PROPERTIESOFTRIANGLE 1. In a triangle ABC , medians AD and DE are drawn. If 4AD = ,

/ 6, / 3DAB ABEπ π∠ = ∠ = then the area of ABCΔ is (1) 8 / 3 (2) 16 / 3 (3) 32 / 3 3 (4) 64 / 3 2. In ABCΔ , cos cos cosA B C+ + =

(1) 1 rR

+ (2) 1 rR

− (3) 1 Rr

− (4) 1 Rr

+

3. If in a triangle 4 4 4 2 2 2 2 2 2sin A sin B sin C sin Bsin C 2sin Csin A 2sin Asin B+ + = + + then its acute angle A is equal to (1) 45° (2) 30° (3) 150° (4) 60°

4 The sides of a triangle are sin , cosα α and 1 sin cosα α+ for some 02πα< < . Then the

greatest angle of the triangle is. (1) 060 (2) 0150 (3) 0120 (4) 090 5 In ABCΔ , the line joining the circum-centre and in-centre is parallel to BC then the value of

cosB+cosC=

1. 1 2. 12

3. 34

4. 32

6. If a, b and c are the sides of a triangle then the minimum value of 2a 2b 2cb c a c a b a b c

+ ++ − + − + −

is

1) 3 2) 9 3) 6 4) 1

7. In ABC,AΔ is acute, b 2,c 3 1= = − , area of ABCΔ is 3 12− then angle B =

1) 0135 2) 015 3) 030 4) 060

8. In triangle ABC, R(b c) a bc+ = where R is the circum radius of the triangle. Then the triangle is

1) isosceles but not right angled 2) right angled but not isosceles 3) right angled isosceles 4) equilateral

9. In a triangle the lengths of two larger sides are 24, 22 respectively. If the angles are in A.P then the third side is

1) 12 2 13+ 2) 12 3) 2 3 2+ 4) 2 3 2−

10. In a triangle ABC, if P, Q, R divide sides BC, CA and AB respectively in the ratio

K :1(in order). If the ratio area PQRarea ABC

is 13

then K =

1) 1 2) 4 3) 12

4) 13

11. The perimeter of a ABCΔ is 6 times the A.M. of the sines of its angles. If the side ' 'a is 1, then the angle ' 'A is

(1) 6π (2)

3π (3)

2π (4) π

12. If the median of triangle ABC through A is perpendicular to AB, then

1) tan 2cot 0A B+ =

2) tan 2 tan 0A B+ =

3)

2 tan A+ cot B = 0

4) 2 tan tan 0A B+ =

13. Let PQR be a triangle of area Δ with a=2 , b= 7/2 and c=5/2; where a,b, and c are

lengths of the sides of the triangle opposite to the angles at P,Q and R respectively. Then

-

2sin p − sin2 p2sin p + sin2 p

equals

1)

34Δ

2)

34Δ

⎛⎝⎜

⎞⎠⎟

2

3)

454Δ

4)

454Δ

⎛⎝⎜

⎞⎠⎟

2

14. In a triangle ABC, D and E are the midpoints of BC and CA respectively. If

AD=5,BC=BE=4, then CA=

1)5 2) 7

3)2 7 4)5 5

15) If ( )22a b cΔ = − − , where Δ is area of triangle ABC then TanA=

1) 1516

2) 815

3) 817

4) 14

16. In ,ABC ADΔ is the altitude from .A If 02 2, 23 , abcb c C ADb c

> = =−

then B∠ =

1) 067 2) 0113 3) 073 4) 046

17. In an equilateral triangle, three coins of radii 1 unit each are kept so that they touch each

other and also the sides of the triangle. Area of the triangle is

1) 4 2 3+ 2) 6 4 3+ 3) 7 3124

+ 4) 7 334

+

18. In triangle ,ABC D is mid point of ,BC if ' 'AD is perpendicular to ' 'AC then cos .cosA C =

1) ( )2 223c aac−

2) 2 2

2c aac− 3)

2 2c aac− 4)

2 2

2c aac+

19. If the angles of a triangle are in the ratio 4 : 1 : 1, then the ratio of the longest side to the

perimeter is

1) 3:(2 3)+ 2) 1 : 6 3) 1: 2 3+ 4) 2 : 3 20. If the median AD of ,ABCΔ makes an angle α with AB then sin ( )A α− =

1) sinbcα 2) sinc

bα 3)

sina

c α 4)

sinc

b α

21:. In triangle ,sin ,sinABC A B and sinC are in A.P., then

1) the altitudes are in H.P. 2) the altitudes are in A.P.

3) the altitudes are in G.P. 4) none of these

22. If , ,f g h are the internal bisectors of a ABCΔ then 1 1 1cos cos cos2 2 2A B C

f g h⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

is equal

to

1) 1 1 1a b c+ − 2) 1 1 1

a b c− + 3) 1 1 1

a b c+ + 4) 1 1 1 1

2⎛ ⎞+ +⎜ ⎟⎝ ⎠a b c

23. In triangle 2 2 2, 2002 ,ABC a c b+ = then cot cotcotA C

B+ is equal to

1) 12001

2) 22001

3) 32001

4) 42001

24. Let 0 1 2 3 4 5A A A A A A be a regular hexagon inscribed in a circle of unit radius. Then the product

of the lengths of the line segments 0 1 0 2,A A A A and 0 4A A is

1) 34

2) 3 3 3) 3 4) 3 32

25. In ,( )( )ABC a b c b c a kbcΔ + + + − = if

1) 0k < 2) 0k > 3) 0 4k< < 4) 4k >

26:. If G is the centroid of a ,ABCΔ then 2 2 2GA GB GC+ + is equal to

1) ( )2 2 23 + +a b c 2) ( )2 2 213a b c+ + 3) ( )2 2 21

2a b c+ + 4) ( )22 2 21

3a b c+ +

27. If the hypotenuse of a right-angled triangle is four times the length of the perpendicular

drawn from the opposite vertex to it, then the difference of the two acute angles will be

1) 060 2) 015 3) 075 4) 030

4p

A

B C

D

p

a

b

28. The side of triangle ABC in A.P. (order being , ,a b c ) and satisfy 2! 2! 1 8 ,1!9! 3!7! 5!5! (2 )!

+ + =a

b

then the value of cos cosA B+ is

1) 107

2) 137

3) 117

4) 127

29. In ,ABCΔ if cos 2cos cos 2A B C+ + = then , ,a b c are in 1) A.P 2) H.P 3) G.P. 4) A.G.P 30. Suppose , ,α β γ and δ are the interior angles of regular pentagon, hexagon, decagon and

dodecagon, respectively, then the value of cos sec cos cosα β γ δ is ______

1) 5 14+ 2) 2 3) 1 4) 5 1

4−

HINTS:

======================================================================== 1. AD 4, BD DC= =

From AG 8 1 8triangle ABG, tan BG AGcot

3 BG 3 3 3 3 3π π= → = = × =

/ 6π

/ 3π

8 / 3

G4 / 3

A

B D C

E

Area of ( )1 1 8 16triangle ADB BG AD 42 2 3 3 3 3

= × = × × =

∴ Area of ABC ( )2 Area of triangle ADB= × 323 3

=

2.sol:

A B C4Rsin sin sinA B C r2 2 2cosA cosB cosC 1 4sin sin sin 1 1

2 2 2 R R+ + = + = + = +

3.sol. Adding 2 22sin sinC B on both sides to make perfect square and using sine rule, we get 4 4 4 2 2 2 2 2 2 2 22 2 2 3a b c b c a c c b b c+ + + − − =

( ) ( )2 22 2 2 3b c a bc⇒ + − =

( )2 2 22 3bc Cos A b c⇒ = 2 34 3 30 ;1502

Cos A CosA A⇒ = ⇒ = ± ∴ =

4. 2 2sin cos 1 sin cos 1

cosA A 1202sin cos 2

α + α− − α α= = − → = °α α

5.sol: In ABCΔ

cos cos cosrA B CR

+ + = rR+ cos B + cosC = r

R

cos cos 1A B∴ + =

6. a b c s s s3s a s b s c s a s b s c

+ + = − + + +− − − − − −

Also,

ss− a

+ ss− b

+ ss− c

3≥ 3

s− as

+ s− bs

+ s− cs

[∵A.M.≥ H.M]

s a s b s c s s s1 9s s s s a s b s c− − −+ + = ⇒ + + ≥

− − −Q

Thus, the minimum value of the expression is 6.

7. ( ) ( )3 1 1 2 3 1 sinA A 30 B C 1502 2− = − ⇒ = ⇒ + =

0tan cot 60 .2 2 2

B C b c A B Cb c

− − −⎛ ⎞ = ⇒ =⎜ ⎟ +⎝ ⎠

0B 135∴ = 8. R(b c) a bc+ =

2RsinA bc= b csinA2 bc+∴ =

Now sinA 1≤ b c 12 bc+⇒ ≤

( )2b c 0 b c sinA 1⇒ − ≤ ⇒ = ⇒ =

0A 90⇒ = and b = c Hence, triangle is right isosceles

9. 2 2 2

0 160 , 24, 22,cos

2 2c a bB a b B

ac+ −= = = = =

c 12 2 13⇒ = ±

10. Area of AQRΔ is 1 b KC. .sinA2 K 1 K 1+ +

( )2K

K 1= Δ

+ when Δ is area of leABCΔ

Similarly area of ( )21KBPR PCQ

KΔ = Δ = Δ

+

∴ Area of ( )

2

2 2

3K K K 1PQRK 2K 1K 1

⎛ ⎞Δ − +Δ = Δ − = Δ⎜ ⎟+ +⎝ ⎠+

2

2

K K 1 1 1K 2(or)K 2K 1 3 2

− +∴ = ⇒ =+ +

11. ( )sinA sinB sinC 1a b c 6 a b c R 13 R

+ +⎛ ⎞+ + = = + + ⇒ =⎜ ⎟⎝ ⎠

a 1sinA A 62R 2π∴ = = → =

0 0 0 0 090 , 90 . 90 23 113B C B C B− = = + ∴ = + =

12: sin sin

DC ACDAC ADC

=∠ ∠

A

B CD

b

a/2 a/2

( ) ( )0 0

/ 2sin 90 sin 90

a bA B

=− +

( )( )

0

0

2sin 90 2coscossin 90

Aa Ab BB

−∴ = = −

+

sin 2cossin cosA AB B= −

tanA+2tanB=0

13:

2sin sin 2 2sin sin cos 1 cos2sin sin 2 2sin 2sin cos 1 cos

P P P P P PP P P P P P− − −= =+ + +

2

2

1 9 316 4× ⎛ ⎞= = ⎜ ⎟Δ Δ⎝ ⎠

14.: ( )2 2 2 2 2 22 2 / 2 50 8 58b c AD BD AD a+ = + = + = + =

A

B C2 D

Eb/2

( )2 2 2 2 2 2 24 2 2 / 2 32 / 2c BE CE BE b b+ = + = + = +

( ) ( )

223 42 28 2 7

2bi ii b b− ⇒ = ⇒ = ⇒ =

15l: ( )( )a b c a b cΔ = + − − + ⇒ ( )( )2 s c s bΔ = − − ⇒ ( )( )

( )14

s b s cs s a− −

=−

⇒ 1 / 24TanA=

12. 2 16 841 4 15 15116

TanA⇒ = = × =−

16. sin sin= ⇒ =ADB AD C BC

2 2 sin=−abc C B

b c

2 2

sin sin sinsin sin

=−A B BB C

( )1 sin= −B C 090⇒ − =B C 0 0 090 23 113⇒ = + =B

A

B CD

17. 02 cot30 31

= = ⇒ = =BDDE BD EC

2 2 3= +BC

( )23 3 4 4 2 3 6 4 34 4

Δ = = × + = +BC

A

B CD E

F G030

18.sol:. 2 2 22 2cos2+ −= ⇒ =b a b c bC

a ab a

2 2 23⇒ − =a c b

2 2 2 2cos cos .2+ −= b c a bA Cbc a

( ) ( )

2 22 2

2 2233

⎛ ⎞− + −⎜ ⎟ −⎝ ⎠= =

a c c a c aac ac

A

B CD

b

2a

090

2a

19.sol:. 0 04 180 30+ + + ⇒ =x x x x angles are 0 0 0120 ,30 ,30 , , .= A B C

3sin 32

sin sin sin 3 3 212

= = =+ + + + ++

a Aa b c A B C

20.sol:. 2sin sinα

=

aADB

sinsinα

= a BAD

sin2sin( ) sin 2sin( )α α

= ⇒ =− −

aAD a CAD

A C A

A

B CD2

a2

a

α α−A

sin sin sin( ) .sin .

sin 2sin( )β α αα α

= ⇒ − =−

a a C CAA b

21.. Altitudes are 1 1,AA BB and 1.CC

1 11 1 1( ') ( ) ( )2 2 2AA a BB b CC cΔ = = =

1 1 12 2 2, ,AA BB CCa b cΔ Δ Δ⇒ = = =

sin ,sinA B and sinC are in A.P. ,a b⇒ and c are in A.P.

1 1,a b

⇒ and 1c

are in H.P.

2 2,a bΔ Δ⇒ and 2

cΔ are in H.P.

1 1,AA BB⇒ and 1CC are in H.P.

A

B C1A

1B1C

22.. ( )2 cos 2bc Af

b c=

+ substituting these values

( )2 cos 2=+ab Ch

a b

( )2 cos 2=+ac Bg

a c

23. cot cot sin( )sincot sin sin cosA C A B

B A C B+ +=

2sin

sin sin cosB

A C B=

2 2

2

44 cos

R bR ac B

=

2 2

2 2 2

2 22 cos

b bac B a c b

= =+ −

2

2 22 2

2002 2001bb b

= =−

24. 0 1 2ΔA A A

( )20 211 1 2.1.1.2

A A ⎛ ⎞= + − −⎜ ⎟⎝ ⎠

( )20 2 0 23 3= =A A A A

1A

2A

3A

4A

5A

0A

11 1 1

1

Similarly 0 4 3=A A

( ) ( )( )Req 1 3 3 3= =

25. ( )( )a b c b c a kbc+ + + − = 2 2( )b c a kbc⇒ + − = 2 2 2 ( 2)b c a k bc⇒ + − = − 2 cos ( 2)bc A k bc⇒ = −

2cos2

kA −⇒ =

Now, A being the angle of a triangle, 1 cos 1 2 2 2A k− < < ⇒ − < − < 0 4k⇒ < <

26 2 2 2 2 2 21 12 2 2 23 3

GA b c a GB a c b= + − = + −

2 2 21 2 23

GC a b c= + −

27.

21 1 4 42 2ab p p ab pΔ = = ⇒ =

Also, 2 2 2 216a b c p+ = = 2 2 2 2( ) 2 8a b a b ab p∴ − = + − = Also, 2 2 2 2( ) 2 24a b a b ab p+ = + + =

1tan cot 12 2 3

A B a b Ca b

− −= =+

0 030 602

A B A B−⇒ = ⇒ − =

28. ( ) ( ) ( )10 10 10 10 10 10 10 101 3 5 1 9 3 7 5

2! 2 1 2 2 1 11!9! 3!7! 5!5! 10! 10! 10! 10!

C C C C C C C C+ + = + + = + + + +

9

10 11 2 8(2 )10! 10! (2 )!

a

b−= = = (given)

3 9,2 10 3, 5a b a b⇒ = = ⇒ = = Also, , ,a b c are in A.P. 7c⇒ =

2 2 2 25 49 9 65 13cos2 2 5 7 2 5 7 14

b c aAbc

+ − + −= = = =× × × ×

2 2 2 9 49 25 11cos2 2 3 7 14

a c bBac

+ − + −= = =× ×

13 11 24 12cos cos14 14 7

A B +⇒ + = = =

29. ( )cos cos 2 1 cos+ = −A C B

( ) ( )12sin cos 4sin2 22−−⎛ ⎞ =⎜ ⎟⎝ ⎠

A CB B

( )cos

2 2sin 2

−⎛ ⎞⎜ ⎟⎝ ⎠

=

A C

B 2 , ,+ = ⇒a c a b c

b are in AP

30. Internal angles of regular polygon of side n is

0

0 360180⎛ ⎞

−⎜ ⎟⎝ ⎠n 0108α∴ =

0120β = substitute values 0144γ = 0150δ =

Re 1∴ =q