JS 15 SOLUTIONS PART III (TRIANGLE)
MISCELLANEOUS EXERCISE PAGE 344(BELOW)-346
1.
2B = A + C
sin B

sin C
(*)
3
3
2 
sin C
From (**)
 C = 45
3
2
. Also A + B + C = 180  B = 60 from (*)
 sin C 
2
1
2
(135 is not possible since B = 60)
 A = 180 – (60 + 45) = 75
2.
r1 




, r2 
, r3 
. If r1, r2, r3 are in A.P.
sa
s b
sc



are in A.P.  s – a, s – b, s – c are in A.P.
,
,
s a s b s c
 -a, -b, -c are in A.P.  a, b, c are in A.P.
3.
b  c  11k , c  a  12k , a  b  13k  2  a  b  c   18k 

a  7k , b  6k , c  5k
We can now easily show
4.
cos A cos B cos C


7
19
25
The given relation is
2 cos
y
y2  1

 x  

2
4

2
y2  1

A B
A B
C 3
cos
 1  2sin 2 
2
2
2 2
A B
A B

, y  cos
 x  cos

2
2 

 2xy – 2x2 = ½

a  b  c  18k
2
 

y
  x    0 
2
 

y2  1
0
4



y2  1
 y = 1, y = -1, y = -1  cos
A B
 1
2
y 2  cos 2

1
A B

2 
A  B  2 which is absurd
Again y = 1  cos
A B
1
2

(the other solutions are not possible)

A B 1

2
2
cos

A B
0
2

A = B Also y = 1, x = y/2 = ½
A B
 60
2
But as A = B we must have A = B = 60 whence C is also 60.
5.
sin A sin B sin C  sin A sin B
(True for all triangle)
Adding cos A cos B on both sides we get; in the given triangle
1  cos  A  B 

cos A cos B  sin A sin B sin C  1 given 

cos  A  B   1 which is not possible unless cos  A  B   1

A B  0

A B
On putting A  B in the given relation we get cos2 A  sin 2 Asin C  1

sin 2 Asin C  sin 2 A
Whence 
6.
A  B  45

sin C  1

a : b : c  1:1: 2

C  90
Taking A as origin median AD as x-axis. Let B be (x1, y1); and C be (x2, y2) then
BC =
 x1  x2 
2
  y1  y2 

2
Also AD = 11  6 3

We easily note that
y1
1
 tan 30 
x1
3
1
2
(*)
 p (say)
y2
 tan135 (slope of AC) = -1
x2
 y1 
x1
3
, y2   x2 . Also
x1  x2
y  y2
 p, 1
0
2
2
On solving these relations we get
2

D lies on x  axis 
x1 
2 3p
3 1
2p
, x2 
3 1
, y1 
2p
3 1
2
, y2  
 2 3p  2 p   4p 
 BC  
 


3  1   3  1 

2
7.
2p
3 1
2
 2p
82 3


 2.
 4  2 3 11  6 3 

2
3 1  4




3 1
2

p  11  6 3
We are given 2B = A + C and A + B + C = . From these two equation B 
putting B 

1
2




2
, AC 
. On
3
3

2
,C 
 A in the given equation we get
3
3


 2

 5
 1
sin  2 A    sin 
 2 A    sin 
 2A 
3

 3

 3
 2
There should be a value of A satisfying the above equality. If we equate all sines
to ½. We get 2 A 
2A 

2
n 
n 
 n   1 ,
 2 A  m   1 ,
3
6 3
6
5
c 
 l    1
3
6

sin      sin  
The first equation above will yield permissible A (i.e. between 0 and 180).
For n = 1 only. This angle is  /4. The second and third yield  /4 for m = 0 and
l = -1 respectively. Thus the only value of A satisfying the conditions is A = 45 whence
B = 60, C = 75.
8.
Let c/b = r < 1 we have to show that altitude
AD 
ar
1 r2
Since  
1
a  AD,
2
(*)
AD 
2
a
 Inequality to be proved is equivalent to
3
2

a
c
b
c2
1 2
b
a

b2  c 2
 2R
a
2
abc
which transforms to
 2
a
b  c2
 2 R sin B 
abc 

 R
 
4 

2
  2 R sin C 
2 R sin A
2
 2R
which is reducible to true inequality sin(B – C)  1
Note: The use of r < 1 can be observed in cross multiplying since b2 – c2 > 0.
9.
Let AB = n, AC = n + 1, BC = n + 2
Further let A = 2C
(Since C is smallest and A is largest) By sine rule we
sin A sin B sin C


n  2 n 1
n
Equating first and third ratios and cancelling sin C  0
we get
2 cos C 1 n  2
 
n2
n
2n

sin B  sin  A  C  etc 
Now as A = 2C, A + B + C = 180 we have B = 180 – 3C
 sin B  sin 3C 
sin 3C sin C

n 1
n
on canceling sin C  0 we get  sin2 C =
2n  1
. But from above
4n
n2
2
sin C  1  cos C  1  
 equating the two values of sin C we get
2
n


2
2
2
n2 – 3n – 4 = 0  n = 4, n = -1
Since -1 is inadmissible we have n = 4 whence the sides are 4, 5 and 6.
10.
Let C1, C2, C3 be centres of the circles. It is evident
that the point O where the tangents meet must at the
incentre I of the triangle C1C2C3. Now sides a’, b’, c’
4
have
of the triangle C1C2C3 are given by
a’ = C2C3 = r2 + r3
b’ = r1 + r3, c’ = r1 + r2
s
a   b  c 
2
whence  
 r1  r2  r3
s  s  a  s  b  s  c   r1r2 r3  r1  r2  r3 
Now 4 = inradius of  C1C2 C3 

11.

4
s
r1 r2 r3  r1  r2  r3 
r1  r2  r3
r1r2 r3
 16 .
r1  r2  r
Let us first show (i)  (ii)
are rational then s is rational (obvious), s  a, s  b, s  c are also rational
If a, b, c,
tan
B

 rational,
2 s  s  b
tan
C

 rational
2 s s  c
(i)  (ii)
Now let us show (ii)  (iii)
Let a, tan
B
C
are rational then we must show that sin A,sin B,sin C are rational.
, tan
2
2
B
2  rational,
sin B 
B
1  tan 2
2
2 tan
C
2  rational
sin C 
C
1  tan 2
2
2 tan
B
C
1  tan tan
A
2
2
Now tan 
2 tan B  tan C
2
2
5
B
C
B
C
A
, tan are rational and tan  tan can not be zero we conclude that tan is
2
2
2
2
2
B
2 tan
2 is also rational. Thus (ii)  (iii)
sin A 
rational

2 B
1  tan
2
Since tan
Finally to show (iii)  (i)
If a,sin A,sin B,sin C are rational then b and c are obviously rationals as
b sin B c sin C
1
and as  bc sin A which is also rational. Thus (iii)  (i)

, 
a sin A a sin A
2
12.
Let OA1 = OA2 = OA3 = r
A1OA2 = A2OA3 = A3OA4

2
.
n
Then  A1 A2   r 2  r 2  2r 2 cos 
2
 2r 2 1  cos    4r 2 sin 2
similarly, A1 A3  2r sin

.
2
 A1 A2  2r sin

2
2
3
 2r sin , A1 A4  2r sin
2
2
On substituting in the given relation we get
1
sin

2
1

sin 

 cos
1
3

3

 2sin  sin
 2sin sin
 2sin  sin
3
2
2
2
2
sin
2

5

3
 cos
= cos   cos 2  cos  cos
2
2
2
2
 cos 2  cos
 2 cos
3
5
 cos   cos
2
2
7

7
3
7 

3 
cos  2 cos
cos
 cos
 cos  cos
0
2
4
2
2
4 
4
4 
6
 cos
7 

2n 2

13.
7 


 2sin sin   0
4 
2
4
since sin


 0, sin
 0,
n
2n
n7


tan B tan C = p  tan B tan   


 B  p
4

 1  tan B 
 p
 1  tan B 
 tan B 
Now make a quadratic in tan B and solve discriminant > 0.
14.
If the triangle ABC is equilateral then tan A + tan B + tan C =
3 follows.
Let us prove the converse. In any triangle tan A + tan B + tan C = tan A tan B tan C
From AM  GM
1
1
x
tan A  tan B  tan C
  tan A tan B tan C  3   x 3 where
3
3
x = tan A + tan B + tan C = tan A + tan B + tan C
x2
 x  x 2  27
27


x3 3
x  0
But we are given
x  3 3  AM  GM
15.
A B
C
  90  . Take cot both sides and cross multiply.
2 2
2

17.
 tan A  tan B  tan C  A  B  C  60
 tan 1
i 1

1
2i 2

  tan 1
i 1
 2i  1   2i  1
2
  tan 1
2
4i
1   2i  1 2i  1

  tan 1  2i  1  tan 1  2i  1
i 1

The sum upto n terms  tan 1  2n  1  tan 1 1

18.
Sum upto  
a, b, c are in A.P

2


4


4
. Hence tan t  tan
 2b  a  c
7

4
1
a
Now cos1 
bc

1  tan 2
 1  tan 2
We have

2
2
abc
abc



tan 2

2  a
 bc
1  tan 2
2

2  abc

 a  (b  c)
1  tan 2  1  tan 2
2
2
Similarly tan 2
19.

1  tan 2
 tan 2
1
2
 tan 2

2
3
2


bca
,
abc
2b
b 2


a  b  c 3b 3
sin 2 A  sin 2 B sin  A  B 

sin 2 A  sin 2 B sin  A  B 
 sin  A  B 

1
 sin  A  B   2

0
2
sin
A

B
sin
A

sin
B




The first possibility gives sin(A – B) = 0
 A = B the second will reduce to cos(A + B) = 0
 A + B = /2  C = /2
A
B sin  A  B 
C
cos
cos


A
B
2
2


2 
2 
2
cot  cot 
A
B
A
B
A
B
2
2
sin
sin
sin sin
sin  sin
2
2
2
2
2
2
cos
20.
 LHS 

C
B
C
A
sin
b cos sin
2
2 
2
2
A
B
sin
sin
2
2
a cos
C
B
C
A
sin
2 R sin B cos sin
2
2 
2
2
A
B
sin
sin
2
2
2 R sin A  cos
 4 R cos
A
C
B
B
C
A
cos sin  4 R cos cos sin
2
2
2
2
2
2
8
 4 R cos
 2
C
A B
C
C
sin
 4 R cos cos


2
2 
2
2
c
C
C
cos 2  c cot
sin C
2
2
C
C

 sin C  2sin cos 
2
2

21. On dividing the given relation by sin 3  ( by sin in each of the individual brackets),
We get 1  (sin A cot   cos A) (sin B cot   cos B ) sin C cot   cos C )
On dividing by sin A sin B sin C we get
cos ecA cos ecB cos ecC  (cot   cot A) (cot   cot B) (cot   cot C )
which reduces to
cot 3  (cot A  cot B  cot C ) cot 2  (cot A cot B  cot B cot C  cot C cot A)
cot   ( cot A cot B cot C  cos ecA cos ecB cos ecC )  0
Now using the common identity cot A cot B  cot B cot C  cot C cot A  1and the
uncommon identity cot A cot B cot C  cos ecA cos ecB cos ecC  cot A  cot B  cot C
The above equation becomes
cot 3   (cot A  cot B  cot C ) cot 2   cot   (cot A  cot B  cot C )  0

(cot 3   1)[cot   (cot A  cot B  cot C )]  0
whence it follows that cot   cot A cot B cot C.
22.
The assertion follows it we could show the validity of the following relations
tan A  tan B  tan C 
p
1 p
, tan A tan B  tan B tan C  tan C tan A 
q
q
and tan A tan B tan C 
p
q
The third one is straight forward since tan A tan B tan C 
sin A sin B sin C
cos A cos B cos C
The first one is also true since in any triangle tan A  tan B  tan C  tan A tan B tan C
Finally the second follows from uncommon triangle identity
9
tan A tan B  tan B tan C  tan C tan A 1  sec A sec B sec C 1 
1
q
Which can be proved by writing cos( A  B  C )  1
expand and divide by cos A cos B cos C.
23.
If tan
A
C 2
B
tan  cot then
2
2 3
2
Multiple by
24.
( s  b)( s  c)
( s  a)( s  b) 2
s ( s  b)


s(s  a)
s ( s  c)
3 ( s  a)( s  c)
(s  a)(s  b)(s  c)s and cancel ( s  b) to get ab  a  c .
We have cos( A  B)  4 / 5 or 2cos 2
A B
A B
3
 1  4 / 5 cos

2
2
10
A B
A B
A B
A B
 A B A B
sin B  sin 

 sin
cos
 cos
sin

2 
2
2
2
2
 2
 cos
C 3
C 1
 sin
2 10
2 10
Similarly sin A  cos
sin B b 1
  
sin A a 2
C 3
C 1
 sin
2 10
2 10
C
C
 sin
2
2
C
C
3cos  sin
2
2
3cos
Hence , area of ABC 
or tan
C
 1C  90
2
1
 6  3 sq. units = 9sq. units .
2
10