Transcript Document

Mathematics
Session
Properties of Triangle - 1
Session Objectives
Session Objective
•Sine Rule
•Cosine Rule
•Projection Rule
•Tangents Rule (Napier’s Analogy)
•Half Angle Formulae
•Area of Triangle
•Circumcircle of a Triangle and Its Radius
•Incircle of a Triangle and Its Radius
•Excircle of a Triangle and Its Radius
Introduction
A
c
A
B
B
b
C
C
a
BAC  A
CBA  B
ACB  C
BC = a
CA = b
AB = c
(a) A  B  C  
(b) a + b > c; b + c > a; c + a > b
Sine Rule
A
B
D
C
In any triangle, sides are proportional to the sines of
the opposite angles, i.e.
a
b
c


.
sin A sinB sinC
IF A,B,C are in A.P. then B = 60o
Sine Rule
Proof :When ABC is an acute angled triangle.
Draw AD perpendicular to BC
In ABD,
A
AD
sinB 
 AD  c sinB ..........(i)
AB
In ACD
AD
sinC 
 AD  bsinC .......... ii  B
AC
b
c
D
a
C
Sine Rule
From (i) and (ii), we get
c sinB = b sinC
c
b


sinC sinB
Similarly
a
c

sin A sinC
a
b
c
Hence


sin A sinB sinC
Proved.
Question
Illustrative Problem
If in a ΔABC,
sinA sin(A - B)
=
,
sinC sin(B - C)
then prove that a2 ,b2 , c2 are in A.P.
Solution :
sinA sin(A - B)
=
,
sinC sin(B - C)
 sinAsin(B - C) = sinCsin(A - B)
 sin  - (B + C) sin(B - C) = sin  - (A +B) sin(A - B)
 sin(B + C)sin(B - C) = sin(A +B)sin(A - B)
Solution
 sin2B - sin2C
= sin2 A - sin2B ....(i)
a
b
c


K
sinA sinB sinC
1 2
1 2
2
 from (i) 2 (b  c )  2 (a  b2 )
K
K
 (b2  c2 )  (a2  b2 )
 a2 ,b2 , c2 are in A.P.
Proved
Cosine Rule or The Law of Cosines
b2  c 2  a 2
In any ABC cos A 
2bc
Proof :-
When ABC is an acute angled 
In BCD
CD
cosC 
 CD  acosC ....(i)
BC
B
c
a
and in ADB
AD
cos A 
 AD  c cos A ...(ii)
AB
C
D
b
A
Cosine Rule or The Law of
Cosines
In BCD,
BC2  BD2  CD2
 BD   AC  AD
2
2
 BD2  AC2  AD2  2AC.AD

B

 AC2  BD2  AD2  2AC.AD
 AC2  AB2  2AC.AD
c
a
 a2  b2  c2  2 bc cos A [form(ii)]
b c a
 cos A 
2bc
2
2
2
C
D
b
A
Class Test
Class Exercise - 1
bc c a ab
In a ABC, if


,
11
12
13
Pr ove that
cos A cosB cosC


.
7
19
25
Solution :
bc c a ab
Let


k
11
12
13
b2  c2  a2
Since, cos A 
2bc
Let us try to find a,b,c in terms of K.
 a = ……….., b = ……...., c = ………..
 b + c = 11k; c + a = 12k; a + b = 13k
Adding we get 2 (a + b + c) = 36k
Solution
 a + b + c = 18k
 a = 7k, b = 6k, c = 5k
b2  c 2  a2 36k 2  25k 2  49k 2
 cos A 

2bc
2  6k  5k
 cos A 
Similarly
1
cos A
1


.........(i)
5
7
35
cosB
1
cosC
1

and

19
35
25
35
cos A cosB cosC



7
19
25
Pr oved.
Projection Rule
In any ABC c = a cosB + b cosA
Proof :When ABC is an acute angled .
In ACD
cos A 
AD
 AD  bcos A ....(i)
AC
C
In BCD
BD
 BD  a cosB ....(2)
BC
Now c = AB = AD + BD
cosB 
 c  b cos A  acosB Proved.
A
D
B
Class Test
Class Exercise - 2
In any ABC, prove that
cos A
cosB
cosC
a2  b2  c 2



.
bcosC  c cosB c cos A  acosC acosB  bcos A
2abc
Solution :
LHS =
cos A
cosB
cosC


bcosC  c cosB c cos A  acosC acosB  bcos A
Solution

cos A cosB cosC


a
b
c
[Applying projection rule]
1 b2  c 2  a2 1 c 2  a2  b2 1 a2  b2  c 2
 
 
 
a
2bc
b
2ca
c
2ab
[Applying cosine rule]
b2  c 2  a2  c 2  a2  b2  a2  b2  c 2

2abc
a2  b2  c 2

 RHS
2abc
Pr oved
Tangents Rule (Napier’s Analogy)
In any ABC tan
Proof :-
BC bc
A

cot
2
bc
2
By sine rule, we have
a
b
c


 k  say 
sinA sinB sinC
 a  k sinA, b  k sinB, c  k sinC
b  c k  sinB  sinC 


b  c k  sinB  sinC 
BC
BC
.sin
2
2

BC
BC
2sin
.cos
2
2
2cos
Tangents Rule (Napier’s Analogy)
 cot
BC
BC
.tan
2
2
BC
 A
 cot    tan
2
2 2

A  B  C  
A
BC
 tan tan
2
2
BC bc
A
 tan

cot
2
bc
2
Proved
Half Angle Formulae
1. Formulae for sin
A
B
C
, sin , sin
2
2
2
In any ΔABC, let a+b+c = 2s, then
A
sin 
2
 s  b  s  c 
bc
Proof :-
b2  c 2  a2
We have cos A 
2bc
b2  c 2  a2
 1
2bc
A
and 2sin
 1  cos A
2
2bc  b2  c 2  a2

2bc
2
Half Angle Formulae


a2  b2  c 2  2bc

a  b  c 
2

2bc
2
2bc
A  a  b  c  a  b  c 
 2 sin

2
2bc
2
a  b  c  2s, we have
 2s  2c  2s  2b 
2 A
 2 sin

2
2bc
A
 sin 
2
 s  c  s  b 
bc
A 4  s  c  s  b 
 sin

2
4bc
A




A

A

  sin  0 


2 2
2
Proved.
2
Half Angle Formulae
A
B
C
2. Formula for cos , cos , cos
2
2
2
In any ABC,let a  b  c  2s,then
A
cos 
2
s s  a
bc
A
B
C
3. Formula for tan , tan , tan
2
2
2
In any ABC,if a  b  c  2s then
A
tan 
2
 s  b  s  c 
s s  a
Class Test
Class Exercise - 5
In a ABC, if acos2
show that cot
C
A 3b
 c cos2  ,
2
2
2
A
B
C
, cot , cot are in AP.
2
2
2
Solution :
C
2 A 3b
 c cos

Given that acos
2
2 2
2
s(s  c)
s(s  a) 3b
 a.
 c.

ab
bc
2

s
3b
2s

a

c


 2
b
Solution
s
3b
 .b 
 2s  3b
b
2
 a + b + c = 3b
a + c = 2b
... (i)
We have to prove
cot
A
B
C
, cot , cot
are in A.P.
2
2
2
A
c
B
i.e., cot  cot  2cot
2
2
2
A
C
L.H.S  cot  cot
2
2
(ii)
Solution
s(s  a)
s(s  c)

(s  b) (s  c)
(s  a) (s  b)

s(s  a)
s(s  c)

s (s  a) (s  b) (s  c)
s (s  a) (s  b) (s  c)



s(s  a) s(s  c)



s
2s  2b


[from (i)]
B
 2cot  R.H.S
2
2s(s  b)


Area of a Triangle
(a)The area of ABC is given by
1
1
1
  bc sin A  ca sinB  ab sinC
2
2
2
Proof :Here BC = a, AB = c, AC = b
A
1
Area of ABC   base  height
2
B
D
C
Area of a Triangle
1
   BC  AD....(i)
2
sinB 
AD
AD
 sinB 
AB
C
In ABD

AD  c sinB
From (i)  
1
 a  c sinB
2
Proved
Area of a Triangle
(b)
Hero’s formula
  s  s  a  s  b  s  c 
Proof :-
1
  bc sin A
2
1 
A
A
 bc  2sin cos   bc
2 
2
2
 s  b  s  c 
s s  a
bc
bc
 s  s  a  s  b s  c 
Class Test
Class Exercise - 6
In any ABC, if   a  b  c  ,
2
8
then prove that tan A  , where
15
  area of triangle.
2
Solution
Given   a2  (b  c)2
   a2  b2  c 2  2bc
 b2  c 2  a2  2bc  
1
 2bc cos A  2bc  bc sin A
2


b2  c2  a2
1
and   bc sin A 
 cos A 
2bc
2


 4bc cosA + bc sinA = 4bc
 4cosA + sinA = 4
Solution
A
A
A

 4  1  2sin2   2 sin cos  4
2
2
2

A
A

cos

4sin
0


2
2

A
A
A
Either sin  0 or cos  4 sin  0
2
2
2
A
1
2 tan
2
A
2
4  8
where sin  0  A  0 or 2 Now, tan A 

A
1 15
2
1  tan2
1
 2sin
A
2
2
which is not possible, since it in an angle of .
A
A
when cos  4 sin  0
2
2
A 1
 tan 
2 4
16
Circumcircle of a Triangle and
its Radius
A
F
B
R
R
O
E
R
D
Circum radius = R
C
Circumcircle of a Triangle and
its Radius
In any ABC
a
b
c
(i)


 2R
sin A sinB sinC
abc
(ii) R 
4
Incircle of a Triangle and its
Radius
A
A
2
F
A
2
r r
E
r i
B
D
In Radius = r
C
Incircle of a Triangle and its
Radius
In any ABC

(i) r 
s
A
B
C
(ii) r   s  a  tan   s  b  tan   s  c  tan
2
2
2
B
C
A
C
B
A
a sin sin
b sin sin
c sin sin
2
2 
2
2 
2
2
(iii) r 
A
B
C
cos
cos
cos
2
2
2
A
B
C
(iv) r  4R sin sin sin
2
2
2
Radii of Ex-circle in Terms of
Sides and Angles
A
B
C
I1
Ex. Radius is = r1
Radii of Ex-circle in Terms of
Sides and Angles
In any ABC



(i) r1 
; r2 
; r3 
sa
sb
sc
A
B
C
(ii) r1  s tan ; r2  s tan ; r3  s tan
2
2
2
B
C
C
A
A
B
acos cos
bcos cos
c cos cos
2
2; r 
2
2 r 
2
2
(iii) r1 
2
3
A
B
C
cos
cos
cos
2
2
2
Radii of Ex-circle in Terms of
Sides and Angles
A
B
C
(iv) r1  4R sin cos cos
2
2
2
A
B
C
r2  4Rcos sin cos
2
2
2
A
B
C
r3  4R cos cos sin
2
2
2
[Proof of these results are same as INCIRCLE
and try yourself]
Class Tests
Class Exercise - 3
In a if b + c = 3a, then the value of
B
C
cot cot
is
2
2
(a) 1
(b) 2
(c)
3
Solution :b + c = 3a
= k (sinB + sinC) = 3k sinA
 2 sin
BC
BC
A
A
.cos
 3.2 sin cos
2
2
2
2
A
BC
BC
A
 cos cos
 3 cos
.cos
2
2
2
2
(d) 3
Solution




A BC   

A B C 
  
2 2 2 2 
 cos
B
C
B
C
cos  sin sin  3
2
2
2
2
 2cos
 cot
B
C
B
C

cos
cos

sin
sin

2
2
2
2 

B
C
B
C
cos  4 sin sin
2
2
2
2
B
C
cot  2
2
2
A
(note : cot  0)
2
Ans : b
Class Exercise - 4
In a ABC, if
cos A cosB cosC


a
b
c
and the side a = 2 units, then area
of the ABC, is
(a) 1 sq. units
(c)
3
sq. units
2
(b) 2 sq. units
(d) 3 sq. units
Solution
Given
cos A cosB cosC


a
b
c
cos A
cosB
cosC



k sin A k sinB k sinC
By sine rule
 cot A  cotB  cotC
A=B=C
 ABC is equilateral
 Area of equilateral  
3

4
4
3
4
 3 sq. units.
 side 
2
Ans : d
Class Exercise - 7
ABC, sides a, b, c are in AP, then
r1, r2 , r3 are in (where r1, r2 , r3 are the ex-
If in a
radii of the ABC, )
(a) AP
(b) GP
(c) HP
(d) None of these
Solution
Since a.b.c are in AP
 b a  c b
 (s  a)  (s  b)  (s  b)  (s  c)
(s  a) s  b s  b s  c








1 1 1 1

  
r1 r2 r2 r3
1 1 1
 ,
,
are in A.P
r1 r2 r3
 r1, r2, r3 are in H.P
Ans : c
Class Exercise - 8
If a circle is inscribed in an equilateral
triangle of side a, then area of the
square inscribed in the circle is
a2
(a)
6
2a2
(c)
5
a2
(b)
3
2a2
(d)
3
Solution
1
3a
s  a  b  c  
2
2

3 2
a
4
r

3 2 2
a

a 

...........(i)
s
4
3a 2 3
If x is the length of a side of the square
inscribed in circle of the triangle, then
A
x2 x2   diameter 
2
a2
 2x   2r   x  2r 
[from(i)]
6
a2
 Area of the square 
Ans : a
6
2
2
2
2
B
C
Class Exercise - 9
If be the length of perpendiculars
drawn from the vertices of a triangle
to the opposite side, then
(a) p1 p2 p3  abc
(c) p1 p2 p3  8R
(b) p1 p2 p3  a2 b2 c 2
(d) p1 p2 p3 
a2 b2 c 2
8R3
Solution
Here AD  p1, BE  p2, CF  p3
and  
1
1
1
ap1  bp2  c p3
2
2
2
2
2
2
 p1 
, p2 
, p3 
a
b
c
 p1 p2p3 
2 2 2 
. .
a b c
A
E


8
abc
3
3

a 2 b 2c 2
8R
3
8  abc 
.

abc  4R 
Ans : d



R
F
abc 
4  
B
D
C
Class Exercise - 10
If in a ABC,
a2  b2
a b
2
2

sin  A  B 
sin  A  B 
,
prove that it is either a right angled
or an isosceles triangle, can it be
right angled isosceles triangle?
Solution
Let
a
b
c


k
sin A sinB sinC
sin  A  B

2
2
sin  A  B
a b
a2  b2




 sin
  sin  A  B
A  sin B  sin  A  B 
k 2 sin2 A  sin2 B
k2
2
2
sin    C sin  A  B
sin2 A  sin2 B
sinC sin  A  B 
sin A  sin B
2
2


sin  A  B sin  A  B 
sin  A  B

sin    C
sin  A  B 
sinC
sin2 A  sin2 B

sin  A  B 
sin  A  B
Solution
sinC
1 

 sin  A  B   2

0

2
 sin A  sin B sinC 
 Either sin  A  B   0 or
 Either A  B  0
 Either A  B or
1
sinC

0
2
2
sinC sin A  sin B
or sin2 A  sin2 B  sin2 C  0
a2
k
2

b2
k
2

c2
k
2
0
 Triangle is either isosceles or right angled.
Proved.
Thank you