMETRIC RATIOS FOR OBTUSE ANGLES:
The trigonometric identities for an obtuse angle are:
sin(1800 - ) = sin
cos(1800 - ) = - cos
When is acute, it lies in the first quadrant; sin and cos both are positive.
When is obtuse, it lies the second quadrant ; sin is positive while cos
is negative.
7. SOLUTIONS OF TRIANGLES: THE SINE RULE,THE COSINE RULE:
In any Triangle, Sum of all three angles
a) A + B + C = 1800
b) The shortest side is opposite the smallest angle and the longest side is opposite
the largest angle,
c) The sum of the length of the two shorter sides is always greater than the length of
the longest side .
A
c b
B a C
A) THE SINE RULE:
i) In a triangle ABC, the Sine Rule states that:
�
��� � =
�
��� � =
�
��� �
A
c b
B a C
ii) The alternative form of the Sine Rule is also useful:
��� �
� =
��� �
� =
��� �
�
iii) We use the Sine Rule when we have:
a) Any 2 angles and 1 side or
b) 2side and 1 angle opposite one of those sides.
In a triangle ABC,
�
��� � =
�
��� � =
�
��� � = 2R
Where, R is the circum radius of the triangle.
The alternative form of Sine Rule as shown below:
a = 2Rsin A ,
b = 2RsinB,
c = 2RsinC
B) THE COSINE RULE:
i)In any triangle ABC, the Cosine Rule states that:
A
c b
B a C
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
Use this to find the remaining side
when given 2 sides and an included
angle.
ii) The alternative form of Cosine Rule is shown below.
Important points:
We use trigonometric ratios ( sine, cosine and tangent ) of acute angles
to solve right – angled triangles .
We use the Sine Rule and the Cosine Rule to solve triangles that are not
right angled .
8. AREA OF A TRIANGLE:
To find the area of a triangle given 2 sides and the included angle, use
A
c b
cos A = bc
ccb
2
222
cos B = ac
bca
2
222
cos C = ab
cba
2
222
Use this ti find an angle if all
three sides are known.
Area of Δ ABC = �
� ab sin C
= �
� bc sin A
= �
� ac sin B
B a C
9. THE BEARINGS:
1. A bearing is an angle that tells the direction of one place from another.
2. Bearings are always
a) Measured from the North.
b) Measured in a Clockwise direction and
c) Written as a Three – Digit number (0000 to 3600)
e.g.-
N N N
B F
0350
A 1150
3350
D
The bearing of B The bearing of D The bearing of F
From A is 0350. From C is 1150. From E is 3350.
10. THE THREE DIMENSIONAL PROBLEMS:
To solve a given Three – Dimensional problem:
a) Reduce in to a problem in a plane, i.e. look for right – angled triangles in the
relevant planes, and then
b) Apply the Pythagoras’ Theorem or trigonometric ratios ( sin , cos or tan )
to these triangles to find the unknowns.
11. RELATION RETWEEN SIDES AND ANGLES OF A TRIANGLE:
a) A triangle consists of three sides and three angles called elements
of the triangle.
In any triangle ABC,
A, B, C denotes the angles of the triangle at the vertices.
A + B + C = 1800
A
b) The sides of the triangle are denoted by a, b, c opposite b c
to the angles A, B and C respectively .
B C
a
c) a + b + c = 2s = The perimeter of the triangle .
12. THE PROJECTION RULE:
In this rule, we show how, one side of a triangle can be expressed in terms of other
two sides. It is called projections rule.
a = b cos C + c cos B ,
b = c cos A + a cos C , c = a cos B + b cos A .
13. THE LAW OF TANGENTS:
In any Δ ABC , Prove that :
a.) � − �
� � � =
2tan
2tan
BA
BA
,
b.) � − �
� � � =
2tan
2tan
CB
CB
,
c.) � − �
� � � =
2tan
2tan
AC
AC
14. EXPRESSIONS FOR HALF ANGLES IN TERMS OF a, b,c:
In any triangle ABC , prove that
a.) sin �
� =
bc
CSbS ,
b.) cos �
� =
bc
aSS ,
c.) tan �
� =
aSS
cSbS
Similarly, we can show that
sin �
� =
ac
csas , cos
�
� =
ac
bss , tan
�
� =
bss
csas
sin �
� =
ac
bsas , cos
�
� =
ac
css , tan
�
� =
css
bsas
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