SOLUTION OF A

TRIANGLE
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SOLUTIONS OF TRIANGLE

1. I nt r oduct ion

There are 6 elements in a ABC, the three sides BC = a, CA = b, AB = c and the three angles
A, B, C.
A + B + C =  = 180Ĉ

1
1. Area of the triangle () = (Base) (height)
2 A

The area of the triangle is a.h/2.

c b
But in triangle BAH, we have sin(B) = h/c. h
Hence the area of the triangle is a.c. sin(B)/2.
B a H C
Similarly we have that the area of the triangle
= b.c.sin(A)/2 = a.b.sin(C)/2

A
The area of a triangle ABC = a.c. sin B = b.c. sin = a.b. sin C
2 2 2

2. Sine rule : In a triangle ABC we have seen that

area = a.c.sin (B)/2 = b.c.sin(A)/2 = a.b.sin(C)/2
 a.c.sin(B) = b.c.sin(A) = a.b.sin (C)
dividing through out by a.b.c., we get
In any triangle ABC we have

a b c
 
sin A sin B sin C

3. Cosine rule : In any triangle ABC we have

a2 = b2 + c 2 ă 2bc cos A
b2 = c2 + a 2 ă 2ca cos B
c 2 = a2 + b 2 ă 2ab cos C

b2  c2  a 2
cos A 
2bc

a2  c2  b2
cos B 
2 ac

a2  b2  c2
cosC 
2ab
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4. Projec tion formula :
a = b cos C + c cos B
b = c cos A + a cos C
c = a cos B + b cos A
5. Semi-Perimeter of  ABC (s) :

a  b c
s
2

 2s = a + b + c
2s ă 2a = b + c ă a
2s ă 2b = c + a ă b
2s ă 2c = a + b ă c
6. Ha lf a ngle fo rmu la e :

A (s  b)(s  c)
sin 
2 bc

A s(s  a)
cos 
2 bc

A ( s  b)( s  c) ( s  b)( s  c)
tan  
2 s(s  a) 

A s( s  a) s( s  a)
cot  
2 ( s  b)( s  c) 

Tip : for sine formula, in numerator the other sides are taken and the side opposite to the
angle is not in the formula.

 = area of triangle ABC = s( s  a)( s  b)(s  c) (HeroÊs formula)

a bc

4R

B B B B C C C C
The expression for sin , cos , tan , cot , sin , cos , tan , cot can be derived using
2 2 2 2 2 2 2 2
symmetry.

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7. Find the value of sin A, using sin A/2 & cos A/2

2 2
sin A = s(s  a)(s  b)(s  c) 
bc bc

8. Napier Analogy :

B C b c A
tan  cot
2 b c 2

3. C ir cles C onn ect ed wit h T r iangle

(i) Circum-circle
The circle passing through the vertices of the triangle ABC is called the circum-circle. Its
radius R is called the circum-radius. In the triangle ABC, A

a b c
R  
2sin A 2sin B 2sin C O
R
abc
R B C
4

(ii) In-circle
The circle touching the three sides of the triangle internally is called the inscribed or the in-
circle of the triangle. Its radius r is called the in-radius of the circle. In the triangle ABC,

 A
r
s

A B C
r  (s  a) tan  ( s  b) tan  (s  c) tan O
2 2 2
r
A B C B C
r  4R sin sin sin
2 2 2

B C A C A B
a sin sin b sin sin c sin sin
r 2 2  2 2  2 2
A B C
cos cos cos
2 2 2

A B C 1
Remark : From r = 4R sin sin sin , we find that r  4R.
2 2 2 8
 2r  R. Here equality holds for the equilateral triangle.

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(iii) Escribed circles
The circle touching BC and the two sides AB and AC produced of ABC externally is called
the escribed circle opposite A. Its radius is denoted by r1. Similarly r2 and r3 denote the radii
of the escribed circles opposite angles B and C respectively.
r 1, r 2, r 3 are called the ex-radii of ABC. Here

 A A B C
r1   s tan  4R sin cos cos , B
s a 2 2 2 2

r1 O1
 B B C A
r2   s tan  4R sin cos cos ,
sb 2 2 2 2
A
C
 C C A B
r3   s tan  4R sin cos cos ,
s c 2 2 2 2

r 1r 2 + r2r 3 + r 3r 1 = 4R + r,

r1r2r3
r1r2  r2r3  r3r1  s2 
r

4. m – n T heor em

If a point D divides the side BC of ABC internally in the ratio m : n and BAD = , DAC =
 and ADC =  then A
(m + n) cot  = m cot  ă n cot   
= n cot B ă m cot C


B m : D n C

The result can be derived using sine rule in ABD and ADC.

Illustration 1

In the figure, ABC is a triangle in which angle C = 90Ĉ C

and AB = 5 cm. D is a point on AB such that AD = 3 cm
and  ACD = 60Ĉ. Find the length of side AC. 60Ĉ

A 3 cm D B
5 cm

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Solution :
Using m ă n theorem,
(3 + 2) cot CDA = 2 cot 30Ĉ ă 3 cot 60Ĉ

3
 cot CDA =
5

Now, using sine rule in CDA,

AC AD

sin CDA sin ACD

3 5 3
 AC  . 5 cm
sin 60 28 7

1. Regular n sides Polygon O

a = side length; r = in-radius; R = circum-radius
R a
a a r
r R 
 and 
2 tan 2sin
n n
a/2 a/2

2. Distance of orthocentre from vertices of triangle :

AD, BE are altitudes and H is the orthocentre of a triangle ABC as shown. As quadrilateral
CEHD is cyclic, (as opposite angles are supplementary)
angle EHA = angle C A
From AHE, AH sin C = AE
 AH sin C = AB cos A = AE
F E
c cos A  c  H
 AH =   cos A
sin C  sin C 
B D C
 AH = 2R cos A
 distances of orthocentre (H) from the vertices A, B & C are :
2R cos A, 2R cos B & 2R cosC respectively.
Note : The triangle formed by joining the points D, E & F is known as pedal triangle.

3. Distance of orthocentre from sides of triangle :

DH = AD ă AH
 DH = AB sin B ă 2R cos A

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 DH = c sin B ă 2R cos A

 DH = 2R sin C sin B + 2R cos (B + C)

 DH = 2R cos B cos C
 The distances of orthocentre (H) from the sides BC, CA & AB are :

2R cosBcosC, 2R cosCcosA and 2R cosAcosB respectively.

4. Distance of circumcentre O from sides : A

BOC = 2A
O
COM = A R

OM = R cos A B M C
The distances of circumcentre from sides BC, CA & AB are :
RcosA, RcosB and RcosC respectively.

I mp or t an t T heor em

The centroid (G), circumcentre (O) & orthocentre (H) in any triangle are collinear. The
centroid divides the line joining orthocentre and circumcentre in 2 : 1 internally.

OG 1 2
i.e.  1
GH 2
H G O

or 1
OG = OH
3

2
& HG = OH
3

The lengths of median

A
1 2 2
AD = b  c  2 bccos A
2

F  
1
BE = c 2  a 2 2a cos B
2

1
CF = a2  b2 2ab cosC B D C
2

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Distance between orthocentre (H) & circumcentre (O)

OH = R 1  8 cosA cosB cosC

1 R
OG = OH = 1  8 cosA cosB cosC
3 3
Distance between circumcentre (O) & incentre (I) / excentre

A B C
OI = R 1  8sin sin sin
2 2 2

A B C
OI1 = R 1  8sin cos cos
2 2 2

A B C
OI2 = R 1  8 cos sin cos
2 2 2

A B C
OI3 = R 1  8cos cos sin
2 2 2 A
Length of angle bisector

2 bc A
AD = cos
b +c 2
B C
D
Illustration 2

A BC
Prove that : ( b  c ) sin  a cos
2 2

Solution :
Since b = 2R sin B and c = 2R sin C

A A
(b  c)sin  2R (sin B + sinC) sin
2 2

B+C B C A
= 4R sin .cos .sin
2 2 2

 A A B C
= 2R  2cos 2 .sin 2  cos 2
 

BC BC
= 2R sin A cos  a cos
2 2

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Illustration 3

Prove that a cos A + b cos B ă c cos C = 2 c cos A cos B

Solution :
L.H.S. = 2R {sin A cos A + sin B cos B ă sin C cos C}
= R {sin 2A + sin 2B ă sin 2C}
= R {2 sin (A + B) cos (A ă B) ă 2 sin C cos C}
= R {2 sin C cos (A ă B) + 2 sin C cos (A + B)} since A + B =  ă C
= 2R sin C {cos(A ă B) + cos (A + B)} = 4R sin C cos A cos B
= 2c cosA cos B since c = 2R sinC

Illustration 4

b c c a a b cos A cosB cosC

If   , prove that :  
11 12 13 7 19 25

Solution :

b c c a a  b b cc a a b a b c A C A+C
    (By ratio proportional i.e. = = )
11 12 13 11  12  13 18 B D B+D

b c a c a b ab c
 ,  ,  , then
11 7 12 6 13 5

a b c
Let    k (say)
7 6 5

b2  c2  a2 (62  52  72 ) 1 7
cos A =  k2 2
 
2 bc k 2(6) (5) 5 35

c 2  a 2  b2 k2 (52  72  62 ) 19
cos B =  
2ca k2 2 (5) (7) 35

a 2  b 2  c 2 k 2 (7 2  6 2  5 2) 5 25
cos C =   
2 k 2 2(7) (6) 7 35

cos A cos B cos C

  
7 19 25

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Illustration 5

In any triangle ABC, prove that (b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c.

Solution :
L.H.S.
= (b + c) cos A + (c + a) cos B + (a + b) cos C
= b cos A + c cos A + c cos B + a cos B + a cos C + b cos C
= (b cos A + a cos B) + ( c cos A + a cos C) + (c cos B + b cos C)
= c + b + a = R.H.S. [By using projection Rule]

Illustration 6

4
In a triangle ABC, a = 6, b = 3 and cos (A ă B) = . Find the angle C.
5
Solution :

A B a b C
Since, tan  cot
2 a b 2

1  cos(A  B) ab C
  cot
1  cos A  B a b 2

4
1
5  6  3 cotC
 4 6 3 2
1
5

C C
 cot  1   45  C  90
2 2

Illustration 7

A B C
If in a triangle ABC, tan , tan , tan , are in harmonic progression, then show that
2 2 2
the sides a , b , c are in arithmetic progression.
Solution :

A B C
tan , tan , tan are in H.P.
2 2 2

A B C
 cot , cot , cot are in A.P.
2 2 2

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B A C
 2cot  cot  cot
2 2 2

s( s  b) s( s  a) s( s  c)
 2  
( s  a)( s  c) ( s  b)( s  c) ( s  a)( s  b)

Multiply throughout by (s  a )(s  b )(s  c )

 2 (s ă b) = (s ă a ) + (s ă c)
 2b = a + c
 a, b, c are in A.P.

7. Solut ion of t r iangles

We have studied that a triangle has six parts or six elements viz three sides and three angles.
From geometry, we know that when any three elements are given of which necessarily a side is
given, the triangle is completely determined i.e., remaining three elements can be determined. The
process of determining the unknown elements knowing the known elements is known as the
solution of a triangle. In practice, there are four different cases for which the solution is discussed
as under.

Case 1 : When all three sides are given.

To solve a triangle given the three sides a, b, c .
To determine angles A, B and C
The angles A, B and C are determined by using the following relations :

A ( s  b)( s  c)
tan 
2 s( s  a )

A (s  b)(s  c)
sin 
2 bc

A s(s  a)
cos 
2 bc

b2  c2  a2
or by cos formulas, cos A 
2bc

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Illustration 8

The sides of a triangle are 20, 30 and 21. Find the greatest angle.
Solution :
The grea test a ngle is opposite to the side whose length is 30.
Let a = 20, b = 30, c = 21
We have to find angle B.

B (s  a )(s  c )  a  b c 
sin   where s  
2 ca  2 

(35.5  20)(35.5  21)

=
20  21

(15.5  14.5)
= = 5351
20  21

= 0.7315

B
sin  0.7315
2

B
 47 (Approximately)
2

B = 94Ĉ
Case 2 : When two sides and the angle included between these are given.

Illustration 9

If b = 251, C = 147, A = 47Ĉ, find the remaining angles.

(Use NapierÊs rule)
Solution :

BC bc A
tan  cot
2 b c 2

B  C 251  147
 tan  cot 23  30´
2 251  147

26
= tan 66 30´ = 0.268 ï 2.2998 {we have used these]
97
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B C
 tan  0.6164
2

B C
  31  39´ (From the trignometric tables)
2

 B ă C = 63Ĉ 18´
But B + C = 133Ĉ (180Ĉ ă A = B + C)
Adding 2B = 196Ĉ 18´
 B = 98Ĉ 9´
 C = 34Ĉ 51´

Case 3 : To solve a triangle having given two angles and a side.

Let the given parts be denoted by B, C, a, then the third angle A can be found from the relation.
A = 180Ĉ ă (B + C)

a sin B
b
sin A

 log b = log a + log sin B ă log sin A

Similarly, c can be found from the equation
log c = log a + log sin C ă log sin A

Illustration 1 0

Solve the triangle ABC, given a = 18, A = 25Ĉ, B = 108Ĉ

Suggested answer :
C = 180Ĉ ă (25 ă 108Ĉ) = 47Ĉ

a sin B
b
sin A

Taking log on both sides, we get

log b = log a + log (sin B) ă log (sin A)
= log 18 + log (sin 108Ĉ ) ă log (sin 25Ĉ)
= log 18 + log (sin 72Ĉ ) ă log (sin 25Ĉ)
= log 18 + log (0.9511) ă log (0.4226)
(from Trigonometric table)
= 1.2553 + 1.9782 ă 1.6259

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= 1.2553 ă 1 + 0.9782 + 1 ă 0.6295

log b = 1.076
Taking antilog, we get b = 40.52.
Similarly using
log c = log a + log (sin C) ă log (sin A),
we can evaluate the value of C as 31.23Ĉ.

Ambiguous Case
Case 4 : Given two sides and the angle opposite to one of the sides

Given the elements b, c and B of a triangle. This case is called Ambiguous case.
Draw the specified side AB (= c) and make the angle B (= ABD) as given. Now to make the
triangle mark an arc of radius equal to b with centre at point A. This arc may intersect or touch
or neither intersect nor touch the line BD. Accordingly we have the following results :

Ć If b < c sin B, there is no triangle. A

B D

Ć Ib b = c sin B and B is acute, there is one right angled triangle and right angle at C.
A

c b

B D
C
Ć If b > c sin B and b < c and B is acute, there are two triangles satisfying the given
conditions. A

c b
b

B D
C2 C1

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Ć If b > c, there is only one triangle.

c b

B D
C

Ć If B is obtuse, there is no triangle except when b > c.

Use sine rule to find angle C. Then A = 180Ĉ ă (B + C). The side a can also be found using sine
rule.

Illustration 1 1

In any ABC, if a = 2, b = 3  1 and ÂCÊ = 60Ĉ, solve the triangle.

Solution :
Two sides and included angle is given (case ă 2)

BA ba C
tan  cot
2 b a 2

3 1  2
= cot30
3 1  2

3 1
= 3
3 3

3 1 tan60   tan 45 
= 
3 1 1  tan60  tan 45 

= tan (60Ĉ ă 45Ĉ) = tan 15Ĉ

BA
  15 
2

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or B ă A = 30Ĉ ...(i)
We know, A + B + C = 180Ĉ     A + B = 120Ĉ ...(ii)
Solving (i) and (ii), we get B = 75Ĉ & A = 45Ĉ
To find side c, we use sine Rule

a c 2 3
 or c  2  6
sin A sin 60  1 2

Thus A = 45Ĉ, B = 75Ĉ and c  6

Illustration 1 2

If A = 30Ĉ, a = 100, c  100 2 , find the number of triangles that can be formed.

Solution :
Here a , c and A are given, therefore we will have to examine whether two triangles are possible
or not. For two triangles
(i) a > c sin A and (ii) a < c

100 > c  100 2 sin 30Ĉ

100 > 50 2 and

a < c i.e. 100  100 2  Two triangles can be formed.

Illustration 1 3
In the ambiguous case, if the remaining angles of the triangle formed with a , b and A be
sin C1 sin C2
B1 , C1 and B2 , C2, then prove that : sin B  sin B  2cos A
1 2

Solution :

b sin A
sin B 1 ă sin B2 = (using sine rule)
a

c1 sin A c2 sin A
sinC1  and sin C2 
a a

c1 sin A c2 sin A
a  a c1  c2 2b cos A
 L.H.S. = b sin A b sin A  L.H.S. =   2cosA
b b
a a

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Illustration 1 4
In a  ABC; a , c , A are given and b1 = 2b 2, where b 1 and b2 are two values of the third side :

then prove that 3a = c 1  8 sin 2 A

Solution :
a2 + b2 + c2 ă 2bc cos A
Consider this equation as a quadratic in b.
b2 ă (2c cos A)b + c 2 ă a2 = 0
 b1 + b2 = 2c cos A
& b1 b 2 = c2 ă a 2
& b1 = 2b 2
 3b1 = 2c cos A and 2b 12 = c 2 ă a 2

2
 2c cosA  2 2
 2  c a 0
 3 

 8c2 cos 2 A = 9c 2 ă 9a 2
 8c2 (1 ă sin2 A) = 9c2 ă 9a2
 9a2 = c2 + 8c2 sin 2 A

 3 a  c 1  8sin2 A

Illustration 1 5

Prove that : r 1 + r 2 + r3 ă r = 4R.

Solution :

      
(r1 + r2) + (r 3 ă r) =  s  a  s  b     
  s c s

 2s  a  b  s  (s  c) 
=   s  a s b     s s c 
( )( )  ( ) 

 c c 
= s a s b  ss c 
(  )(  ) (  ) 

c
= s( s  a)( s  b)( s  c)  s( s  c)  (s  a)( s  b)

 cab abc abc

= ss a s b s c  2   4R
(  )(  )(  )  

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Alternatively :

A B C B C A
r 1 + r2 = 4R sin cos cos + 4R sin cos cos
2 2 2 2 2 2

C  A B A B
= 4R cos sin 2 cos 2  cos 2 sin 2 
2  

C A +B C
= 4R cos sin    4R cos2
2  2  2

A B C A B C
r 3 ă r = 4R cos cos sin ă 4R sin sin sin
2 2 2 2 2 2

C  A B A B
= 4R sin cos 2 cos 2  sin 2 sin 2 
2  

C A + B  C
= 4R sin cos    4R sin 2
2  2  2

 2 C C
 r 1 + r2 + r 3 ă r = 4R  cos  sin2   4R
 2 2

Illustration 1 6

1 1 1 1 a 2  b2  c 2
Prove that :    
r12 r22 r32 r2 2

Solution :

1
r12

1
r22

1
r32

r
1
2

1
 2 ( s  a) 2
 ( s  b) 2  ( s  c) 2  s2 

=
1
2
4 s 2
 2 s( a  b  c)  a2  b2  c2 

=
1
 2 4s 2
 2s(2s)  a 2  b2  c 2 
a2  b2  c2
=
2
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Illustration 1 7

In a triangle ABC, the vertices A, B, C are at distance of p , q , r from the orthocentre

respectively. Show that a q r + brp + cpq = a bc .
Solution :
Let H be the orthocentre of triangle ABC
From question, HA = p, HB = q, HC = r A

p
F E

H
q r
From figure,
HBD = EBC = 90Ĉ ă C
B C
D
HCD = FCB = 90Ĉ ă B
 BHC = 180Ĉ ă (HBD + HCD)
= 180Ĉ ă [90Ĉ ă C + 90Ĉ ă B]
= B + C = 180Ĉ ă A
Similarly, AHC = 180Ĉ ă B and AHB = 180Ĉ ă C
Now, Area of BHC + Area of CHA + Area of AHB
= Area of ABC

1 1 1
 .q.r .sin BHC + .r. p.sin CHA + . p.q.sin AHB = 
2 2 2

 1
 bc.sin A
2

1 1 1
 qr .sin(180  A) + rp sin(180  B)  pq sin (180  C)  
2 2 2

1 1 1
 qr .sin A + rp sin B + pq .sin C = 
2 2 2

1 a 1 b 1 c
 qr .  pr .  pq . 
2 2R 2 2R 2 2R

abc  abc 
 aqr + brp + cpq = 4R.  = 4. .   abc   
4  4R 

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Illustration 1 8

If the bisector of the angle C of a triangle ABC cuts AB in D and the circum-circle in E,
prove that CE : DE = (a + b) 2 : c 2.
Solution :
 AD is internal bisector of C.

AD b
 
DB a
C
AD + DB a  b
or, 
DB a
C/2 C/2
ac
or, BD = {remember this result}
ab
D
Since angles of the same segment are equal. A B

C
 ABE = ACE =
2 E
and BEC = BAC = A
applying sine rule in triangle BEC,

 C
a sin  B + 
CE

BC
 CE =  2 ...(i)
sin CBE sin BEC sin A

applying sine rule in triangle BDE,

 C
acsin  
DE

BD
 DE =  2
C sin A (a  b)sin A ...(ii)
sin
2

From (i) and (ii),

 C
a sin  B + 
CE
  2
( a  b)
DE C
ac sin
2

 C
( a  b)sin  B + 
CE
  2
or, DE C ...(iii)
csin
2

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 C  C C
sin B +  sin B + 2cos
 2
  2 2
Now, C C C
sin 2sin .cos
2 2 2

sin  B + C  sin B
=
sin C

sin A +sin B a  b
=  ...(iv)
sin C c

From (iii) and (iv), we have

2
CE a  b a  b (a  b)
 . 
DE c c c2

Illustration 1 9

A B C
 cot  cot
cot
(a  b  c) 2
2 2 2
Prove that : in any  ABC 2 2 2

a  b  c cot A + cot B + cot C

Solution :

cos A 2 bccos A b2  c 2  a 2
cot A =  
sin A 2 bcsin A 4

b2  c2  a2 a2  b2  c2
  cot a   4

4

A
2cos2
1  cos A 2  cot A

Also sin A A A 2
2sin cos
2 2

2 bc(1  cos A) A
 cot
2bc sin A 2

A 2bc  2bc cos A

  cot 2 
4
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=
 2bc   (b2  c2  a2 )
4

a2  b2  c2  2ab  2bc  2ca (a  b  c )2

= 
4 4

A
 cot 2 (a  b  c )2
 
 cot A a2  b2  c2

Illustration 20

If p 1 , p 2, p3 are the length of the altitudes of a triangle ABC, prove that p 1 ă2 + p 2ă2 + p 3ă 2

(cotA + cot B + cot C)

=


Solution :

1 1 a
ap1    
2 p1 2 

1 a 2  b 2  c2
  p2 
4 2
1

2bc cos A + 2ca cosB + 2ab cos C

= 2
4

1  cos A 
=
2 2  abc  a 

abc  cos A 
=
2 2   2R sin A 

 cot A  abc 
=  R 
  4 

=
 cotA since abc = 4R.


SOLUTION OF A TRIANGLE
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Illustration 21

If a 2 , b 2, c 2 are in A.P., prove that cot A, cot B and cot C are also in A.P.
Solution :
b2 ă a2 = c2 ă b2
sin 2 B ă sin2 A = sin2 C ă sin2 B
or sin (B + A) sin (B ă A) = sin (C + B) sin (C ă B)
or sin C (sin B cosA ă cos B sin A) = sin A (sin C cos B ă cos C sin B)
Divide each term by sin A sin B sin C
 cot A ă cot B = cot B ă cot C
 cot A, cot B, cot C are in A.P.

Illustration 20

Prove that
(b + c ă a ) {cot (B/2) + cot (C/2)} = 2a cot (A/2).
Solution :

B+C
sin
L.H.S. = 2(s  a ) 2
B C
sin sin
2 2

2(s  a ).cos(A/2) 2a cos(A/2) cos(A / 2) A

= = = 2a sin A/2  2a cot 2
1/ 2
 (s  c)(s  a) (s  a) (s  b)   (s  b) ( s  c) 
1/ 2  
 .   
 ca ab  bc

Illustration 23
1 1 1
Prove : 2a bc cos A cos B cos C = (a + b + c)S.
2 2 2
Solution :

1 1 1
2 abc cos A cos B cos C
2 2 2

 s( s  a)   s( s  b)   s( s  c) 
= 2abc  .  .  
 bc   ca   ab 

= 2 s s( s  a) ( s  b) ( s  c)  ( a  b  c)S

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Illustration 24

1 1 1
tan A tan B tan C
2 2 2 1
Prove   
(a  b )(a  c ) (b  c )(b  a ) (c  a )(c  b ) S
Solution :

1  (s  b) (s  c ) 
Putting tan A   s s a  etc.,
2  (  ) 

we have, L.H.S.

  ( s  b) ( s  c)   ( s  c) ( s  a)  ( s  a) ( s  b)  
      
  s( s  a)   s ( s  b)   s( s  c)  
 
=  ( a  b) ( a  c) ( c  a) ( b  a) ( c  a) ( c  b) 

 
 

(c  b ) (s  b ) (s  c )  (a  c ) (s  c ) (s  a )  (b  a ) (s  a ) (s  b )
=
( a  b) ( b  c) ( c  a) [ s ( s  a) ( s  b) ( s  c)
Nr = ă  {(b ă c) {s2 ă s (b + c) + bc}
= ă s2  (b ă c) + s (b2 ă c 2 ) ă bc (b ă c)
= 0 + 0 + (b ă c) (c ă a) (a ă b)
Note that on simpliication,
bc ( b ă c ) + ca (c ă a) + ab (a ă b) = ă (b ă c) (c ă a) (a ă b)]
Substituting in (1), we get

Nr 1 1
L.H.S. = r
 
D [ s( s  a) ( s  b) ( s  c)] S

Illustration 25

A
Given S = a 2 ă (b ă c )2 in a triangle ABC whose area is S, then find the value of tan
2

Solution :

1
S bc sin A  ( a  b  c) (a  b  c)
2

A A 2(s  b).2(s  c ) A
or sin cos   4sin 2
2 2 bc 2

A 1
 tan 
2 4
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Illustration 26

a 3 cos (B ă C) + b 3 cos (C ă A) + c 3 cos (A ă B) = 3a bc .

Solution :
a3 cos (B ă C) + b3 cos (C ă A) + c3 cos (A ă B)
= k3 sin3 A cos (B ă C) + k3 sin3 B cos (C ă A) + k3 sin 3 C cos (A ă B)
= k3 [sin2 A sin (B + C) cos (B ă C) + sin 2 B sin (C + A) cos (C ă A) + sin2 C sin (A + B) cos (A ă B)]

1 3
= k [sin2 A [sin 2B + sin 2C) + sin 2 B (sin 2C + sin 2A) + sin2 C (sin 2A + sin 2B)]
2

= k3 [sin 2 A sin B cos B + sin 2 A sin C cos C + sin 2 B sin C cos C + sin2 B sin A cos A +
sin 2 C sin A cos A + sin2 C sin B cos B]
= k3 [sin A sin B (sin A cos B + cos A sin B) + sin B sin C (sin B cos C + cos B sin C)
+ sin C sin A (sin C cos A + cos C sin A)]
3
= k [sin A sin B sin (A + B) + sin B sin C sin (B + C) + sin C sin A sin (C + A)]
= k3 [sin A sin B sin C + sin B Sin C sin A + sin C sin A sin B]
= 3k sin A.k sin B.k sin C = 3abc.

Illustration 27

1 5 1 20 1
In a triangle ABC, if tan A = and tan B = , find tan C , and prove that in this triangle
2 6 2 37 2
a + b = 2b .
Solution :

C  A B A B 
We have tan  tan  90    cot  2  2 
2  2   

A B 6 37
cotcot  1 . 1
2 2  5 20
= B A 37 6
cot  cot 
2 2 20 5

222  100 122 2  61 2

=   
5  61 5  61 5  61 5

A C ( s  b) ( s  c)   ( s  a) ( s  b) 
Again tan . tan   .  
2 2  s (s  a )   s (s  c) 

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5 2 s b
Hence . 
6 5 s

 3s ă 3b = s or 2s = 3b
or a + b + c = 3b or a + c = 2b

Illustration 28
If p 1 , p2, p 3 are the altitudes of a triangle from the vertices A, B, C and  be the area of

1 1 1 2a b C
triangle, prove that    cos 2
p1 p2 p3 ( a  b c 
) 2

Solution :
Since p1 , p2, p 3 are perpendiculars from the vertices A, B, C to the opposite sides, we have

1 1 1
 ap1  bp2  cp3
2 2 2

1 1 1 a b c
Hence p  p  p   
1 2 3 2 2  2 

a  b  c a  b  c  2c 2 s  2 c
=  
2 2 2

s  c ab s (s  c )
=  .
 s ab

ab 21 2ab 21
= s cos 2 C = (a + b + c)  cos 2 C

Illustration 29

If ,  ,  are the length s of the altitu des of a tr iangle ABC, prove tha t

(cot A  cot B  cot C)

     
2 2 2


Solution : Since ,  ,  are the lengths of altitudes of ABC, we have

1 1 1
 a   b  c  ...(1)
2 2 2

1 1
Also absin C  bc sin A
2 2

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1
= casin B   ...(2)
2

Hence from (1), we have

1 1 1 a2 b2 c2 a2  b 2  c 2
      ...(3)
2 2 2 4 2 4 2 4 2 4 2

1 1 a 2  b2  c2
And (cot A + cot B + cot C) =
  4

a2  b2  c2
=
4 2

ă2 + ă2 + ă2 = (cot A cot B + cot C)/.

Illustration 30

Let O be a point inside a triangle ABC such that  OAB = OBC=  OCA =  then show that
(a) cot  = cot A + cot B + cot C
(b) cosec2  = cosec2 A + cosec2 B + cosec2 C
Solution :
OCB = C ă  and
A
BOC = 180Ĉ ă  ă (C ă ) = 180Ĉ ă C.

Similarly AOB = 180Ĉ ă B
Now from OAB, we have
O
OB AB c
  
sin  sin (180   B) sin B 
B C

c sin 
so that OB  ...(1)
sin B

Again from OBC, we get

OB BC a
 
sin (C - ) sin (180   C) sin C

a sin (C   )
 OB  ...(2)
sin C

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 MARKS 29

From (1) and (2), we get

or k sin C sin  sin C = k sin A sin B sin (C ă )
or sin C sin sin (A + B) = sin A sin B sin (C ă )
or sin C sin  sin A cos B + sin C sin cos A sin B
= sin A sin B cos  ă sin A sin B cos C sin 
Dividing by sin A sin B sin C sin , we get
cot B + cot A = cot  ă cot C
or cot = cot A + cot B + cot C
squaring
cot2  = cote2 A + cot2 B + cot2 C + 2 cot A cot B + 2 cot B cot C + 2 cot C cot A
or cosec2 ă 1 = cosec2 A ă 1 + cosec2 B ă 1 + cosec 2 C ă 1 + 2
[ In a , cot A cot B + cot B cot C + cot C cot A = 1]
or cosec2  = cosec 2 A + cosec2 B + cosec 2 C.

Illustration 31

a b c
The sides of a triangle are such that 2 2
 2 2

1m n m n (1  m )(1  n 2 )
2

m
Prove that A = 2 tană1 , B = 2 tană1 (mn) and   mnbc
n m2  n2

Solution :
From the given ratios we have

ab ab c
2 2
 2 2

(1  m ) (1  n ) (1  m ) (1  n ) (1  m ) (1  n2)
2

a  b 1  m 2 a  b 1  n2
 , 
c 1  m2 c 1  n2

Now from sine rule, we know that

a b a sin A
  
sin A sin B b sin B

A B A B
sin cos
a b sin A + sin B 2 2
  
a  b sin A ă sin B A+B A B
cos sin
2 2
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a b A B A ă B
  tan   cot  
a b  2   2 

or a + b = k si

AB A B
cos 2 sin 2
2 1m , 2 1n
A + B 1  m2 A  B 1  n2
cos sin
2 2

Apply C and D on both.

A B A B
tan tan  m2 , cot tan  n2
2 2 2 2

Multiplying and dividing them, we get

2A m2 2B
 tan  2
, tan  m 2n 2
2 n 2

m
 A = 2 tană1 , B = 2 tană1 (mn)
n

1 1 2tan (A/2)
 bc sin A  bc.
2 2 1 + tan 2 (A/2) etc.

SOLUTION OF A TRIANGLE