Download RELATIONS BETWEEN SIDES AND ANGLES OF A TRIANGLE

Relations Between Sides and Angles of a Triangle
MODULE - IV
Functions
19
Notes
RELATIONS BETWEEN SIDES AND
ANGLES OF A TRIANGLE
In an earlier lesson, we have learnt about trigonometric functions of real numbers, relations
between them, drawn the graphs of trigonometric functions, studied the characteristics from
their graphs, studied about trigonometric functions of sum and difference of real numbers, and
deduced trigonometric functions of multiple and sub-multiples of real numbers. We also studied
about inverse trigonometric functions, some of their properties and solved problems based on
their concepts.
In this lesson, we shall try to establish some results which will give the relationship between sides
and angles of a triangle and will help in finding unknown parts of a triangle.
OBJECTIVES
After studying this lesson, you will be able to :
l
derive sine formula, cosine formula and projection formula
l
apply these formulae to solve problems.
EXPECTED BACKGROUND KNOWLEDGE
l
l
l
Trigonometric and inverse trigonometric functions.
Formulae for sum and difference of trigonometric functions of real numbers.
Trigonometric functions of multiples and sub-multiples of real numbers.
19.1 SINE FORMULA
In a ∆ABC, the angles corresponding to the vertices A, B, and C are denoted by A, B, and C
and the sides opposite to these vertices are denoted by a, b and c respectively. These angles
and sides are called six elements of the triangle.
MATHEMATICS
159
Relations Between Sides and Angles of a Triangle
MODULE - IV Result 1 : Prove that in any triangle, the lengths of the sides are proportional to the sines of the
Functions angles opposite to the sides,
a
b
c
=
=
sinA sinB sinC
i.e.
Proof : In ∆ABC, in Fig. 19.1 [(i), (ii) and (iii)], BC = a, CA = b and AB = c and ∠ C is acute
Notes angle in (i), right angle in (ii) and obtuse angle in (iii).
(i)
(ii)
(iii)
Fig. 19.1
Draw AD prependicular to BC
In ∆ABC,
(or BC produced, if need be)
AD
= sinB
AB
or
AD
= sinB
c
In
∆ADC,
or,
AD
= sinC
b
⇒
AD = c sin B
.....(i)
AD
= sinC in Fig 19.1 (i)
AC
If Fig. 19.1 (ii), or
⇒
AD = b sin C
....(ii)
AD
π
= 1 =sin
=sinC
AC
2
AD = b sin C
and in Fig. 19.1 (iii),
AD
= sin ( π −C) =sinC
AC
or
AD
= sinC
b
or
AD = b sin C
Thus, in all three figures, AD = b sin C
From (i) and (ii), we get
c sin B = b sin C
⇒
160
b
c
=
sinB sinC
....(iii)
MATHEMATICS
Relations Between Sides and Angles of a Triangle
MODULE - IV
Functions
Similarly, by drawing prependicular from C on AB, we can prove that
a
b
=
sinA sinB
.....(iv)
From (iii) and (iv), we get
a
b
c
=
=
sinA sinB sinC
.....(A)
Notes
(A) is called the sine-formula
Note : (A) is sometimes written as
sinA sinB sinC
=
=
a
b
c
....(A')
The relations (A) and (A') help us in finding unknown angles and sides, when some others are
given.
Let us take some examples :
Prove that acos
Example 19.1
Solution : R.H.S. = ( b +c ) sin
We know that,
A
2
a
b
c
=
=
=k (say)
sinA sinB sinC
⇒
a = k sin A, b = k sin B, c = k sin C
⋅
R.H.S. = k (sinB +sinC ) sin
∴
= k 2⋅ sin
Now
B−C
A
= ( b +c ) sin , using sine-formula.
2
2
A
2
B+C
B −C
A
⋅ cos
⋅ sin
2
2
2
B+C
A
= 90 −°
2
2
∴ sin
B+ C
A
= cos
2
2
∴
R.H.S.
(Q
= 2 k cos
A
B−C
A
⋅cos
⋅ sin
2
2
2
= k ⋅ sinA ⋅ cos
= a ⋅ cos
MATHEMATICS
A + B + C = π)
B−C
2
B− C
= L.H.S
2
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Relations Between Sides and Angles of a Triangle
MODULE - IV
Functions
Example 19.2 Using sine formula, prove that
a (cosC − cosB ) = 2 ( b −c ) cos 2
A
2
Solution : We have
a
b
c
=
=
=k
sinA sinB sinC
Notes
⇒
∴
(say)
a = k sin A, b = k sin B, c = k sin C
2
R.H.S = 2k (sinB −sinC) ⋅cos
= 2k ⋅2cos
A
2
B+ C
B −C
A
⋅ sin
⋅ cos 2
2
2
2
= 4ksin
A
B−C
A
B− C
A
⋅ sin
⋅ cos 2 = 2asin
⋅ cos
2
2
2
2
2
= 2asin
B+C
B −C
⋅ sin
2
2
= a ( cosC − cosB)
= L.H.S.
Example 19.3 In any triangle ABC, show that
asinA − bsinB = csin (A −B )
Solution : We have
a
b
c
=
=
=k (say)
sinA sinB sinC
L.H.S.
= ksinA ⋅ sinA− ksinB⋅ sinB
= k sin 2 A −sin 2 B


= ksin ( A + B ) ⋅ sin ( A − B)
A + B = π −C ⇒ sin ( A + B) =sinC
∴
L.H.S. = ksinC ⋅sin ( A− B)
= csin ( A − B ) = R.H.S.
Example 19.4 In any triangle, show that
a (b cosC − ccosB) = b2 −c2
Solution : We have,
162
a
b
c
=
=
=k (say)
sinA sinB sinC
MATHEMATICS
Relations Between Sides and Angles of a Triangle
MODULE - IV
Functions
= k sinA ( k sinBcosC − k sinCcosB)
L.H.S.
= k 2 ⋅ sinA sin ( B − C ) 
Q sin A = sin ( B + C ) 
= k 2 ⋅sin (B+ C )⋅ sin ( B
− C)
= k 2 (sin 2 B −sin 2 C )
Notes
= k 2 sin 2 B −k 2 sin 2 C
= b 2 −c2 =R.H.S
CHECK YOUR PROGRESS 19.1
1.
Using sine-formula, show that each of the following hold :
(i)
A −B
2 = a −b
A+ B a +b
tan
2
(iii)
asin
tan
(v)
2.
(ii)
bcosB + ccosC = acos ( B − C )
B−C
A
b+ c
B +C
B −C
= ( b − c ) cos
= tan
⋅cot
(iv)
2
2
b −c
2
2
acosA + bcosB +ccosC =2asinBsinC
In any triangle if
a
b
=
, prove that the tiangle is isosceles.
cosA cosB
19.2 COSINE FORMULA
Result 2 : In any triangle, prove that
(i)
cosA =
b 2 + c 2 −a 2
2bc
(ii) cosB =
c 2 + a 2 −b 2
a 2 + b 2 −c 2
(iii) cosC =
2ac
2ab
Proof :
(i)
(ii)
(iii)
Fig. 19.2
MATHEMATICS
163
Relations Between Sides and Angles of a Triangle
MODULE - IV Three cases arise :
Functions (i)
When ∠C is acute
(iii)
(ii)
When ∠C is a right angle
When ∠C is obtuse
Let us consider these one by one :
Notes
Case (i) When ∠C is acute
AD
= sinC
AC
AD = b sin C
⇒
 DC

Q b = cosC
Also BD = BC −DC =a −bcosC
From Fig. 19.2 (i)
c 2 = ( bsinC) 2 + ( a − bcosC)2
= b 2 sin 2 C +a 2 +b 2 cos2 C 2abcosC
−
= a 2 +b 2 −2abcosC
⇒
a 2 + b 2 −c 2
cosC=
2ab
Case (ii) When ∠C = 90°
c 2 = AD2 + BD2 = b2 + a2
As
C = 90° ⇒ cos C = 0
∴
c 2 = b 2 + a 2 − 2ab ⋅ cosC
⇒
cosC =
b 2 + a 2 −c 2
2ab
Case (iii) When ∠C is obtuse
AD
= sin (180−° C) =sinC
AC
∴
Also,
∴
AD = bsinC
BD = BC +CD =a bcos
+
(180−
° C)
= a −bcosC
c 2 = ( bsinC ) + ( a − b cosC )
2
2
= a 2 + b 2 − 2ab cosC
⇒
164
cosC =
a 2 + b 2 −c 2
2ab
MATHEMATICS
Relations Between Sides and Angles of a Triangle
MODULE - IV
Functions
a 2 + b 2 −c 2
∴ In all the three cases, cosC =
2ab
Similarly, it can be proved that
c 2 + a 2 −b 2
b2 + c 2 − a 2
and
cosA =
2ac
2bc
Let us take some examples to show its application.
cosB =
Notes
Example 19.5 In any triangle ABC, show that
cosA cosB cosC a 2 + b2 +c2
+
+
=
a
b
c
2abc
Solution : We know that
cosA =
∴ L.H.S. =
b 2 + c 2 −a 2
c 2 + a 2 −b 2
a 2 + b 2 −c 2
, cosB =
, cosC =
2bc
2ac
2ab
b 2 + c 2 − a 2 c 2 +a 2 −b2 a2 +b2 −c2
+
+
2abc
2abc
2abc
=
1
[ b 2 + c 2 − a 2 + c 2 + a 2 − b 2 + a 2 +b 2 − c 2 ]
2abc
=
a 2 + b 2 +c 2
=R.H.S.
2abc
Example 19.6 If ∠ A = 60 °, show that in ∆ABC
(a + b +c)(b +c a)
− =
3bc
Solution : cosA =
b 2 + c 2 −a 2
2bc
A = 60° ⇒ cosA = cos60 °=
∴ (i) becomes
1
2
1 b2 + c2 −a 2
=
2
2bc
⇒
b 2 + c2 − a2 = bc
or
b 2 + c 2 +2bc −a 2 =3bc
( b + c )2 − a 2 = 3bc
or
or
.....(i)
( b + c +a ) (b +c −
a ) = 3bc .
MATHEMATICS
165
Relations Between Sides and Angles of a Triangle
MODULE - IV Example 19.7 If the sides of a triangle are 3 cm, 5 cm and 7 cm find the greatest angle of a
Functions triangle.
Solution : Here
a = 3 cm, b = 5 cm, c = 7 cm
We know that in a triangle, the angle opposite to the largest side is greatest
Notes
∴ ∠C is the greatest angle.
cosC =
∴
a 2 + b 2 −c 2
2ab
=
cosC =
∴
9 + 25 −49 −15 −1
=
=
30
30
2
−1
2
⇒ C=
∴ The greatest angle of the triangle is
2π
3
2π
or
3
120°.
Example 19.8 In ∆ABC, if ∠ A = 60 °, prove that
Solution : cosA =
b 2 + c 2 −a 2
2bc
or
cos60 °=
∴
b 2 + c2 − a2 = bc
R.H.S. =
166
1 b2 + c2 − a2
=
2
2bc
b 2 + c 2 =a 2 +bc
or
b
c
+
=1 .
c + a a +b
.....(i)
b
c
ab + b 2 +c2 +ac
+
=
c + a a +b
(c + a ) ( a +b )
=
ab + ac +a 2 +bc
(c + a ) (a +b )
=
a ( a + b ) + c (a +b )
(a + c ) (a +b )
=
( a + c ) ( a + b)
=1
(a + c) ( a + b )
[Using (i)]
MATHEMATICS
Relations Between Sides and Angles of a Triangle
MODULE - IV
Functions
CHECK YOUR PROGRESS 19.2
1.
In any triangle ABC, show that
(i)
b2 − c 2
a2
sin2A +
c2 − a 2
b2
sin2B +
a 2 − b2
c2
sin2C =0
Notes
(ii) (a 2 − b 2 + c 2 ) tanB = ( b2 −c2 +a 2 ) tanC =(c 2 −a 2 +b 2 ) tanA
k
a 2 + b2 +c2
(iii) (sin2A + sin2B +sin2C ) =
2
2abc
where
2.
a
b
c
=
=
=k
sinA sinB sinC
(iv) ( b2 − c 2 ) cotA + (c 2 − a 2 ) cot B +( a 2 − b 2 ) cotC =0
The sides of a triangle are a = 9 cm, b = 8 cm, c = 4 cm. Show that
6 cos C = 4 + 3 cos B.
19.3 PROJECTION FORMULA
Result 3 : In ∆ABC, if BC = a, CA = b and AB = c, then prove that
(i) a = b cosC +c c o s B
(ii) b = c c o s A +a c o s C
(iii) c = a c o s B +b c o s A
Proof :
(i)
(ii)
(iii)
Fig. 19.3
As in previous result, three cases arise. We will discuss them one by one.
(i) When ∠C is acute :
In ∆ ADB ,
BD
= cosB
c
In ∆ ADC ,
DC
= cosC
DC = b cos C
⇒
b
a = BD + DC = c cos B + b cos C
∴
MATHEMATICS
⇒
BD = ccosB
a = c cos B + b cos c
167
Relations Between Sides and Angles of a Triangle
MODULE - IV (ii) When
Functions
∠C =90 °
BC
⋅ AB = cosB ⋅ c
AB
= ccosB + 0
a = BC =
( Q cos90° = 0 )
= ccosB +b cos90 °
= ccosB +b cosC
Notes
(iii) When ∠C is obtuse
In ∆ ADB ,
BD
= cosB ⇒ BD = c cos B
c
In ∆ ADC ,
CD
= cos ( π − C ) = − cosC
b
CD = −bcosC
⇒
In Fig.19.3 (iii),
BC = BD − CD
a = ccosB −(− bcosC )
= ccosB +b cosC
Thus in all cases, a = bcosC + ccosB
Similarly, we can prove that
b = ccosA +acosC and c = acosB +b c o s A
Let us take some examples, to show the application of these results.
Example 19.9 In any triangle ABC, show that
( b + c ) cosA + (c +a ) cosB +( a +b ) cosC =a +b c+
Solution : L.H.S. = bcosA + ccosA + ccosB + acosB + acosC + b cosC
= ( b cosA + a c o s B) + ( ccosA + a cosC) + ( ccosB + b cosC )
= c + b +a
= a +b +c =R.H.S.
Example 19.10 In any ∆ ABC , prove that
cos2A
a
Solution : L.H.S. =
=
2
−
cos2B
b
2
=
1
a
2
−
1
b2
1 − 2sin 2 A 1 − 2sin 2 B
−
a2
b2
1
a2
1
1
1
− 2 −2k 2 +2k 2 = 2 − 2
a
b
a
b
= R. H. S.
=
168
1
2sin 2 A 1 2sin 2 B
−
−
+
a2
b2
b2
2
 sinA sinB 
=
=k
∴

a
b

MATHEMATICS
Relations Between Sides and Angles of a Triangle
Example 19.11
In ∆ ABC , if a c o s A = bcosB , where a ≠ b prove that ∆ ABC is a
MODULE - IV
Functions
right angled triangle.
Solution : a c o s A = b c o s B
∴
2
2
 2 2 2
 2
a  b + c −a  = b  a +c −b 

2bc


2ac

or
a 2 ( b2 + c 2 −a 2 ) = b 2 (a 2 +c 2 −b 2 )
or
a 2 b2 + a 2 c 2 − a 4 = a 2 b 2 +b2 c 2 −b4
Notes
or
c 2 ( a 2 − b 2 ) = ( a 2 −b2 )( a 2 +b2 )
⇒
c 2 = a 2 + b2
∴ ∆ ABC is a right triangle.
Example 19.12 If a = 2, b = 3, c = 4, find cos A, cos B and cos C.
Solution :
cosA =
=
and
b 2 + c 2 −a 2
2bc
9 + 16 −4 21 7
=
=
2 × 3 ×4
24 8
cosB =
c 2 + a2 − b2 16 + 4 −9 11
=
=
2ac
2 × 4 ×2 16
cosC =
a 2 + b 2 − c2 4 + 9 − 16 −3 −1
=
=
=
2ab
2 × 2 × 3 12 4
CHECK YOUR PROGRESS 19.3
1.
If a = 3, b = 4 and c = 5, find cos A, cos B and cos C.
2.
The sides of a triangle are 7 cm, 4 3 cm and
triangle.
If a : b : c = 7 : 8 : 9, prove that
cos A : cos B : cos C = 14 : 11 : 6.
3.
13cm . Find the smallest angle of the
5.
If the sides of a triangle are x 2 + x +1 , 2x + 1 and x 2 − 1 . Show that the greatest angle
of the triangle is 120°.
In a triangle, b cos A = a cos B, prove that the triangle is isosceles.
6.
Deduce sine formula from the projection formula.
4.
MATHEMATICS
169
Relations Between Sides and Angles of a Triangle
MODULE - IV
Functions
LET US SUM UP
It is possible to find out the unknown elements of a triangle, if the relevent elements are given by
using
Notes
Sine-formula :
(i)
a
b
c
=
=
sinA sinB sinC
Cosine foumulae :
(ii)
cosA =
b 2 + c 2 −a 2
2bc
cosB =
c 2 + a 2 −b 2
2ac
a 2 + b 2 −c 2
cosC =
2ab
Projection formulae :
a = bcosC +c c o s B
b = ccosA +acosC
c = acosB +b c o s A
SUPPORTIVE WEB SITES
l
l
http://www.wikipedia.org
http://mathworld.wolfram.com
TERMINAL EXERCISE
In a triangle ABC, prove the following (1-10) :
1.
a sin (B − C) + bsin (C −A ) +csin (A −B) =0
2.
a c o s A + b c o s B + ccosC =2asinBsinC
3.
170
b2 − c 2
a2
⋅ sin2A+
c2 − a 2
a2 − b2
⋅ sin2B
+
⋅ sin2C
=
b2
c2
0
MATHEMATICS
Relations Between Sides and Angles of a Triangle
4.
1 +cosBcos ( C −A)
=
b2 + c 2 1 + cosAcos ( B −C)
c2 + a 2
5.
c − bcosA c o s B
=
b − ccosA cosC
6.
a − bcosC s i n C
=
c − bcosA sin A
7.
( a + b + c)  tan A + tan B  = 2c cot C

2
2
2
8.
sin
9.
(i)
b c o s B + ccosC =acos (B −C)
(ii)
a c o s A + bcosB = ccos (A −B )
MODULE - IV
Functions
Notes
A−B a−b
C
=
cos
2
c
2
B
B
2
+(c +a) sin 2
2
2
10.
b 2 = (c −a) cos 2
11.
In a triangle, if b = 5, c = 6, tan
12.
In any ∆ ABC , show that
2
A
1
=
, then show that a = 41 .
2
2
cosA b − acosC
=
cosB a − b cosC
MATHEMATICS
171
Relations Between Sides and Angles of a Triangle
MODULE - IV
Functions
ANSWERS
CHECK YOUR PROGRESS 19.3
Notes
1.
cosA =
4
5
cosB =
3
5
cosC = zero
2.
172
The smallest angle of the triangle is 30°.
MATHEMATICS