1. The document discusses three "master triangles" - the equilateral triangle, right triangle, and right isosceles triangle. It provides the key properties and relationships for each triangle. 2. Examples are provided showing how the master triangles can be used to solve other triangle problems by applying their specific properties. If a problem's conditions match the properties of one of the master triangles, it can be used to find the solution. 3. The document contains several triangle word problems and their solutions demonstrated through applications of the appropriate master triangle. This establishes the master triangles as useful tools for solving a wide range of triangle questions.

1. 1

we all know very well that a master key unlocks many locks,Similarly our just one
Master triangle can solve many questions belonging to properties and solutions of
triangles.
Properties and Solutions of Triangles
We
 It is the only triangle in which the
special points i.e. Centroid,Circumcenter, Incenter and Orthocen
Equilateral Triangle (A Magical Triangle ) : -
have made 3 Master Triangles, Equilateral , Rightangled and Rightangled isoscales.
,

0 0 0
ter are coincide.
also it is the only triangle in which Median, Internal angle bisector, Altitute are
also coincide.
For this triangle we take the sides as a = 1,b = 3 c = 1
and angles as A = 30 , B = 60 and C = 90 .
Right - angled Triangle : -
Right
 
0
,
3
Let a = b = c = 1 A= B = C = 60 Area= Δ =
4
Radius of circumcirc
  

(Equilateral Triangle)
0 0 0
For this triangle we take sides as a = 1,b = 1 c = 2
and angles as A = 45 , B = 45 and C = 90 .
- angled Isoscales : -
Master Triangle - MT -1 1
1 2 3
abc 1
le R
4Δ 3
Δ 1
Radius of in circle r
s 2 3
3
Radii of ex circles are r r r
2
3
Length of perpendicular (altitutes) AD BE CF
2
  
   
   
   
* The above master triangle can be used most widely to all the problems which
belong to a famil
-
-
Q In a triangle ABC the correct statement is :
y of triangles i.e. problems belong to all triangles.
* It can also be applied to all those problems for which the data of this triangle
satisfy the given conditions in the questions.
[
A B+C A B C
(a) (b c) cos = 2asin (b) (b c)cos = a sin
2 2 2 2
B C A B+C A
(c) (b c) sin =acos (d) ( b+c )sin =acos
2 2 2 2
S.
       
        
       
       
        
       
Here above problem belongs to all triangles,so
]IITJEE2005
Let the triangle be an equilateral,so according to data of MT 1,
a b c 1 and A B C 60º after putting the value of a,b,c then only option
(b)will balance and s o it is the right choice.
     
MT - 1 can be applied here.
–
A
B CD
E
P
F
1

2. 2
 
 
0 0 0
a 1,b 3 ,c 2
A 30 ,B 60 ,C 90
3 1
r .
2
R 1.
1 3
Δ= . 3 .
2 2
   
   

 
 
 
Right Angled Triangle
Master Triangle 2 MT - 2
* The above master triangle can be used most widely to all the problems which
belong to a family of triangles i.e. pro
Q. If sides of a triangle are in ratio 1: 3: 2, then ratio between its orresponding angles is :
(
(a
ΙΙΤJEE 2004)
blems belong to all triangles.
and if the data of this triangle satisfy the given conditions in the questions.
     ) 3 2 1 b 3 1 2 c 1 2 3 d 1 3 : 2
S. The data of MT–2, a 1,b 3 and c 2 satisfy the given conditions
so triangle is right angled & A 30º, B 60º, C 90º. A:B:C 30 60:90 1 : 2 3
Q In a Δ A
      
  
       
 BC, If r and R are in- radius and circum-radius then 2(r R)
(a) a b (b) b c (c) c a (d) a b c
ST. Let the triangle be a right angled isoscales, According to MT–3,
a 1, b
 
    
 
AIEEE2005
 
 
1 1 1 1
1, c 2 & r 1 , R L.H.S. 2 r R 2 1 2
2 2 2 2
put the values of a, b & c in the options,(a) will give R.H.S
a c
Q. In ABC if angles are in arithmatic Pr ogressions then value of sin 2C sin 2A is :
c a
(a) 1/2 (b) 3
 
          
 
 
IITJEE2010
0 0 0
0 0
2 2 2
/2 (c) 1 (d) 3
S. Angles are in A.P. so by data of MT-2,A 30 ,B 60 ,C 90 and a 1,b 3,c 2
a c 1 2 3
sin 2C sin 2A sin180 sin 60 2 3.
c a 2 1 2
a b c
Q. In ABC if A : B : C 1 : 2 : 3,then the value of is equals to :
ab bc ac
2 2 8
(a) (b) (c)
3 1 3 1 3 3
     
     
 
 
 
  

0 0 0
2 2 2
8
(d)
2 3 3 1
S. Angles of MT-2 are 30 : 60 : C 90 1 : 2 3 and so we can apply MT-2,
a b c 1 3 4 8
and we can take a 1,b 3,c 2.
ab bc ac 1. 3 3.2 1.2 3 3 2

  
   
     
    

2
1
r
A
BC

3. 3
 
 
0 0 0
1
R ,
2
a 1, b 1, c 2 1
r 1 ,
2A 45 ,B 45 ,C 90
1 1
Δ 1 1
2 2
 
   
  
   
    
Right Angled Isoscales
Master Triangle - 3 MT - 3
* The above master triangle can be used most widely to all the problems which
belong to a family of triangles i.e. p
Q. In a ABC if sinA sinB sinC 1 2 and cosA cosB cosC 2 ,then triangle is:
(a) Equilateral (b) Isosceles (c)
       
roblems belong to all triangles.
and if the data of this triangle satisfy the given conditions in the questions.
2 2 2
Right angled (d) Right angle isosceles
S. data of MT–3, A 45º, B 45º andC 90º satisfies given both conditions,
it is right angled isosceles and hence option (d) is right choice.
Q. In ABC if b c 3a ,then the val
  

  

0 0 0
0 0 0
ue of cot B cot C cot A is equals to :
ab
(a) 4 (b) (c) (d) 0
4
S. Since data of MT–3, a 1, b 1 and c 2 satisfies given condition,
therefore A=45 ,B=45 and C 90
then cot B cot C cot A cot 45 cot 90 cot 45 0
2c
Q. In a ABC, if
 
 

  

     

 
0 0 0 0
0
os A cos B 2cos C a b
then the value of A is :
a b c bc ca
(a) 60 (b) 90 (c) 30 (d)75
S. Since for A 90º,B 45º and C 45º and a 2, b 1 and c 1
2.0 1 2 1 3 3
The given condition 2 satisfies ,
12 2 2 2 2
therefore A 90 .
Q. In any tria
    
     
     
 
2 2 2 2
2 2 2
C C
ngle ABC, (a b) cos (a b) sin is equals to :
2 2
(a) a (b) b (c) c (d) 1
S. :- Let the triangle be right angled isoscales.
Therefore, according to M
   
      
   
Master Triangle
2 2 2 2 2
T–3 a 1, b 1, c 2 and A 45º, B 45º,C 90º.
C C 1
L.H.S. (a b) cos (a b) sin 0 4 2 c (since c 2)
2 2 2
     
   
            
   
Alart: - Here MT -1 can't be applied,as it will give more than one options same.
2
1
r
A
B
C
1

4. 4
    
3 3 3
0 3 0 3 0 3 0 0 0 0
3 0 3 0
a b c
Q. In a ΔABC, if sin A sin B sin C = 3sinA sinB sinC, then b c a =
c a b
(a) 0 (b) a b c (c) a b c ab bc ca (d) 1
S. Let A B C 60 so sin 60 sin 60 sin 60 3sin60 .sin60 .sin60
3sin 60 3sin 60 Satisfy the given co
 
     
     
  ndition
a b c 1 1 1
the triangle is an equilateral so take a b c 1 b c a 1 1 1 0
c a b 1 1 1
cosA cosB cosC
Q In triangle ABC if and if a=2 then area of triangle will be :
a b c
3
(a)1 (b) 2 (c) (d) 3
2
      
 
2 2
2 2
S. The data of MT–1 satisfies the given condition. so the triangle is An equilateral
3 3
area of the equilateral triangleis Δ a (2) 3
4 4
a a a
Q. In ABC, if cos , cos & cos ,
b c b c b c
then tan tan
2 2
    
      
  
   
 
  
 
2
0
2 2 2 2 0
2 2
+ tan =
2
(a) 3 (b) 1 (c) 3 3 (d) 9
1
S. Let the ABC is an equilateral then cos cos cos 60
2
tan tan + tan = 3tan 30 = 1
2 2 2
(a b c)(b c a) (c a b) (a b c)
Q. In triangle ABC, =
4b c
(a)
  
   
  
             
       
      
     
       
2 2 2 2
sin A (b) cos A (c) tan A (d) cot A
S. Let the triangle be an equilateral. Therefore, according to MT–1
3
a b c 1 and A B C 60º so L.H.S.
4
after putting the value of A in the options, (a) will
      
2 2 2
2 3
0
give R.H.S.
.
1 A 1 B 1 C
Q In triangle ABC , cos cos cos
a 2 b 2 c 2
s s s
(a) s (b) (c) (d)
abc abc abc
S. Let the triangle is an equilateral and a b c 2 & A B C 60
L.H.S. 3
     
       
     
     

 
2 3 9
cos 30º 3. =
4 4
put values of a, b, c and s in the options,then c will give the required result.

Alart: - Here MT -1 can't be applied,as it will give more than one options same.
A
B CD
E
P
F
1
A
B CD
E
P
F
1

5. 5
 
2 2 2 2 2 2 2 2 2 2 2 2
A– B+C
Q. In a triangle ABC, 2ac sin =
2
(a) a b – c (b) c a b (c)b – c – a (d)c – a – b
S. Let the triangle is a right angled. Therefore, according to MT 2
a 1, b = 3 ,c 2 & A 30º,B 60º,C 90º.
then L.
  

    
2 2 2
1
H.S. 2 1 2 sin 30º 4 2.
2
put values of a, b, c in the options, then (b) will match with L.H.S.
Q. In a triangle ABC, If c a b , then 4s(s
      
 
Alart: - Here MT -1 can't be applied,as it will give more than one options same.
 
4 2 2 2 2 2 2
2 2 2
2 2 2
2 2 2
a)(s b)(s c)
(a) s (b) b c (c) c a (d) a b
S. :- c a b the triangle is right angled and c is hypoteneous.
1 a b
L.H.S. 4s(s a) (s b) (s c 4Δ ×a b = 4 a b
2 4
Therefore, option d i

   
 
 
           
 
Master Triangle 
2 2
s the right option.
Q. In triangle ABC if is the area of the triangle then a sin 2C c sin 2A
(a) Δ (b) 2Δ (c) 3Δ (d) 4Δ
S. :- Let the triangle be an equilateral so according to MT–1 ,
a b c 1, A B
  
    
Master Triangle
2
2 2
2 2 2
3 3 3
C 60°, Δ= (1) , Δ 3 4Δ
4 4 4
3
LHS a sin2C c sin2A 1 sin120° 1 sin120°= 2sin120° 2 3 4Δ .
2
a b c A B C
Q. In a ABC,value of the exp ression sin sin sin is :
sin A sin B sin C 2 2 2
(a)2 (b) (c) (d
2
     
         
       
         
      

 
0
2 2 2
)
4
S. :- let the triangle be an equilateral,then according to data of MT-1
3
a b c 1, A B C 60 ,
4
a b c A B C 2 2 2 1 1 1
sin sin sin
sin A sin B sin C 2 2 2 2 2 23 3 3
Q. In a ABC, if G is th

       
        
               
        

Master Triangle
e centroid of the triangle and GA,GB,GC makes angles , ,
with each other then cot A cot B cot C cot cot cot
(a) 1 (b) 0 (c)3 (d)3 3
S. :- Let the triangle be an equilateral
then according to data of MT-1.
A B C
  
        
 
Master Triangle
 
0 0
0 0 0 0 0 0
0 0
60 then 120
cot 60 cot 60 cot 60 cot120 cot120 cot120
3(cot 60 cot120 ) 3 3 3 0
     
     
     
A
B C
G

6. 6
Q. Let f,g,h are the lengths of perpendiculars from circumcentre of ABC on the sides
a b c abc
a,b and c and if k then the value of k is :
f g h fgh
1 1
(a) 1 (b) (c) (d) 2
2 4
S. :- Let the triangle be an equilateral.
so accor ding to data

  
Master Triangle
2
1 2 3
2 2 2
2 3 1 1
of MT-1,
1
a b c 1and f g h
2 3
a b c abc
k
f g h fgh
2 3 2 3 2 3 k 2 3 2 3 2 3
k 1/4
Q. Let in a ΔABC with side 'a ',AD is a median of If AE and AF are medians of ΔABD
a
and Δ ADC and AD m ,AE m ,AF m then
8
(a) m m 2m (b) m
     
   
     
 
   
  2 2 2 2 2 2 2 2 2
2 3 2 3 1 2 3 1
1 2 3
2 2
2 3
m 2m (c) m m m (d) m m m
S. :- Let the triangle is an equilateral with side 'a'.
therefore according to MT-1
3 a
AD = m , AE = AF, m m
2
a 3 a a 13 a
ED ,AE = AF m m
4 2 4 4
Afte
     
  
   
            
Master Triangle

 
2
1 2 3
1 2 3
1
a
r putting values of m ,m ,m in options then only (a) will give .
8
Q. If , , are the int ernal bi sec tors of angles A,B and C of a ABC, then :
A
cos co
2
Statement : I

 
 
  
Reasoning Based problems can also be solved by Master Triangles
l l l
l
 
1 1
2 2 2
2 2 2
1 2 3
A A
s cos
1 1 12 2
2
a b c
a b c
Statement : II bc 1 , bc 1 , bc 1
b c a c b a
ST. Let the triangle is an equilateral then according to data of MT-1.
A B C
   
   
        
 
          
                                
  
l l
l l l
 
 
 
0
1 2 3
2
3
60 and a b c 1 and
2
3 3 3
1 1 12 2 2Statement : I 2 3 6(false)
1 1 13 3 3
2 2 2
3 1 3 3
Statement : II 1 1 1 , which is true.
4 1 1 4 4
other two are also true so d is right option.
   
 
       
 
  
         
l = l = l
A
DΕ
B C
F
A
B CD
E
P
F
1
A
B CD
E
P
F
1

7. 7
Q. In a triangle ABC, points D and E are taken on side BC such that BD = DE = EC.
If ADE = AED = , then:
(a) tan =3tanB (b) 3t
  

One or more than one currect answers types problems can also be solved by Master Triangles
 
2
0
0
2 2
6tan
an =tanC (c) =tanA (d) B = C
tan 9
3 2
S. Let tan = 3 3 and tanB tan60 3
1 6
tan =3tanB so
6tan 6 3 3
= =tan60 3 = 3 tanA so
tan 9 3 3 9
clearly B = C so

  
 
   
 
 
 
  
 
(a) is true.
(c) is true.
(d) is true. Hence (a) ,(c) and (d) are
Q. The int ernal bi sec tor of A of ABC meets side BC at D. A line drawn through D and
perpendicular to AD intersects side AC at E and AB at F,If a,b and c are the sides of
triangle then which of the followi
 
the correct answers.
0
ngs is true :
2bc A 4bc A
(a) AE is H.M.of b & c. (b)AD cos (c) EF sin (d) AEF is isoscales.
b c 2 b c 2
S. Let the triangle is an equilateral therefore according to the data of MT-1,
3
A B C 60 & a b c 1,AD ,AE AF AB BC AC
2

  
 
          
[ IITJEE 2006,5M]
 
 
   
 
0 0
0
1
2bc 2 1 1
then (a) 1 AE
b c 1 1
3 2 1 1 3
(b) cos 30 cos 30 .
2 1 1 2
4 1 1
c 1 sin 30 1 .
1 1
(d) As AD is perpendicular EF and DE DF
and AD is the bi sec tors therefore the AEF can be isoscales. .

 
  
 
 
  

 
 



True .
True
True
True
Hence (a),(b),(c),(d
Q. Let AD, BE and CF are the perpendiculars from the angular points of a ABC upon
the opposite sides, then which of the following is not true :
Perimeter of DEF
(a)
Perimeter of ABC




) are correct answers.
r
. (b) Area of DEF = 2 cosAcosBcosC.
R
R
(c) Area of AEF = cos2A. (d) Circum radius of DEF = .
4
S. Let the triangle be an equilateral then according to data of MT 1.
Perimeter of DEF 3 2 1 r
(a)
Perimeter of ABC 3 2 R
(b)
 
  

  


-
2
3 0
2
3 1 3 3 3
of DEF = 2 cosAcosBcosC =2 cos 60
4 2 4 16 16
3 1 3 1 R 1 1
(c) of AEF= cos2A (d) R of DEF =
4 2 4 2 4 2 3 4 3
.
 
     
 
   
           
   
Hence options (c) & (d) are correct
A
B C
  EFD
1/3
1/6
A
B CD
E
P
F
1

8. 8
 
1 2 3
2 2 2
1 2 3
Q If P ,P ,P are altiudes of a triangle ABC from thevertices A,B,C and if is area of
the triangle then
1 1 1
i The value of is :
P P P
sin A sin B si
(a)

 
 
: -Paragraph Based problems can also be solved by Master Triangles
 
 
1 2 3
0
1 2 3
n C cot A cot B cot C tan A tan B tan C
(b) (c) (d) 0
1 1 1
ii The value of is :
P P P
1 2 1
(a) R (b) (c) (d)
r r R
S. i Let the triangle is an equilateral triangle then according to MT 1.
3
A=B =C=60 . a=b=c=1.P P P .
2
   
  
 

  
 
 
2 2 2
1 2 3
1 2 3
3 1 1
,r = , R = .
4 2 3 3
1 1 1 4 4 4
L.H.S. 4.
3 3 3P P P
After putting the values of A,B,C and in options
then d will give R.H.S.
1 1 1 2 1
ii L.H.S. ×3= 2 3 .
P P P r3
Q. Let DEF is the triangle formed by joining the points of contacts of th
 
      

    
e incircle with
sides of ABC also let R,r and Δ are radius of circumcircle, radius of incircle, and
area of the triangle ABC, then
(i). The sides of ΔDEF are :
A B C A B
(a) 2r cos ,2rcos , 2rcos (b) r cos , rcos
2 2 2 2 2

         
        
         
C
, rcos
2
r A r B r C
(c) r sin2A,r sin2B, r sin2C (d) cos , cos , cos
2 2 2 2 2 2
(ii). Area of ΔDEF is :
rΔ rΔ rΔ 2rΔ
(a) (b) (c) (d)
4R 2R R R
S. :- Let the triangle be an equilateral then according to data of MT-1,
A
 
  
 
     
     
     
Master Triangle
0
0
2
1 1 3
=B=C=60 and a =b=c=1,R = ,r = , area = Δ =
43 2 3
1 1 1
The sides of ΔDEF are , , .
2 2 2
1 1 1 1
put values A=B=C=60 & r = in options then only option (a) will give , , .
2 2 22 3
3 1 3 3
also of ΔDEF , put r,Δ,R in options then only (b)will give .
4 2 16 16
 
   
 
A
B CD
E
P
F
1