mathematics short tricks books samples for jee(main)

Q. The value of (1 cot cosec )(1 tan sec ) is equals to :

(a) 3 (b) 0 (c)

S. Here the value of the expression is independe

1 * (d) 2

nt of θ,therefore givenexpression is

an identity i

Magical Conceptual Short Tricks

0

n θ, so we can put any suitable value of θ to minimise the calculations.

here let θ 45 then value of given expression is (1

Q. Let k=4cos x cos2x cos3x cos x cos

1 2)(1 1 2) (2

2x cos3x thenk is equals to :

(a) 1 (b) 0

2)(2 2)=2.

*(

0 0 0 0 0 0

2

0

2

S. Again value of the expression is independent of x, so the expression is anidentity in x,

so let x 0 then k=4cos0 cos0 cos0 cos0 cos0 cos0

c) 1 (d) 2

Q. The value of the expression

4 1 1 1

cos c (

1

os

2

0

2

S. Here options are in terms of so the value of the expression is independent

) 2cos cos cos( ) is ?

* (a) sin (b

of .

so we can put any su

)cos (

itable value of to minimise the calculations.

pu

c)sin2 (d)cos2

t =0

2 2

2 2 2 2 2

then, cos cos ( ) 2cos cos cos( )

cos 0 cos (0 ) 2cos0cos cos(0 ) 1 cos 2cos cos 1 cos s

Q. If tanA tanB x and cot A cotB y then cot(A B)

1 1 1 1 (a) *(b)

in .

(c)x y x y

0 0

0 0 0

1 2 1 2S. Let A 60 &B 30 then x 3 & y 3 ,

3 3 3 3

so L.H.S. cot(60 30 ) cot30 3

Now put the values of x and y in the options then option (

x y

b) will give

(d) x y

7Q.

s 3 .

If X sin +sin s12 12

0 0 0 0

3 7 3in ,Y cos +cos cos

12 12 12 12

X Ythen

Y X

(a) 2sin2 (b) 2cos2 * (c) 2tan2 (d) 2cot

3 3S. Let 15 then X sin120 +sin0 sin60 = 0

2

2

0 0 0

0

32

1 1 3 1 2Y cos120 +cos0 cos60 1 1, then L.H.S.

2 2 1 3 3

Now put 15 in the options then option (c) will gives 2/ 3 .

Q. If cos cos cos 0 and if cos3 cos3 cos3 kcos cos cos then k

(a) 1 (b) 8 *(c) 12 (d)

0 0

0 0 0 0 0 0

S. cos cos cos 0 so let 0, 120 & 120 satisfying given condition

cos3 cos3 cos3 kcos cos cos

cos0 cos360 cos360 kcos0 cos120 cos120

1 1 1 k.1.( 1/2).( 1/2) k 1

9

2

The beauty of these short tricks is that many problems of this kind can be solved easily.

:S.

Let = 15º then LHS = tan15º+2tan30º+4 tan60º + 8cot1

Q. The valu

20º

1 1 =2 3+2 × +4 3 +8

3 3

e of tan +2tan2 +4tan4 +8cot 8 is :

(a) tan (b) tan2 * (c) cot (d) cot2

Magical Method of Substitution and Balancing

2 2 2 2 2 2 2

0

2

p 1Q. If tanA and if =6 is acute angle then (pcosec2 q sec2 )

q 2

(a) p +q * (b) 2 p +q (c) 2 p q (d)

=2+ 3 = cot 15º =cot

S. : Let A 45 then q p

2 2 2 2and LHS pcosec15 psec15 p 2 2 p

3

p q

1 3 1

Magical Method of Substitution

2 2 2 2

0

2 2

8abQ. If a sinx sin y, b cos x cos y, c tanx tan y then

(a b ) 4a

* (a) c (b) c (c) 2c (d)

now put q p in options then b wil

2c

l match with LHS.

S. : let x y 45 ,


2 2 2 2

0 0

Q. If cos 2cos then tan tan2 2

1 1 1 1a b c d

3 3 3 3

therefore a 2, b 2, c 2

8 2 2then 2 c

( 2 2 ) 4 2

S. : cos 2cos so let 60 & 0

tan


n n1 2 3

0

1 2 3

n1 2 3

n/ 2

0

2 n/

Q. If 0 , , ,................. and if tan tan tan ...........................tan 12

then

1ta

cos cos cos ............

n tan 30 tan 30

...............cos

(a) 2 * (b) 2 (c)

2 2 3

0n1 2 3

n/4 n/

0

4

0 0

:S.

Let ............. 45 (Satisfying both given conditions)

Required value cos45 cos4

2 (d

5 cos45 ............n terms

1 1 1.....................

2 2 2

) 2


k k4 6

n/2n/2n

k k 4 4 6 6k

k

4 6

1Q. Let f (x)= (sin x cos x),x R and k 1,then f (x) f (x)

k1 1 1 1

(a) * (b) (c) (

1 1.....n tertms 2

2( 2)

1 1 1S. f (x)= (sin x cos x) f (x) f (x) (sin x cos x) (sin x cos x)

k 4 6

As va

d

lue of above expr

)4

ession

12 6 3

is i

4 4 6 64 6

ndependent of x soput x 0 in above expression

1 1 1 1f (x) f (x) (sin 0 cos 0) (sin 0 cos 0)

4 6 4.

26

1

1


0 0 00

2

:

sin5 sin2 sinQ. The value of the expression is equals to :

cos5 2cos3 2cos cos

* (A) tan (B) cos (C) c

sin150 sin60 sin30Let 30 then LH

ot (

S = cos

D si

.

n

S

)


0 0 2 0 0

0

0 0

2 2 2

:

Q. If (cos cos

1

150 2c

) (sin

os90 2cos 30 cos30 3

Now

sin ) ksin

put 30 in options then (a) w

, then k isequals to:2

(A) 0 (B) 1 (C) 3 *(D) 4

ill match with L.H.S.

S.

Let 90 , 0 then (0

Concept of Identity

2 2

4 4 2

22 2

2 2 2 2 2 2 2 2

4 4 2

Q. Which of the followings is not equals to unity :

(A) cos sin 2sin

sin cos(B) (1 cot ) (1 tan )

2 2

(C) sin cos cos sin s

11) (1 0) k k 4

2

in si

S.

n cos cos

* (D)sin cos 2sin

Concept of Identit

0 0 0 0 0

4 4 2

22 2

2 2 2 2

the value of theabove expressions should be 1.

put 45 ( sin45 cos45 and also tan45 cot45 ).

1(A) cos sin 2sin 0 2 1

2

sin cos 1 1 1 1(B) (1 cot ) (1 tan ) .2 .2 1

2 2 4 4 2 2

(C) sin cos cos sin si

:

n

y

2 2 2

3 3

2

3

4 4 2

1 2

1 1 1 1sin cos cos 1

4 4 4 41

(D)sin cos 2sin 0 2 1(not equals to unity).2

S.

The value o

4Q. sin sin sin is equals to :

sin3 3 3

4 3 3(A) (B) * (C) (D) no

f

ne3 4 4

:

Concept of Identity

0

3 0 3 0 3 0

0

above expressions is independent of so put 30 .

sin 30 sin 150 sin 270 1 1 3then value of given expression 1 .

8 8 4sin90


cot 1S. 3cot cot cot3

cot 1 tanQ. If then is equals to :

cot cot3

2cot cot3 2tan3 tan

3 tan tan3

1 2 2 1(A) *(B) (C) (D)

3 3 3 3

Q. If 13 cos 12cos( 2 ), then va

cot cot3 3

tan 1 1 2

tan3 1tan tan3 31 1tan 2

cos( 2 ) 13 cos( 2 ) cos

lue of cot( ) 25tan is :

* (A) 0 (B) 1 (C) 3 (D) 4

Q. I

25 2cos( ).cosS. 25

cos 12 cos( 2 ) cos 1 2sin( ).sin( )

cot( ).cot 25 cot( ) 25tan cot ( ) 25t

f x y 3 cos4 a

0

n

a

d

n

0

4 4 3 3 2 2

:S.

The value of the expression is independent of , so put 0 .

x y 3 1 x y 2...

x y 4sin2 the

(i) and

n:

(A) x y 9 (B) x y 9 (C)x y 2(x y ) * (D) x 2

0

y

x y .

Magical Method of SubstitConcept of Identity ution and Balancing

0 0 0 0 0 0 0 0

0 0 0 0 0 0b(cos1 sin1 )(cos2 sin2 )(cos3 sin3 )........

.....(ii), by (i) and (ii) x y 1

Now put x y 1 in the options the

......(cos45 sin45 )

cos1

n optio

cos2 cos3 cos4 .......

n (D) is sat

.....cos44 cos

isf

5

.

4

ied

Q. If = a ,

whe

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0

(cos1 sin1 ) (cos2 sin2 ) (cos3 sin3 ) (cos44 sin44 ) (cos44 sin44 )S. ..........

cos1 cos2 cos3 cos44 cos4

(A) 22 (B ) 23 (C) 24 *(D) 2

5

(1+tan1°)(1+tan2°)............. .

5

.

re a and b are prime numbers and a b then a b

0

0

.......(1+tan43°)(1+tan44°)(1+ tan45°)

1 44 2 43 3 42 ......... 45

So first we find out if 45 then (1+tan )(1+tan ) ?

let 0 and 45 then(1+tan )(1+tan ) (1) 2 2

(1+tan1°)(1+tan4

By method of substitution

b

22 pairs

22 k 23 b

4°)(1+tan2°)(1+tan43°).............(1+tan22°)(1+tan23°). (1+1) = a

2 2 = 2 2 = a a 2 andb 23 and hence a b 25

*Do you know what are the methods of Substitutions & Balancing to Solves problems of Trigonometry? *Do you know what are Master A.P.,G.P.& H.P. and How these Solve the problems of Progressions ?

*Do you know the Master Quadratic Equations to Solves problems of Quadratic Equations ? *Do you know the Master Triangles to Solve problems of Solutions Of Triangles ?


A triangle has three sides and three angles.

The sides opposite to angles A, B and C. are denoted

by a,b and c.

the sum of its angles is A+B+C =180°and

the sum of its angles is a+b+c

Properties and Solutions of Trian gles

= 2s (perimeter),

In any triangle sum of any two sides is greater than third side a+b >c,c+b >a,a+c>b*

Most of the questions of this chapter can be solved veryeasily bymaster triangles,

for that we have invented three Master Triangles.

As we all know that a master key unlocks many l

Master Tri

ocks. Simil

angles

arly our Master triangles

can solve many questions belong to family of triangles.

Q In a triang

* The above master triangle can be used most widely to all the problems which

belong to a family of triangles i.e. problems belong to all triangles.

* If the data of this triangle satisfy the given conditions in the questions.

le ABC correct statement is : [ ,

B+CA A B C(a) (b c) cos = 2a sin (b) (b c) cos = a sin

2 2 2 2

B+CB C A A(c) (b c) sin =a cos (d) (b+c) sin =a cos

2 2 2 2

IITJEE 2005 1M]

S.

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)

wil

–

Here above problem belongs to all triangles,so MT -1 can be applied here.

l balance and so it is the right choice.

A

F

B CD

E

P

R

-

-

0

1 2 3

3Let a = b = c = 1 A= B = C = 60 Area= Δ =

4

abc 1Radius of circumcircle R

4Δ 3

Δ 1Radius of in circle r

s 2 3

3Radii of ex circles are r r r

2

3Length of perpendicular (altitutes) AD BE CF

2

Equilateral Triangle

Master Triangle 1 MT -1

CB

b

A




and if the data of this triangle satisfy the given conditions in the questions.

Some of the exa

Q. If sides of a triangle are in ratio 1: 3: 2, then ratio between its orresponding angles is :

(

(a) 3 2 1 b 3 1 2 c 1

S

2 3 d 1 3 : 2

. The data

ΙΙΤJEE 2004,1M)

mples are given below : -

of MT–2, a 1,b 3 and c 2 satisfy the given conditions

so triangle is right angled & A 30º, B 60º, C 90

Q. If the angles A,B and C of a triangle are in arithmatic Progressions a

º. A:B:C 30 60:90 1

nd a,b and c are

: 2 3

sides

0 0 0

0 0

S. Angles are in A.P. so according t

a copposite to angles A,B and C then value of the expressi

o data of MT-2,

A 30 ,B 60 ,

on sin2C sin2A is :c a

C 90 and a 1,b

1 3(a) (b) (c)1 (d) 3

2

3,c 2

a c 1 2sin2C sin2A sin180 sin60

c a 2 1

2

IITJEE2010

2 2 2

2 2 2

32 3

2

S. A : B : C 1: 2 : 3, therefore we can apply a

a b cQ. In ABC if A : B : C 1: 2 : 3,then the value of is equals to :

ab bc ac

2 2 8 8(a) (b) (c

bove master triangle here.

a b c 1 3 4a 1,b 3,c

) (d)3 1

2 and he

3 1 3 3 2 3

nceb b ac

3

a

1

c

8

3 2 3 2 3 3 2

Many questions can be solved by this above Master Triangle

Right Angled Triangle


0 0 0 A 30 ,B 60 ,C 90

a 1,b 3,c 2

3 1r .

2

R 1.

31Δ= . 3 .

2 2

2

1

r

A

BC




and if the data of this triangle satisfy the given conditions in the questions.

Some of the exa

S. Since data of above triangle( MT–3) i.e. A

Q. In a Δ ABC if sinA +sinB+sinC =1+ 2 and cosA +cosB+cosC = 2 ,then triangle is :

a Equilateral b Isosceles c Right angled d Right angle isosceles

mples are given below : -

45º, B 45º and C 90º

when put in given conditions of the questions

sinA sinB sinC 1 2

1 1sin45º sin45º sin90º 1 2 1 1 2 1 2 1 2(satisfy)

2 2

1 1cos45º cos45º cos90º 2 0 2 2 2(satisfy)

2 2

therefore, it is right an

2 2 2 2 2 2

0

gled isosceles.

S. Since data of above triangle( MT–3) , a 1, b 1 and c 2

satisfies given condition b c 3a i.e.1 ( 2) 3.1 3 3

therefore A=45 ,B=4

2 2 2Q. In ΔABC if b +c = 3a then cotB+cotC - cotA =

ab(a) 4Δ (b) Δ (c) (d) 0

4Δ

0 0

0 0 0

5 and C 90

then cotB cotC cot A cot45 cot90 cot45 0

S. Since data of above triangle( MT–3)

A 90º,B 45º and C 45º and a

0 0 0 0

2cosA cosB 2cosC a bQ. In a ΔABC, if + + = + then the value of ÐA is :

a b c bc ca

(a) 60 (b) 90 (c) 30 (d)75

0

2, b 1 and c 1

2.0 21 1 3 32 satisfies ,therefore A 90 .

12 2 2 2 2

Right Angled Isoscales


0 0 0

a 1, b 1, c 2

A 45 ,B 45 ,C 90

1R ,

2

1r 1 ,

2

1 1Δ 1 1

2 2

2

1

r

A

BC

1

3 3 3

0

a b c

Q. In a Δ ABC, if sin A sin B sin C = 3sinAsinB sinC, then b c a =

c a b

a 0 b a b c c a b c ab bc ca

S. : The data of MT–1 , A B C 60 satisfies the given condition.

d 1

MT -1

Questions which have single currect answers :

cosA cosB cosC

Q In triangle AB

let a=b=c, therefore thetriangle is an equ

C if and if a = 2 then area of triangle wi

ilater

ll be :

al sotake a b c 1

a b c 1 1

a b c

a 1 b 2

1

b c a 1 1

1

0

c a

b 1 1 1

2

2 2

S. : The data of MT–1 satisfies the given condition. so the triangle is

3 3 an equilateral area o

3 c d 3

2

a a aQ. In ABC, if cos , cos & cos then tan

b c

f the equilateral triangleis Δ a (2) 34

b c c 2

4

b

MT -1

0

2 2 2 2 0

2 2

S. : Let the ABC is an equilateral and a b c 1,

1cos cos cos 60

2

tan +tan =2 2

a 3 b 1 c 3 3 d 9

(a b c)(b c a) (c a bQ. In triangle ABC,

tan tan +tan = 3tan 0 =12 2 2

3

MT -1

2 2

2 2 2 2

S. : Let the triangle be an equilateral. Therefore, accordi

) (a b c)=

4b c

a sin A b cos A

ng to MT–1

3a b c 1 and A B C 60º so L.H.S.

4

after

c tan A d cot A

putting the value of A

in th

MT -1

2 2 2

2 3

0

1 A 1 B 1 CQ In triangle ABC , the value of cos cos cos is equals to ?

a 2 b 2 c 2

s s sa s b

e options, a will give R.H.S

S. : le

t a b c 2 &

A B C 60 ,L

c d abc

H

c

.

abc ab

MT -1

2 9.S. 3cos 30º =

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.


S. : Let the triangle be right angled isoscales.

Therefore, according to MT–3 a

2 2 2 2

2 2 2

C CQ. In any triangle ABC, (a - b) cos +(a + b) sin is equals to : -

2 2

a a b b c c d

3

1

MT -

2 2 2 2 2

1, b 1, c 2 and A 45º, B 45º,C 90º.

C C 1L.H.S. (a b) cos (a b) sin 0 4 2 c (since

AQ. In a triangle ABC,

c 2)2 2 2

2ac sin


2 2 2 2 2 2 2 2 2 2 2 2

S. : Let the triangle is a right angled. Therefore, according to MT 2

a 1, b = 3 ,c 2 & A

– B+C =

2

(a

30º,

) a b – c (b) c a b (c) b – c – a (

B 60º,C 90º.

1then

d) c –

L.H.S. 2 1 2 sin 30º 4

a – b

MT -2

2 2 2

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 a)(s b)(s c)

(a) s (b


2 2 2 2 2 2

2 2 2

2 2 22 2 2

S. : c a b the triangle is right angled and c is hypoteneous.

1 a bL.H.S. 4s(s a) (s b) (s c 4Δ ×a b = 4 a b

2 4

Therefore, option d is the right o

) b c (c) c a (d

ption.

Q. In triang

) a

b

MT -2

2

2

2

2

S. : Let the triangle be an equilateral so according

le ABC if is the area of the triangle then a sin2C c sin2A

(a) Δ (b) 2Δ

to MT–1 ,

3 3 3a b c 1, A B C 60°, Δ= (1) , Δ 3 4Δ

4

(

4 4

LHS

c) 3Δ (d

a

) 4

2

Δ

sin C

MT -1

2 2 2

2

2

2 2

8a b cQ. In a triangle,if (a b c)(a b c)(b c a)(c a b) , then the triangle is

a b c

[Note: All symbols used have usual meaning

3c sin2A 1 sin120° 1 sin120°= 2sin120° 2 3 4Δ .

2

in ABC.]

(a) isosceles (b) right angled (c)

S. : The sides of MT 2 i.e. a 1, b = 3 ,c 2 satisfy the given relation of

the t

equilateral (d) ob

riangle so it is

tuse angl

right an e .

ed

gl d

MT -2


Q. In a ABC, if G is the centroid of thetriangle and GA,GB,GC makes angles , ,

with each other then cot A cotB cotC cot

S. : Let the triangle be an equilater

cot cot

(a)1 (b

al

then accor

) 0 (c

ding

)3 (d)3 3

to data o

MT -1

0 0

0 0 0 0 0 0

0 0

2 2 2

f MT-1.

A B C 60 then 120

cot60 cot60 cot60 cot120 cot120 cot120

3(cot60 cot120 ) 3 3 3 0

a b c A B CQ. In a ABC, the value of the expression sin sin sin is :

sinA sinB sinC 2 2 2

(a

0

2 2 2

S. : let the triangle be an equilateral, then according to data of MT-1

3a b c 1, A B C 60 ,

4

a b c A B C 2 2 2 1 1 1sin sin sin

sinA sinB sinC 2 2 2 2

)2 (b) (c) (d)2 4

Q. Let f,g,h are th

2 2

e

3 3 3

MT -1

lengths of perpendiculars from circumcentre of ABC on the sides

a b c abca,b and c and if k then the value of k is :

S. : Let thetrianglebe anequilateral so according to data of MT-1

f g h fgh

1 1(a)1 (b ) (c) (d) 2

2 4

1a b c 1and f g h

2 3

th

MT -1

Q. In a triangle ABC with fixed base BC, the vertex A moves such that

B C Acot +cot =2cot and If a,b a

a b c abc

nd c de

1erefore k 2 3 2 3 2 3 k 2 3 2 3 2 3 k

f g

note the lengths of the sides of2 2 2

h fgh 4

a triangle

opposite to the angles A, B and C, then:

(a) locus of points A is an ellipse.

(b) b + c = 2a

(c) b + c = 4a.

(d) locus of point A is a pair of st

S. : The data of M

raight l

aster Tri

ines.

angle-1 sMT -1

atisfy the given condition so

let the triangle be an equilateral. b+c=2a

AB AC BC 2a dis tance between two fixed points locus of vertex A is ellipse.

A

B C

G

2

1 2 3

2 2 2 2 2 2 2 2 2 2 2 22 3 1 1 2 3 2 3 1 2 3 1

Q. Let in a Δ ABC with side 'a ',AD is a median of If AE and AF are medians of Δ ABD

aand Δ ADC and AD m ,AE m ,AF m then

8

a m m 2m b m m 2m

S. :

Let

c m m m d m m

the triangle i a

m

s n eMT -1

1 2 3

2 2

2 3

1 2 3

2

quilateral with side 'a'.

therefore according to MT-1

3 aAD = m , AE = AF, m m

2

3 a 13 aa aED , AE = AF m m

4 2 4 4

After putting the values of m , m , m in the options

athen (a) will give ,therefore (a) is the correct ch

8

Q. Let DEF is the triangle formed by joining the points of contacts of the incircle with

sides of ABC, also let R,r and Δ are radius of circumcircle, radius of incircle, and

area of the triangle ABC

oice

, the

.

n

(i)

Questions based on Paragraph

. The sides of ΔDEF are :

A B C A B C*(a ) 2rcos ,2rcos ,2rcos (b) r cos ,rcos , rcos

2 2 2 2 2 2

r r rA B C(c) rsin2A,rsin2B, rsin2C (d ) cos , cos , cos

2 2 2 2 2 2

(ii). Area of ΔDEF is :

rΔ(a) (b)

4R

0

S. : Let the triangle be an equilateral then according to data of MT-1,

1 1 3A=B=C=60 and a=b=c=1, r = , R= , area of triangle = Δ =

42 3 3

1(i) The length of

rΔ rΔ 2rΔ(c) (d)

2R

the sides of ΔDEF are DE EF FD2

Now put the values of r,A,B a

R

d n

R

n C i

MT -1

2

the options then

A 1 60 1only option (a)will gives 2rcos 2. .cos = ,

2 2 22 3

B 1 C 1similarly 2rcos and 2rcos .

2 2 2 2

3 1 3(ii) Area of ΔDEF=Δ = .

4 2 16

Nowput the values of r,R, in the options then only option (a

3)will gives .

16

(a) is right choice.

A

DΕB C

F

A

B C

EF

R r

D 1/2

1/2


2

Q.In ΔABC,the rario of angles A,B,C is 1:2 :3,then which of the following is/(are)correct ?

(a) Circumradius of ΔABC = c. * (b) a : b : c= 1 : 3 : 2.

3(c) Perimeter of Δ ABC= 3+ 3. *(d) Area of ΔABC = c .

8

S. : A :B : C 1MT -2

one or more currect answers

:2 :3 so let A 30º,B 60º,C 90º.

3according to MT 2, a 1, b = 3 ,c 2, = , R 1

2

(a) Circumradius of ΔABC = c 1 3 R 1 and c 3 .( Not true)

(b) a : b : c= 1 : 3 : 2.

Not necessorly always true as(c) Perimeter of Δ ABC = 3+ 3

A 45º,B 60º,C 75ºa

(True)

2

Q.Two parallel chords are drawn onthesame sides of the center of a circle of radius R.

It is found that they subtands an angle

lso possible

3 3 3 3 3(d) Area of ΔABC = c = .4 =

s of and 2 at the cente

8 2 8 2 2(True)


r of a circle,thenthe

perpendicular dis tance between the chords is equals to ?

3(a) 2R.sin sin *(b) R 1 cos 1 2cos

2 2 2 2

3* (c) 2R.sin sin (d) R 1 cos 1 2cos

4 4 2 2

S. COD and AOB 2 and required dis tance MN ON OM

Let 60 then 2 120

ON 3cos30 ON R cos 30 R.

R 2

OM 1Similarly cos 60 OM R

R 2

3 1 Rd MN R 3 1

2 2 2

put 60 in options,then

(a) 2R.sin 90 .sin30 2R 1

1R d

2

3 2 3 R(b) (1 cos 30 ) (1 2cos30 ).R 1 1 R ( 3 1) d

2 2 2

( 3 1) R( 3 1)1(c) 2R. sin45 sin 15 2R d

22 2 2

[ 3 1]R3 3(d) 1 cos30 1 2cos30 R 1 1 2 R d

2 2 2

(True)

(True)

M

Nd

RO

BA

C D


Q. In an acute angled triangleABC, let AD,BE and CF be the perpendicular from A,B and

C uponthe opposite sides of the triangle.(All symbols used have usuaul meaning in a

triangle )

(i).The ratio of product of the

Questions based on Paragraph

1

3

side lengths of the triangles DEF and ABC, is equal to ?

3(a bc ) 1 A B C(a) (b) (c) cos AcosB cos C (d) sin sin sin

4(a+b+c) 4 2 2 2

(ii). The circumradius of the triangle DEF can be equal to ?

abc a R r(a) (b) (c) (d) cose

8Δ 4sinA 2 8

0

S. Let the Triangle is an equilateral triangle then according to MT-1.

3 1 1A=B =C=60 ,a=b=c=1, Δ= ,r = , R = .

4 2 3 3

1DEF will also be an equilateral triangle with side .

2

Product of the sides lengths of the DEF(i).

A B Cc cosec cosec

2 2 2

Produc

0

1 1 1. .

12 2 2 =t of the sides lengths of the ABC 1.1.1 8

now putting the values of A=B=C=60 and a=b=c=1

1in options then onlyoption (d) will give so it is the right choice.

8

1(ii).The circumradius of the DEF =The inradius of the ABC

2 3

now pu 0tting values of A=B=C=60 and a=b=c=1and r,R in options

1then onlyoption (a),(b),(c)and(d) will give so all the opti

Q. In a triangle ABC, poi

ons are currect

nts D and E are taken

.2

on side BC such

3


2

0

that BD = DE = EC.

If ADE = AED = , then:

6tanθ(a) tan = 3tanB (b) 3tan = tanC (c) = tan A

S. Let ΔABC be an equilateral and let AB=BC=AC=1 and A=B=C=60 .

3 21then AF= 3 2 and DF= ,tanθ =

(d) B = C t

=3 36

a

1 6

(

n θ 9

a) t

0

0

0

02 2

0

an = 3tanB 3 3= 3tan60 3 3= 3 3

(b) 3tan = tanC 3.3 3= tan60 9 3 3

6×3 36tanθ 18 3(c) = tanA =tan60 = 3

18tan θ 9 3 3 9

(d) B = C tan60 tan60

(True).

(True).

(True).

Hence (a) ,(c) and (d) are the correct answers.

A

B C

EF

R r

D 1/2

1/2

A

B C

EF

R r

D 1/2

1/2

A

B C

EFD

1/3

1/6


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