 A Help Guide On 
PROPERTIES OF TRIANGLES
A Help Guide By OP Gupta (Indira Award Winner)
Q01. In a triangle ABC, if a = 18, b = 24 and c = 30, find cos A, cos B and cos C.
Sol. We have a = 18, b = 24 and c = 30
b2  c2  a 2
242  302  182 4
 cos A 
 cos A 

2bc
2  24  30
5
3
Similarly, cos B  and cos C  0 .
5
Q02. In a ABC , if a = 18, b = 24 and c = 30, find sin A, sin B and sin C.
Sol. We have a = 18, b = 24 and c = 30
sin A sin B sin C
sin A sin B sin C






 k (say)
a
b
c
18
24
30
 sin A  18k, sin B  24k and sin C  30k .
Since 302 = 182 + 242 that means ABC is a right angled triangle such that C = 90o ,
which is the angle opposite to the biggest side c.
1
 sin C  sin 90o  30k  1  30k  k =
30
1 3
1 4
So, sin A  18  = , sin B  24   .
30 5
30 5
 A B 
cos 

ab
2 


Q03. For any triangle ABC, prove that
.
C
c
sin
2
Sol. See Q17 (e).
 A B 
sin 

a b
 2 .
Q04. For any triangle ABC, prove that

C
c
cos
2
a  b k sin A  k sin B

Sol. LHS :
[By Sine rule
c
k sin C
A B
AB
sin A  sin B 2 cos 2 sin 2



sin C
sin C

C
A
B


C
A B
2 cos  90o   sin
sin sin
2
2


2
2
[ A  B  C  180o



C
C
C
C
2sin cos
sin cos
2
2
2
2
 AB 
sin 

 2   RHS .


C
cos
2
A
 B C  b c
Q05. For any triangle ABC, prove that sin 
cos .

a
2
 2 
Sol. See Q17 (h).
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Top Solved Questions - Class 11  Compiled By OP Gupta (Indira Award Winner | M.+91-9650350480)
Q06. For any triangle ABC, prove that a  b cos C  c cos B   b2  c 2 .
Sol. See Q17 (o).
Q07. For any triangle ABC, prove that a  cos C  cos B   2  b  c  cos 2
Sol.
See Q17 (j).
Q08. For any triangle ABC, prove that
Sol.
A
.
2
b 2  c 2 sin  B  C 
.

a2
sin  B  C 
See Q17 (f).
BC
BC
 a cos
.
2
2
BC
BC
B C
 π A
Sol. LHS : (b  c) cos
[ A  B  C  π
 a cos
 (b  c) cos     a cos
2
2
2
2 2
A
 BC 
 a cos 
   b  c  sin . Now see Q17 (k).
2
 2 
Q10. For any triangle ABC, prove that a cos A  b cos B  c cos C  2a sin B sin C .
Sol. LHS : a cos A  b cos B  c cos C
[By Sine rule
 k sin A cos A  k sin B cos B  k sin C cos C
k
  2 sin A cos A  2 sin B cos B  2 sin C cos C 
2
k
  sin 2 A  sin 2 B  sin 2C 
2
k
  2 sin( A  B) cos( A  B)  2 sin C cos C 
2
 k  sin(π  C ) cos( A  B )  sin C cos C 
Q09. For any triangle ABC, prove that (b  c) cos
 k sin C  cos( A  B )  cos[  ( A  B )]
 k sin C  cos( A  B )  cos( A  B ) 
 k sin C  2 sin A sin(  B ) 
 2  k sin A  sin B sin C
 2a sin B sin C  RHS .
cos A cos B cos C a 2  b2  c 2



Q11. For any triangle ABC, prove that
.
a
b
c
2abc
Sol. See Q17 (p).
Q12. For any triangle ABC, prove that  b2  c 2  cot A   c 2  a 2  cot B   a 2  b2  cot C  0 .
Sol.
See Q17 (v).
Q13. For any triangle ABC, prove that
b2  c2
c2  a 2
a 2  b2
sin
2
A

sin
2
B

sin 2C  0 .
a2
b2
c2
Sol. See Q17 (s).
Q14. A tree stands vertically on a hill side which makes an angle of 15o with the horizontal. From
a point on the ground 35m down hill from the base of the tree, the angle of elevation of the
top of the tree is 60o . Find the height of the tree.
Sol. Let PQ = h (in metres) be the tree on the hill QR.
In triangle AQR ,
QR
 sin15o  QR  35sin15o
AQ
and, AR  AQ cos15o  35cos15o .
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Top Solved Questions - Class 11  Compiled By OP Gupta (Indira Award Winner | M.+91-9650350480)
In triangle APR,
PR
 tan 60o  PR  3 AR
AR
 PQ  QR  3  35cos15o 
 h  35sin15o  3  35cos15o 

 h  35 
 h  35
3 cos15o  sin15o

3 cos[45o  30o ]  sin[45o  30o ]


3 1
3 1 
 h  35  3 


2 2
2 2 

 3  3  3 1
 h  35 

2 2


 h  35 2 .
Hence the height of the tree is 35 2m .
Q15. Two ships leave a port at the same time. One goes 24kmph in the direction N 45o E and other
goes 32kmph in the direction S 75o E. Find the distance between the ships at the end of
3hours.
Sol. Let two ships A and B start from the port O with speed 24kmph and 32kmph in the direction
of OA and OB respectively.
When OA = N 45o E and OB = S 75o E.
AOB  AOE  EOB  45o  15o  60o .
The distance raveled by ship A in 3 hours

 3  24  72km = OA
The distance raveled by ship B in 3 hours
= 3  32  96km = OB
OA 2  OB2  AB2
In OAB, cos O 
2.OA.OB
2
72  962  AB2
o
 cos 60 
2  72  96
 AB  86.5km .
Q16. Two trees A and B are on the same side of a river. From a point C in the river, the distance
between the trees A and B is 250m and 300m respectively. If the angle C is 45o , find the
distance between the trees. [ Use 2  1.414 ].
Sol. As shown in the figure, A and B are the trees and
C is the point in the river.
So AC = 250m, BC = 300m and C  45o .
BC 2  CA 2  AB2
2.BC.CA
2
300  250 2  c 2
 cos 45o =
2  300  250
2  300  250
 3002  250 2  c 2 
2
 c  215.49m
 The distance between the two tress is 215.5m (Approx.)
In ABC, cos C =
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Top Solved Questions - Class 11  Compiled By OP Gupta (Indira Award Winner | M.+91-9650350480)
Q17. In any ABC , prove that :
a 2  c 2 sin  A  C 

(a)
b2
sin  A  C 
(b) b cos B  c cos C  a cos  B  C 
(c) a sin A  b sin B  c sin  A  B 
 A B 
cos 

ab
 2 

(e)
C
c
sin
2
 A B
tan 

ab
 2 
(g)

A B 
ab
tan 

 2 
A
B
tan  tan
c
2
2

(i)
a  b tan A  tan B
2
2
B

C
A


(k) a cos 
  b  c  sin

2
 2 
C
C
2
2
(m)  a  b  cos 2   a  b  sin 2  c 2
2
2
2
(o) a  b cos C  c cos B   b  c
2
(d)
a 2  b 2 1  cos  A  B  cos C

a 2  c 2 1  cos  A  C  cos B
(f)
b 2  c 2 sin  B  C 

a2
sin  B  C 
A
 BC  bc
(h) sin 
cos


a
2
 2 
(j) a  cos C  cos B   2  b  c  cos 2
A
2
C
B

(l) 2  b cos 2  c cos2   a  b  c
2
2

c  b cos A cos B
(n)

b  c cos A cos C
cos A cos B cos C a 2  b2  c 2



(p)
a
b
c
2abc
A
2
A
B
C
(r)  b  c  cot   c  a  cot   a  b  cot  0
2
2
2
2
2
2
2
2
2
b c
c a
a b
sin 2 A 
sin 2 B 
sin 2C  0
(s)
2
2
a
b
c2
(t) a (sin B  sin C )  b(sin C  sin A)  c(sin A  sin B)  0
2
(q) a 2   b  c   4bc cos 2
(u) 2  bc cos A  ca cos B  ab cos C   a 2  b 2  c 2
(v)  b2  c 2  cot A   c 2  a 2  cot B   a 2  b2  cot C  0
Solution :
a 2  c 2 k 2 sin 2 A  k 2 sin 2 C

(a) LHS :
[By Sine rule
b2
k 2 sin 2 B
sin 2 A  sin 2 C
sin(A  C)sin(A  C)




2
sin 2 [π  (A  C)]
sin B
sin(A  C)sin(A  C)
sin(A  C)




 RHS .
2
sin (A  C)
sin(A  C)
(b) LHS : b cos B  c cosC  k sin Bcos B  k sin Ccos C [By Sine rule
k
k

  2 sin Bcos B  2 sin C cos C 

  sin 2B  sin 2C
2
2
k
2B  2C
2B  2C 

 k sin(B  C) cos(B  C)

  2sin
cos
2
2
2 
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Top Solved Questions - Class 11  Compiled By OP Gupta (Indira Award Winner | M.+91-9650350480)

 k sin Acos(B  C)

 k sin(π  A) cos(B  C)

 a cos(B  C)  RHS .
(c) LHS : a sin A  b sin B  k sin A sin A  k sin Bsin B
[By Sine rule
2
2

 k [sin A  sin B]

 k [sin(A  B) sin(A  B)]

 k sin(π  C)sin(A  B)

 k sin Csin(A  B)

 c sin(A  B) = RHS .
1  cos  A  B  cosC 1  cos  A  B cos  π  (A  B) 
(d) RHS :

1  cos  A  C  cos B 1  cos  A  C  cos  π  (A  C) 


1  cos  A  B cos  A  B
1  cos  A  C  cos  A  C 
1  cos2 A  sin 2 B
1  cos2 A  sin 2 C
k 2 sin 2 A  k 2 sin 2 B

 2 2
k sin A  k 2 sin 2 C
a 2  b2

 2
 LHS .
a  c2
a  b k sin A  k sin B

(e) LHS :
c
k sin C




k [sin A  sin B]
k sin C
AB
 π C
sin    cos
2
2 2


C
C
sin cos
2
2
AB
cos
2  RHS .


C
sin
2
2
2
b c
k 2 sin 2 B  k 2 sin 2 C

(f) LHS :
a2
k 2 sin 2 A
k 2 [sin 2 B  sin 2 C]


k 2 sin 2 A
sin(B  C)sin(B  C)


sin 2 (B  C)
a  b k sin A  k sin B

(g) LHS :
a  b k sin A  k sin B

sin A  sin B

sin A  sin B

 cot
AB
AB
tan
2
2


1  [cos2 A  sin 2 B]
1  [cos2 A  sin 2 C]


sin 2 A  sin 2 B
sin 2 A  sin 2 C
[Multiplying Nr & Dr by k 2
[By Sine rule
AB
AB
cos
2
2

C
C
2sin cos
2
2
C
AB
cos cos
2
2

C
C
sin cos
2
2
2sin


[By Sine rule


sin(B  C)sin(B  C)
sin 2 [π  (B  C)]
sin(B  C)

 RHS .
sin(B  C)

AB
AB
sin
2
2

AB
AB
2sin
cos
2
2
 A B
tan 

 2   RHS .

 A  B
tan 

 2 
2 cos


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Top Solved Questions - Class 11  Compiled By OP Gupta (Indira Award Winner | M.+91-9650350480)
(h) RHS :


A k sin B  k sin C
A
bc
cos 
cos
a
2
k sin A
2
sin B  sin C
A
cos
sin A
2
BC
BC
sin
2
2 cos A

A
A
2
2sin cos
2
2
A
BC
sin sin
2
2

A
sin
2
2cos

BC
π A
cos    sin
2
2 2 



A
sin
2
BC

 sin
 LHS .
2
A
B
B
A
sin cos  sin cos
2
2
2
2
A
B
A
B
tan  tan
cos cos
2
2 
2
2
(i) RHS :
A
B
A
B
B
A
tan  tan
sin cos  sin cos
2
2
2
2
2
2
A
B
cos cos
2
2
 A B
A
B
B
A
sin   
sin cos  sin cos
2 2
2
2
2
2




A
B
B
A
A B
sin cos  sin cos
sin   
2
2
2
2
2 2
 π C
C
sin   
cos
2 2
2




A
B

A

B

sin 
sin 


 2 
 2 
C
C
2 sin cos
sin C
2
2




C  A B
 π A  B  A B
2 sin sin 
2 sin  

 sin 

2
2   2 
 2 
2
sin C
sin C




sin A  sin B
 A B  A B
2 cos 
 sin 

 2   2 
k sin C


[Multiply Nr & Dr by k
k sin A  k sin B
c


 LHS .
ab
(j) LHS : a  cosC  cos B   k sin A  cos C  cos B



CB
C  B

 k sin A  2sin
sin
2
2 

A
A 

 π A  C  B
 k  2sin cos   2sin    sin
2
2 
2 

2 2 
A
A
A
C B

 2k  2sin cos cos sin
2
2
2
2 

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Top Solved Questions - Class 11  Compiled By OP Gupta (Indira Award Winner | M.+91-9650350480)

 2k cos2
A
 π C+ B  C  B 
2sin  
sin

2
2 
2 
2
A
C+B
C  B
2 cos
sin

2
2
2 
A

 2k cos2  sin C  sin B 
2
A

 2 cos 2  k sin B  k sin C 
2
A

 2  b  c  cos2  RHS .
2
A
A
(k) RHS :  b  c  sin   k sin B  k sin C  sin
2
2
A

 k  sin B  sin C  sin
2
BC
BC
A


 k  2sin
cos
sin

2
2 
2


 2k cos 2

 k cos
BC 
 π A  A
2sin    sin 

2 
2
2 2
BC 
A
A
2cos sin 

2 
2
2
BC

 k cos
sin A
2
BC

 k sin A  cos
2
BC

 a cos
 LHS .
2
C
B
C 
B


(l) LHS : 2  b cos2  c cos2   b  2 cos2   c  2 cos2 
2
2
2 
2



 b 1  cosC   c 1  cos B

 k cos
 b  b cos C  c  c cos B
[By Projection formula, a  b cosC  c cos B
 b  c  b cos C  c cos B
 a  b  c  RHS .
C
C
2
2
(m) LHS :  a  b  cos 2   a  b  sin 2
2
2
C
C
C
C
C
C

 a 2 cos2  b2 cos2  2ab cos2  a 2 sin 2  b2 sin 2  2ab sin 2
2
2
2
2
2
2
C
C
C
C
C
C




 a 2  cos2  sin 2   b 2  cos 2  sin 2   2ab  cos 2  sin 2 
2
2
2
2
2
2



2
2

 a (1)  b (1)  2ab cos C



 c 2  RHS .
c  b cos A a cos B  b cos A  b cos A
(n) LHS :

b  c cos A c cos A  a cos C  c cos A
a cos B cos B



 RHS .
a cos C cosC

[By Projection formulae
(o) LHS : a  b cos C  c cos B  ab cos C  ac cos B
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Top Solved Questions - Class 11  Compiled By OP Gupta (Indira Award Winner | M.+91-9650350480)






 a 2  b2  c2 
 a 2  c 2  b2 
 ab 

ac



2ab
2ac




2
2
2
2
2
2
a b c
a c b


2
2
2
2
2
2
a b c
a  c 2  b2


2
2
2
2
2
2
a  b  c  a  c 2  b2

2
2
2
2(b  c )

2
2
 (b  c 2 )  RHS .
cos A cos B cos C 1  c 2  b2  a 2  1  a 2  c 2  b2  1  a 2  b2  c 2 


 
 
 

a
b
c
a
2bc
2ac
2ab
 b
 c

2
2
2
2
2
2
2
2
2
c b a
a c b a b c




2abc
2abc
2abc
2
2
2
2
2
2
2
c  b  a  a  c  b  a  b2  c2


2abc
2
2
2
a b c


 RHS .
2abc
A
A
2
(q) RHS :  b  c   4bc cos 2  b2  c 2  2bc  4bc cos2
2
2
A 


 b 2  c 2  2bc  2 cos 2  1
2


2
2

 b  c  2bc cos A

 a 2  LHS .
A
A
(r) Consider,  b  c  cot   k sin B  k sin C  cot
2
2
BC
BC
A
B  C
A


π A
sin
cot

 k  2 cos

 k  2cos    sin
cot


2
2 
2
2 
2

2 2 

A
A
B  C  cos 2
BC


 π B  C 

 k  2sin sin

 k  2sin
cos  

2
2  sin A
2
2  

2

2


 B C   B C 
 B C   B C 

 k  2sin    sin    

  k  2sin    sin    
 2 2   2 2 
 2 2   2 2 


(p) LHS :
B
C

  k  cos  cos 

2
2

A
C
B

So,  b  c  cot  k  cos  cos  …(i)
2
2
2

B
A
C

Similarly  c  a  cot  k  cos  cos  …(ii)
2
2
2

B
B
A

And,  a  b  cot  k cos  cos  …(iii)
2
2
2

Now adding (i), (ii) & (iii), we get :

C
B

 k  cos  cos 
2
2

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Top Solved Questions - Class 11  Compiled By OP Gupta (Indira Award Winner | M.+91-9650350480)
A
B
B
C
B
A
C


  c  a  cot   a  b  cot  k  cos  cos   k  cos  cos 
2
2
2
2
2
2
2


B
A

k cos  cos 
2
2

C
B
A
C
B
A


 k  cos  cos  cos  cos  cos  cos 
2
2
2
2
2
2

A
B
B
  b  c  cot   c  a  cot   a  b  cot  0 .
Hence Proved.
2
2
2
b2  c2
b2  c 2
sin
2A
[2sin A cos A]
=
(s) Consider,
a2
a2
 b 2  c 2   b2  c 2  a 2 

= 2(ak ) 


2
2bc
 a 

 b  c  cot
 (b2  c 2 )( b2  c 2 )  a 2b 2  a 2c 2 
=k

2abc



i.e.,
 b 4  c 4  a 2 b2  a 2 c 2 
b2  c2
sin
2A
=
k

 …(i)
a2
2abc


Similarly,
 c 4  a 4  c 2b 2  a 2 b 2 
c2  a 2
sin
2B

k

 …(ii)
b2
2abc


 a 4  b 4  c 2 a 2  c 2 b2 
a 2  b2

sin
2C
k

 …(iii)
c2
2abc


Now adding (i), (ii) & (iii), we get :
 b 4  c 4  a 2b 2  a 2 c 2 
b2  c2
c2  a2
a 2  b2



sin
2A
sin
2B
sin
2C
k


a2
b2
c2
2abc


And,
 c 4  a 4  c 2 b 2  a 2b 2 
 a 4  b 4  c 2 a 2  c 2b 2 
k 

k



2abc
2abc




k
b 4  c 4  a 2b2  a 2 c 2  c 4  a 4  c 2 b2  a 2b 2  a 4  b4  c 2 a 2  c 2b 2 
2abc 
k


 0  0
2abc
b2  c 2
c2  a2
a 2  b2

sin
2A

sin
2B

sin 2C  0 .
Hence Proved.
a2
b2
c2
(t) LHS : a (sin B  sin C)  b(sin C  sin A)  c(sin A  sin B)

 a sin B  a sin C  b sin C  b sin A  c sin A  c sin B

 k sin A sin B  k sin A sin C  k sin Bsin C  k sin Bsin A  k sin Csin A  k sin Csin B

 0  RHS .
(u) LHS : 2  bc cos A  ca cos B  ab cos C 


  b2  c 2  a 2 
 a 2  c 2  b2 
 b2  a 2  c 2  
 2  bc 

ca

ab





2bc
2ac
2ba





 
2
2
2
2
2
2
2
2
2

b c a a c b b a c

 a 2  b 2  c 2  RHS .
cos A
(v) Consider,  b2  c 2  cot A   b2  c 2 
sin A

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Top Solved Questions - Class 11  Compiled By OP Gupta (Indira Award Winner | M.+91-9650350480)
  b2  c 2 

1  b2  c 2  a 2 


2bc
ka 

b 4  c 4  a 2 b2  a 2 c 2
i.e.,  b  c  cot A 
…(i)
2kabc
c 4  a 4  c 2b 2  a 2b 2
Similarly,  c 2  a 2  cot B 
…(ii)
2kabc
a 4  b 4  c 2 a 2  c 2b 2
And,  a 2  b 2  cot C 
…(iii)
2kabc
Now adding (i), (ii) & (iii), we get :
b 4  c 4  a 2 b2  a 2 c 2
2
2
2
2
2
2
 b  c  cot A   c  a  cot B   a  b  cot C 
2kabc
4
4
c  a  c 2 b 2  a 2b 2 a 4  b 4  c 2 a 2  c 2b 2


2kabc
2kabc
4
4
2 2
2 2
4
4
2 2
2 2
4
4
2 2
2 2
b c a b a c c a c b a b a b c a c b


2kabc
0


2kabc
2
2
  b  c  cot A   c 2  a 2  cot B   a 2  b2  cot C  0 .
Hence Proved.
2
2
 Dear Student,
I would urge you for a little favour. Please notify me about any error(s) you notice in this (or
other Maths) work. It would be beneficial for all the future learners of Maths like us. Any
constructive criticism will be well acknowledged. Please find below my contact info when you
decide to offer me your valuable suggestions. I’m looking forward for a response.
Also I would wish if you inform your friends about my efforts for Maths so that they may also
benefit.
Let’s learn Maths with smile :-)
 For any clarification(s), please contact :
MathsGuru OP Gupta
[Maths (Hons.), E & C Engg., INDIRA Award Winner]
Contact Nos. : +91-9650 350 480 | +91-9718 240 480
Mail me at : [email protected]
@theopgupta
Official Web-page : www.theOPGupta.com
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