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Solutions of Triangles

1. ELEMENTS OFATRIANGLEIn a triangle ABC the angles are denoted by capital letters A, B and C and the length of the sidesopposite to these angles are denoted by small letters a, b and c. Semi perimeter of the triangle is

given by s = a b c

2

and its area is denoted by .

2. SINE RULE

In a triangle ABC,a b c

sin A sin B sin C = 2R (where R is circum radius)

Note : Area of triangle=2

1bc sinA =

2

1ac sinB =

2

1ab sinC.

Drill Exercise - 1

1. In any triangle ABC, prove thatCsinBsin

)CBsin(a2

+AsinCsin

)ACsin(b2

+BsinAsin

)BAsin(c2

= 0

2. If in a ABC,Csin

Asin = )CBsin(

)BAsin(

, prove that a2, b2, c2 are in A. P..

3. ABCD is a trapezium such that AB and CD are parallel and CB is perpendicular to them. IfADB = 60º, BC = 4 and CD = 3, then find the length of side AB.

4. If the sides of a triangle are in arithmetic progression, and if its greatest angle exceeds the least angle

by, show that the sides are in the ration 1 – x : 1 : 1 + x, where x =

cos7

cos1.

5. Through the angular points of a triangle are drawn straight lines which make the same angle with theopposite sides of the triangle; prove that area of the triangle formed by them is to the area of theoriginal triangle as 4 cos2 : 1.

3. COSINE RULE

In a triangle ABC,

(i) cos A =bc2

acb 222 (ii) cos B =

ca2

bac 222

Theory Notes - Solutions of Triangles

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(iii) cos C =ab2

cba 222

Drill Exercise - 2

1. In any ABC, prove thata

Acos +

b

Bcos +

c

Ccos =

abc2

cba 222 .

2. Let ABC be a triangle such that 2b = (m + 1)a and cos A =m

)3m)(1m(

2

1 , where m(1, 3).

Prove that there are two values of the third side one of which is m times the other.

3. In a triangle ABC,C = 60º, then prove thatca

1

+

cb

1

=

cba

3

4. If in a triangle ABC,Bcos2Acos

Ccos2Acos

=Csin

Bsin prove that the triangle is either isosceles or right

angled.

5. A ring, 10 cm, in diameter, is suspended from a point 12 cm, above its centre by 6 equal stringsattached to its circumference at equal intervals. Find the cosine of the angle between consecutivestrings.

Illustration 1:Find the angles of the triangle whose sides are 3 + 3 , 2 3 and 6 .

Solution:

Let a = 3 + 3 , b = 2 3 , c = 6 cos A =

212

3639612

bc2

acb 222

=22

31

212

366

= cos 1050 A = 1050

Applying Sine formula :Bsin

b

Asin

a , we get

sin B = 0b 2 3sinA sin 105

a 3 3

=1

2 B = 450

A = 1050, B = 450, C = 300

4. PROJECTION FORMULAE(i) a = b cos C + c cos B (ii) b = c cos A + a cos C (iii) c = a cos B + b cos C

Illustration 2:If A = 450, B = 750, prove that a + c 2 = 2b.

Solution:


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As A = 450, B = 750 we have C = 600

2b = 2 (a cos C + c cos A) = 2(a cos 600 + c cos 450) = a + c 2 = L.H.S.

5. NAPIER’S ANALOGY (TANGENT RULE)

(i)2

Acot

cb

cb

2

CBtan

(ii) tan2

Bcot

ac

ac

2

AC

(iii) tan2

Ccot

ba

ba

2

BA

6. HALF ANGLE FORMULAE

(a) (i) sin

bc

csbs

2

A (ii) sin

s c s aB

2 ca

(iii) sin

ab

bsas

2

C

(b) (i) cos2

A =

bc

ass (ii) cos

2

B=

ca

bss

(iii) cos s s cC

2 ab

(c) (i) tan2

A =

s b s c

s s a

(ii) tan

2

B= bss

ascs

(iii) tan css

bsas

2

C

Drill Exercise - 3

1. In a ABC, if a = 13, b = 14, and c = 15, find the followings

(i) (ii) sin2

A(iii) cos

2

A(iv) tan

2

A

2. The sides of a triangle are x2 + x + 1, 2x + 1 and x2 –1; prove that the greatest angle is 120º.

3. In anyABC, prove thatCsin

Bsin =

Ccosab

Bcosac

4. Ifare the lengths of the altitudes of aABC, prove that

2

1

+ 2

1

+ 2

1

= CcotBcotAcot

where is the area of ABC.


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5. In a ABC, prove that cot2

A + cot

2

B + cot

2

C =

acb

cbacot

2

A.

7. AREA OF TRIANGLE

Area of triangle = s s a s b s c( )( )( )

7.1 (i) sin A = 2 2Δs s a s b s c

bc bc (ii) sinB= 2 2Δ

s s a s b s cca ca

(iii) sin C = 2 2Δs s a s b s c

ab ab

8. m-n THEOREMLet D be a point on the side BC of a ABC such that

BD : DC = m : n and ADC = , BAD = and

DAC = . Then

(i) (m + n) cot = m cot – n cot (ii) (m + n) cot = n cot B – m cot C

9. CENTROID AND MEDIANS OF A TRIANGLEThe line joining any vertex of a triangle to the mid point of the opposite side of the triangle is called themedian of the triangle. The three medians of a triangle are concurrent and the point of concurrency ofthe medians of any triangle is called the centroid of the triangle. The centroid divides the medianin the ratio 2 : 1.

Illustration 3:Find the lengths of the medians and the angles made by the medians with the sides of a triangle ABC.

Solution :

AD2 = AC2 + CD2 – 2AC. CD cos C = b2 +4

a 2

– ab cosC,

and c2 = b2 + a2 – 2ab cos C.

Hence 2AD2 – c2 = b2 –2

a 2

,

so that AD = 222 ac2b22

1 = Acosbc2cb

2

1 22

Similarly,

D

A

F

B

E

C

G

BE = 222 ba2c2

2

1 , and CF = 222 cb2a2

2

1


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If BAD and ,CAD we have

sin DC a

sinC AD 2AD

sin =2 2 2

a sin C a sin C

2AD 2b 2c a

Similarly

sin =222 ac2b2

Bsina

Drill Exercise - 4

1. If the medians of aABC make angleswith each other, prove thatcot + cot + cot + cotA + cotB + cotC = 0

2. In an isosceles right angled triangle a straight line is drawn from the middle point of one of the equalsides to the opposite angle. Show that it divides the angle into parts whose cotangents are 2 and 3.

3. D, E and F are the middle points of the sides of the triangle ABC; prove that the centroid of thetriangle DEF is the same as that of ABC, and that its orthocentre is the circumcentre of ABC.

4. Prove that the median through A divides it into angles whose cotangents are 2 cot A + cot C and 2

cotA + cotB, and makes with the base an angle whose cotangent is2

1 (cot C ~ cot B)

5. Prove that the distance between the middle point of BC and the foot of the perpendicular from A is

a2

c~b 22

.

10. CIRCUM CIRCLEThe circle which passes through the angular points of a ABC, is called its circumcircle. The centreof this circle i.e., the point of concurrency of the perpendicular bisectors of the sides of the ABC,is called the circumcenter.

D

A

B

F

C

EO

A A

a/2 a/2


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Radius of the circumcircle is given by the following formulae

R =a b c abc

2sin A 2sin B 2sin C 4

Illustration 4:

If in a ABC, O is the circumcenter and R is the circumradius and R1, R

2 and R

3 are the circumradii

of the triangles OBC, OCA and OAB respectively, then prove that 3321 R

abc

R

c

R

b

R

a .

Solution:Clearly, in the OBC, BOC = 2A, OB = OC = R, BC = a.

2R1 =

A2sin

a{using sine rule in BOC)

Similarly, 2R2 =

C2sin

cR2and

B2sin

b3

A

O

B C

321 R

c

R

b

R

a = 2(sin2A + sin2B + sin 2C)

= 2.4 sin A sin B sin C,

= 8 3R

abc

R2

c.

R2

b.

R2

a .

Illustration 5:If the distances of the sides of a ABC from its circumcenter be x, y and z respectively, then prove

that xyz4

abc

z

c

y

b

x

a .

Solution :

Let M be the circumcenter. MD BC. So BD = DC =2

aand BMD = A.A.

In BDM,MD

BD= tan A or

x2a

= tan A, i.e.,x2

a= tan A,A,

Similarly, y2

b= tan B,

z2

c= tan C

y

A

M

B D C

z

A x

EF

tan A + tan B + tan C =a b c

2x 2y 2z

and tan A. tan B. tan C = z2

c.

y2

b.

x2

a

But in a triangle ABC, tan A + tan B + tan C = tan A. tan B. tan C


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xyz4

abc

z

c

y

b

x

a

11. ORTHOCENTER AND PEDAL TRIANGLE OF A TRIANGLE.In a triangle the altitudes drawn from the three vertices to the opposite sides are concurrent and thepoint of cuncurrency of the altitudes of the triangle is called the orthocenter of the triangle. Thetriangle formed by joining the feet of these perpendiculars is called the pedal triangle i.e.

DEF is the pedal triangle of ABC.

D

A

F

B

E

C

P

90 – C0

Illustration 6:Find the distance of the orthocenter from the sides and angular points of a triangle ABC.

Solution : PD = DB tan PBD = DB tan (900 – C)

= AB cos B cot C =Csin

ccos B cos C= 2R cos B cos C

SimilarlyPE = 2R cosA cosC and PF = 2R cosA cosB

Again AP = AE sec DAC = c cos A cosec C

=Csin

c cos A = 2 R cos AA

D

A

F

B

E

C

P

90 – Cso, BP = 2R cos B and CP = 2R cos C

Illustration 7:Find the distance between the circumcenter and the orthocenter of a triangle ABC

Solution :Let O be the circumcenter and P be the orthocenter of the ABCIf OF be perpendicular to AB, we have

0OAF 90 C

Also C90PAL 0

A

B CKD

F

LOP

OAP C – B

Also OA = R and PA = 2R cosA

OP2 = OA2 + PA2 – 2OA. PA cos OAP


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= R2 + 4R2 cos2 A – 4R2 cosA cos (C – B)= R2 – 4R2 cosA [cos(B + C) + cos (C – B)] = R2 – 8R2 cos A cos B cosC

OP = R CcosBcosAcos81

12. BISECTORS OF THE ANGLESIf AD bisects the angle A and divide the base into portions x and y, we have, by Geometry,

b

c

AC

AB

y

x cb

a

cb

yx

b

y

c

x

x =ac

b c and y =

ab

b cAlso let be the length of AD

we have ABD + ACD = ABC

,Asinbc2

1

2

Asinb

2

1

2

Asinc

2

1

y

A

B CDx

i.e.,2

Acos

cb

bc2

2A

sin

Asin

cb

bc

Drill Exercise - 5

1. Show that the distances of the orthocentre from the sides of a triangle ABC are

AcosCcosR2&AcosCcosR2,CcosBcosR2 .

2. In anyABC, prove that a cosA + b cosB + c cosC = 4R sinA sinB sinC.

3. If p1, p

2 p

3 are respectively the perpendiculars from the vertices of a triangle to the opposite sides,

prove that p1p

2p

3 = 3

222

R8

cba.

4. In aABC, if 8R2 = a2 + b2 + c2, show that the triangle is right angled.

5. AD, BE and CF are the perpendiculars from the angular points of a triangle ABC upon the oppositesides : prove that the diameter of the circumcircles of the triangle AEF, BDF, and CDE are respectivelya cot A , b cot B and c cot C and that the perimeters of the triangles DEF and ABC are in the ratio r : R.

13. INCIRCLEThe circle which can be inscribed within the triangle so as to touch each of the sides of the triangle iscalled its incircle. The centre of this circle i.e., the point of concurrency of angle bisectors of thetriangle is called the incentre of the ABC.

D

A

B

F

C

E

I

C/2B/2

r

r

r90 – B/20


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Radius of the Incircle is given by the following formulae

r =s

= (s – a) tan

2

A= (s – b) tan

2

B= (s – c) tan

2

C= 4R sin

2

Asin

2

Bsin

2

C.

Illustration 8:Find the distance between the circumcenter and the incentre.

Solution :Let O be the circumcenter and I be the incentre of ABC.Let OF be perpendicular to AB and IE be perpendicular to AC.

.C90OAF 0

OAFIAFOAI

O

A

F

B

E

I

C

= 2

BC

2

CBAC

2

AC90

2

A 0

Also,

AI =2

Csin

2

BsinR4

2A

sin

r

2A

sin

IE

OAIcosAI.OA2AIOAOI 222

= R2 + 16R2 sin2

2

Bsin2

2

C– 8R2 sin

2

Bsin

2

Ccos

2

BC

2

Csin

2

Bsin161

R

OI 222

2

– 8sin

2

Csin

2

Bsin

2

Ccos

2

Bcos

2

Csin

2

B

= 1 – 8 sin

2

Csin

2

Bsin

2

Ccos

2

Bcos

2

Csin

2

B

= 1 – 8 sin2

Asin

2

Csin

2

B. . . (i)

2

Asin

2

Csin

2

Bsin81ROI .

Illustration 9:If the distances of the vertices of a triangle ABC from the points of contacts of the incircle with sides

be , and , then prove that

2r

Solution:Let the incircle touches the side AB at P, where AP = . Let I be the incentre.

From the right-angled IPA,

2

Acotr;

2

Atan

r

similarly, = r cot

2

Ccotrand

2

B


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In ABC, we have the identity

cot2

Ccot

2

Bcot

2

Acot

2

Ccot

2

Bcot

2

A

r.

r.

rrrr

A

P

B

E

C

I

or 3r

1

r

1

r2 =

.

Illustration 10:Show that the line joining the in-centre to the circumcenter of a triangle ABC is inclined to the side BC

at an angle 1 cos B cos C 1tan

sin C sin B

.

Solution:Let I be the in-centre of O be the circumcenter of the triangle ABC. Let OL be parallel to BC. Let

IOL . IM = r OC = R, NOC A

IL IM LM IM ON r R cos A

tanBOL BM BN BM NC r cot R sin A2

A B C4R sin sin sin R cos A

2 2 2A B C B

4R sin sin sin .cot R sin A2 2 2 2

B

O LI

C

A

NM

cos A cos B cos C 1 cos A cos B cos C 1

sin A sin C sin B sin A sin C sin B

1 cos B cos C 1

tansin C sin B

.

14. THE DISTANCES BETWEEN THE SPECIAL POINTS

(i) The distance between circumcenter and orthocenter is = R. CcosBcosAcos81

(ii) The distance between circumcenter and incentre is = Rr2R 2 .


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(iii) The distance between incentre and orthocenter is CcosBcosAcosR4r2 22 .

Drill Exercise - 6

1. In a triangle ABC, the incircle touches the sides BC, CA and AB at D, E, F respectively. If radius ofincircle is 4 units and BD, CE and AF be consecutive natural numbers, find the sides of the triangleABC.

2. Show that the distances of the incentre from vertices A,B & C are

2

Bsin

2

AsinR4,

2

Asin

2

CsinR4,

2

Csin

2

BsinR4 respectively..

3. In aABC, prove that ratio of the area of the incircle to that of the triangle is : cot2

Acot

2

Bcot

2

C.

4. Prove that )rR(2CcotcBcotbAcota .

5. If the incentre & the circumcentre of a triangle are equidistant from the side BC,prove that 1CcosBcos .

15. ESCRIBED CIRCLESThe circle which touches the side BC and the two sides AB and AC produced is called the escribedcircle opposite the angle A. Its centre and radius will be denoted by I

1 and r

1 respectively.

Radii of the excircles are given by the following formulae

(i) r1 =

2

Ccos

2

Bcos

2

AsinR4

2

Atans

as

(ii) r2 =

2

Ccos

2

Bsin

2

AcosR4

2

Btans

bs

B

A

M

E1

CD1

F1

I1L

(iii) r3 =

2

Csin

2

Bcos

2

AcosR4

2

Ctans

cs

.

16. EXCENTRAL TRIANGLEThe triangle formed by joining the three excentres I

1, I

2 and I

3 of

ABC is called the excentral or excentric triangle. Not that "

(i) Incentre I ofABC is the orthocenter of the excentralI

1I

2I

3.

(ii) ABC is the pedal triangle of theI1I

2I

3.

(iii) The sides of the excentral triangle are 4 R cos2

A, 4 R cos

2

B and


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4 R cos2

C and its angles are

2

–2

A,

2

–2

B and

2

–2

C.

(iv) II1 = 4 R sin

2

A : II

2 = 4 R sin

2

B : II

3 = 4 R sin

2

C

Illustration 11 :If the exradii r

1, r

2 and r

3 of a ABC are in HP, show that its sides a, b and c are in A.P..

Solution:

We know that r1 =

csr,

bsr,

as 32

r1, r

2, r

3 are in HP

cs

,bs

,as

are in AP

s – a, s – b, s – c are in AP a, b, c are in AP

Drill Exercise - 7

1. Prove the following :

Asin2

Asinr

Acos42A

tan)cb(

Asin4

Acosa

r

1

r

1

r

1

r

1

r

1

r

1

4

sr

r4

)rr)(rr)(rr(

)BAsin(c2

ba

2

Csec

2

)BA(sec)ba(

4

1

Ccos4

rrrr)rrrr(

4

1

rs4

abcR

321

22

2321

22

321321

2. Prove the following :

3/13/5

3/2222

2222

1321

)C2sinB2sinA2(sin2

)abc(

)CcotBcotA(cot4

cba

)ba()BAsin(2

BsinAsin)B2sincC2sinb(

4

1)CsinBsinA(sinRr

)cs)(bs(bc2

Acos

2

Atan)as(s

2

Acotrrrrrr

3. Show that the radii of the three escribed circles of a triangle are the roots of the equation,x3 – x2 (4R + r) + xs2 – rs2 = 0

4. If R1, R

2 and R

3 be the diameter of the excircles of a ABC (opposite to the vertices A, B and C

respectively), then prove thatcba

RRR

R

c

R

b

R

a 321

321

.


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5. Prove that )cba(R16rrrr 222223

22

21

2 .

17. SOLUTION OF TRIANGLESWhen any three of the six elements (except all the three angles) of a triangle are given, the triangle isknown completely. This process is called the solution of triangles.

(i) If the sides a, b and c are given, then cos A =2 2 2b c a

2bc

. B and C can be obtained in the

similar way.

(ii) If two sides b and c and the included angle A are given, then using

tanB C b c A

cot2 b c 2

, we getB C

2

.

AlsoB C

2

= 900 –

A

2, so that B and C can be evaluated.

The third side is given by a =bsin A

sin B.

(iii) If two sides b and c and the angle B (opposite to side b) are given, then sin C =c

bsin B, A =

1800 – (B + C) and a =bsin A

sin Bgive the remaining elements. If b < c sin B, there is no triangle

possible (fig 1). If b = c sin B and B is an acute angle, then there is only one triangle possible(fig 2). If c sin B < b < c and B is an acute angle, then there are two values of angle C (fig 3).If c < b and B is an acute angle, then there is only one triangle (fig 4).

A

B D

cb c sinB

A

B D

cb c sinB

(Fig 2)(Fig 1)


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A

BD

c

b c sinB

C1C2

c sinB

C1

b

B

b

c

C2

(Fig 3)(Fig 4)

A

b

This case is, sometimes, called an ambiguous case.

Illustration 12:In any triangle ABC, the sides are 6 cm, 10 cm and 14 cm. Show that the triangle is obtuse-angledwith the obtuse angle equal to 1200.

Solution:Let a = 14, b = 10, c = 6

The largest angle is opposite the largest side.

cos A =2 2 2

0b c a 100 36 196 1A 120

2bc 120 2

Illustration 13:If in a triangle ABC, a = (1 + 3 ) cm, b = 2 cm and C = 600, then find the other two angles and the

third sideSolution:

cos C =2 2 2a b c

2ab

.

2 21 3 4 c1

2 2 1 3 .2

c2 = 6 c = 6

Also,sin A sin B sin C

a b c

3sin A sin B 2

21 3 6

sin B =

1

2 B = 450

A = 1800 – (450 + 600) = 750

Illustration 14 :Given the base of a triangle, the opposite angle A, and the product k2 of the other two sides, show

that it is not possible for a to be less than 2k sinA

2.

Solution:Given b.c = k2


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Now cosA =2 2 2b c a

2bc

or 2k2 cosA = b2 +

222k

ab

A

B Ca

c bor b4 – (a2 + 2k2 cosA). b2 + k4 = 0Since b2 is real, (a2 + 2k2) (a2 + 2k2 cosA – 2k2) 0

2 2 2 2 2 2A A

a 2k .2cos a 2k .2sin 02 2

2 2 2 2 2 2A A

a 4k cos a 4k sin 02 2

2 2 2 A

a 4k sin 02

[since a2 + 4k2 cos2A

2 is always positive]

A A

a 2k sin a 2k sin 02 2

A A

a 2k sin or a 2k sin2 2

(since 2ksin(A/2) is real)

But a must be positive. a – 2k sinA

2 is rejected

HenceA

a 2 k sin2

.

DRILL EXERCISE - 8

1. A right triangle has c = 64,A = 61º andC = 90º. Find the remaining parts.

2. Solve the triangle in which b = 100, c = 100 2 and B = 30º.

3. In aABC if a, b and A are given, then prove that two triangles are formed such that the sum of their

areas is2

1b2 sin2A.

4. The lengths of two sides of a triangle are 12 cm and 12 2 cm respectively, and the angle opposite theshorter side is 30º; prove that there are two triangle satisfying these conditions, find their angles and

show that their areas are in the ratio 13 : 13

5. In aABC, if a, b and A are given, then there are two triangles with third sides c1 and c

2 such that

c1 – c

2 = 2 Asinba 222

18. INSCRIBED & CIRCUMSCRIBED POLYGONS


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(Important Formulae)

I. Area of Polygone of n sides inscribed in a circle of radius r =n

2sinnr

2

1 2

II. Area of Polygone of n sides inscribing a circle of radius r =n

tannr2

1 2

III. Side of Inscribed polygone =n

sinr2

.

IV. Side of Circumscribed polygone =n

tanr2

.

Illustration 15 :Find the radii of the inscribed and the circumscribed circles of a regular polygon of n sides with eachside a and also find the area of the regular polygon.

Solution:Let AB, BC and CD be three successive sides of the polygon and O be the centre of both the incircleand the circumcircle of the polygon

2BOC

n

1 2

BOL2 n n

If a be a side of the polygon, we have

a = BC = 2BL = 2RsinBOL = 2Rsinn

ar cot

2 n

.

Now the area of the regular polygon = n times the area of the

OBC21 1 a na

n OL.BC n . cot .a cot2 2 2 n 4 n

.

A O D

R R

B L C

DRILL EXERCISE - 9

1. If a, b, c, d are the sides of a quadrilateral described about a circle then prove that

2

Csinbc

2

Asinad 22 .

2. Two regular polygons of n & 2n sides have the same perimeter, show that their areas are in the ratio

ncos1:

ncos2

.

3. If 2a be the side of a regular polygon of n sides, R & r be the circumradius & inradius, prove that

n2cotarR

.

4. With reference to a given circle, 11 B&A are the areas of the inscribed and circumscribed regular

polygons of n sides, 22 B&A are corresponding quantities for regular polygons of 2n sides. Prove

that 2A is a geometric mean between 11 B&A and 2B is a hormonic mean between 12 B&A .


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Answer Key

Drill Exercise - 1

3.334

325

Drill Exercise - 2

5.338

313

Drill Exercise - 3

1. (i) 84 (ii)5

1(iii)

5

2(iv)

2

1

Drill Exercise - 6

1. a = 13, b = 15, c = 14

Drill Exercise - 8

1. A = 29º , a = 64 cos 29º, b = 64 sin 29º

2. a = 50 )26( , A = 15º , C = 135º

4. 45º and 105º; 135º and 15º

SOLVED OBJECTIVE EXAMPLESExample 1:

If D is the mid point of the side BC of a triangle ABC and AD is perpendicular to AC, then(a) 3b2 = a2 – c2 (b) 3a2 = b2 – 3c2

(c) b2 = a2 – c2 (d) a2 + b2 = 5c2

Solution:From the right angled CAD, we have

a / 2

A

c

B

b

CD

900

a / 2

A – / 2cos C =

ab2

cba

a

b2

2/a

b 222

a2 + b2 – c2 = 4b2 a2 – c2 = 3b2.


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Example 2:There exists a triangle ABC satisfying

(a) tanA + tanB + tanC = 0 (b)7

Csin

3

Bsin

2

Asin

(c) (a + b)2 = c2 + ab (d) none of these

Solution:(a) In a triangle ABC, we know that tan A + tan B + tan C = tan A tan B tan C. Since none oftan A, tan B, tan C can be zero, (a) is not possibleIf (sin A)/2 = (sin B)/3 = (sin C)/7, then by the laws of sines

a b c

2 3 7

which is not possible, as the sum of two sides of a triangle is greater than the third side

If (a + b)2 = c2 + ab, then2 2 2a b c

2ab

=

1

2 = cos C =

3

, which is possible

Hence (c) is the correct answer.

Example 3:If the tangents of the angles A and B of a triangle ABC satisfy the equation abx2 – c2x + ab = 0, then(a) tan A = a/b (b) tan B = b/a(c) cos C = 0 (d) sin2 A + sin2 B + sin2 C = 2

Solution:From the given equation, we gettan A + tan B = c2 / ab and tan A tan B = 1.

Since tan (A + B) =BtanAtan1

BtanAtan

We get A + B =2

and hence C =

2

.

a

A

c

B

b

C

Therefore, triangle ABC is right angled at C. Hence,tan A = a/b, tan B = b/a, cos C = 0, sin A = a/c, sin B = b/c and sin C = 1, so that

sin2 A + sin2 B + sin2 C = 2111c

ba1

c

b

c

a2

22

2

2

2

2

[ a2 + b2 = c2]

Hence, all options are correct.

Example 4:If in a triangle ABC sin A , sin B and sin C are in A.P., then the altitudes are in(a) A.P. (b) H.P.(c) G.P. (d) none of these

Solution:If p

1, p

2, p

3, are altitude from A, B, C respectively,


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then =2

1ap

1 =

2

1bp

2 =

2

1cp

3 p1 =

a

2, p

2 =

b

2, p

3 =

c

2

By the law of sines

a b c

sin A sin B sin C = k (say)

p1 =

2Δk sin A

, p2 =

2Δk sin B

, p3 =

Csink

2

Now, sin A, sin B, sin C are in A.P. p1, p

2, p

3 are in H.P.

Example 5:In a triangle ABC, medians AD and CE are drawn. If AD = 5, DAC = / 8 andACE = /4, then the area of the triangle ABC is equal to

(a)25

9(b)

25

3

(c)25

18(d)

10

3

Solution:Let O be the point of intersection of the medians of triangle ABC. Then the area of ABC is three

times that of AOC. Now, in AOC, AO =3

2AD =

3

10. Therefore, applying the sine rule to

AOC, we get

4/sin

8/sin.

3

10OC

4/sin

AO

8/sin

OC

area of AOC =2

1. AO.OC. sin AOC

=

82sin.

4/sin

8/sin.

3

10.

3

10.

2

1

D

B

O

E

A C/8 /4

= 9

25

18

50

4/sin

8/cos8/sin.

9

50

area of ABC = 3.3

25

9

25

Example 6:In a triangle ABC, if tan (A/2) = 5/6 and tan (B/2) = 20/37, the sides a, b and c are in(a) A.P. (b) G.P.(c) H.P (d) none of these

Solution:


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We have tan2

C = tan

2

BA900

= cot2

BA =

2/Bcot2/Acot

12/Bcot2/Acot

=

6 37. 1 222 100 122 25 20

6 37 120 185 305 55 20

Also tan2

Atan

2

C =

css

bsas

ass

csbs

s

bs

5

2.

6

5 sbs3 b3s2

a + b + c = 3b a + c = 2bWhich shows that a, b and c are in A.P.

Example 7:If in a triangle ABC, a = 5, b = 4 and cos (A – B) = 31/32, then the third side c is equal to(a) 6 (b) 8(c) 4 (d) none of these

Solution:

cos (A – B) =2

BAtan1

2BA

tan1

32

31

2BA

tan1

2BA

tan1

2

2

2

2

63 tan2

63

1

2

BAtan1

2

BA

Now

tan 2

Ccot

45

45

63

1

2

Ccot

ba

ba

2

BA

tan2

C =

9

63

Also, cos C = 8

1

144

18

81/631

81/631

2/Ctan1

2/Ctan12

2

c2 = a2 + b2 – 2ab cos C = 25 + 16 – 2.5.4. (1/8) = 36 c = 6Hence (a) is the correct answer.

Example 8:In a triangle ABC, if r

1 = 2r

2 = 3r

3, then a : b is equal to

(a)5

4(b)

4

5


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(c)7

4(d)

4

7

Solution:From the given relation, we have

s tan2

A = 2s tan

2

B = 3s tan

2

C

k2

)2/ctan(

3

)2/Btan(

6

)2/Atan( (say)

Also, since A/2 + B/2 + C/2 = 900, we get

tan 12

Atan

2

Ctan

2

Ctan

2

Btan

2

Btan

2

A

6k. 3k + 3k. 2k + 2k. 6k = 1 36k2 = 1 k = 1/6

sin A = 2 2

2 tan A / 2 12k1

1 tan A / 2 1 36k

andsin B = 5

4

k91

k6

2/Btan1

2/Btan222

Hence, by the law of sines, sin A/a = sin B/b, we have

4

5

Bsin

Asin

b

a

a : b = 5 : 4

Example 9:Let AD be a median of the ABC. If AE and AF are medians of the triangles ABD and ADC

respectively and AD = m1, AE = m

2, AF = m

3, then

8

a2

is equal to

(a) 21

23

22 m2mm (b) 2

322

21 m2mm

(c) 21

23

22 m2mm (d) none of these

Solution:

In ABC, AD2 = m12 =

4

a

2

bc 222

A

B CFDE

In ABD, AE2 = m2

2 =

2

2 2a

AD c 22 4


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AF2 = m32 =

2

2 2a

AD b 22 4

m22 + m

32 = AD2 +

8

am2

8

a

4

amm

8

a

2

cb 221

2221

21

222

22 2 2

2 3 1

am m 2m

8

Example 10:If I is the incentre of a triangle ABC, then the ratio IA : IB : IC is equal to

(a) cosec2

A : cosec

2

B : cosec

2

C(b) sin

2

A : sin

2

B : sin

2

C

(c) sec2

A : sec

2

B : sec

2

C(d) none of these

Solution:Here BD : DC = c : bBut BD + DC = a;

BD = a.cb

c

In ABD, Bsin

AD

2A

sin

BD

A

B CD

I

AD =2

Aeccos

cb

2

2A

sin

Bsin.

cb

ca

Also, a

cb

cb/ca

c

BD

AB

ID

AI

AI =2

Aeccos

sAD.

cba

cb

Similarly BI =B C

cos ec , CI cos ecs 2 s 2

A B CIA : IB : IC cos ec : cos ec : cos ec

2 2 2

Example 11:

In a ABC, the value ofcba

CcoscBcosbAcosa

is equal to

(a)r

R(b)

r2

R


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(c)R

r(d)

R

r2

Solution:a cosA bcosB ccosC 2R sin AcosA 2R sin BcosB 2R sin CcosC

a b c 2s

= C2sinB2sinA2sin.s2

R = CsinBsinAsin4.

s2

R = 3 2

4R abc abc.

2s 8R 4sR

But R =4

abc, r =

s

So, the value =R

r

R.r

.4

R4

2

Example 12:The area of a circle is A

1and the area of a regular pentagon inscribed in the circle is A

2. Then

A1 : A

2 is

(a)10

cos5

(b)

10sec

5

2

(c)10

eccos5

2 (d) none of these

Solution:

In the OAB, OA = OB = r and AOB =5

3600

= 720

ar (AOB) =2

1. r . r. sin 720 =

2

1r2 cos 180

O C

D

E

A B

r

A1 : A

2 =

10sec

5

2

18cosr5

r202

2

Example 13:In a triangle ABC a = 5, b = 4 and c = 3. ‘G’ is the centroid of the triangle. Circumradius of triangleGAB is equal to

(a) 132 (b) 1312

5

(c) 133

5(d) 13

2

3

Solution:


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AG =3

2A AA

1, BG =

3

2BB

1

AG =3

12 2 22b 2c a

A

BA1

B1

C

G

and BG =222 bc2a2

3

1

AG =22 c4b

3

1BG,a

3

1 as a2 = b2 + c2

AG = 133

23616

3

1BG,

3

5

Also, AB = c = 3 and GAB ABC1

23

If ‘R1’ be the circumradius of triangle GAB then

R1 =

2.4

1.3.13

3

2.

3

5

4

ABBGAG

GAB

=12

135 units.

Example 14:A variable triangle ABC is circumscribed about a fixed circle of unit radius. Side BC always touchesthe circle at D and has fixed direction. If B and C vary in such a way that (BD). (CD) = 2 then locusof vertex A will be a straight line(a) parallel to side BC (b) right angle to side BC(c) making an angle /6 with BC (d) making an angle sin–1 (2/3) with BC

Solution:BD = (s – b), CD = (s – c) (s – b)(s – c) = 2 s(s – a) (s – b) (s – c) = 2 s(s – a)

2 = 2 s(s – a) 1s

)as(2

s2

2

(radius of incircle of triangle ABC)

s

a = constant.

Now =2

1aH

a, where ‘H

a’ is the distance of ‘A’ from BC.

aaH1

s 2 s

= 1 H

a =

a

s2= constant

Locus of ‘A’ will be a straight line parallel to side BC.

Example 15:In the adjacent figure AB is the diameter of circle, centered at ‘O’. If COA = 600. AB = 2r,,AC = d and CD = , then


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A

B D

C

O(a) 3 r d (b) r 2 d 2 (c) r 3 d 3 (d) 2 r d

Solution:

AC = d, OA = OB = r , CD = BD = , COA =3

AC2 = OA2 + OC2 – 2AOOC. cos3

d2 = 2r2 – 2r2 .1

2 = r2

A

B D

C

O

Also, BOD = COD =2

3.2 3

BD

tan r 3 d 33 OB r

Hence the correct answer is (c)

SOLVED SUBJECTIVE EXAMPLES

Example 1:The lengths of sides of a triangle are three consecutive natural numbers and its largest angle is twicethe smallest one. Determine the sides of the triangle.

Solution:Let the lengths of the sides be n, n + 1, n + 2, where n N.

From the question, the largest angle opposite to the side n + 2 is 2 while the smallest angle opositeto the side n is .

Now cos =

2n1n2

5n6n

2n1n2

n2n1n 2222

= 2n2

5n

2n1n2

5n1n

and cos 2 =

2 22n n 1 n 2

2n n 1

= 1nn2

3n2n2

= n2

3n

1nn2

3n1n

But cos 2 = 2 cos2 – 1; so 12n2

5n2

n2

3n2

or

12n2

5n

n2

3n2

2

or (n – 3)(n+2)2 = n{(n + 5)2 – 2 (n + 2)2}

or (n – 3)(n2 + 4n + 4) = n (– n2 + 2n + 17)or n3 + n2 – 8n – 12 = – n3 + 2n2 + 17nor (n – 4)(2n2 + 7n + 3) = 0 n = 4 or 2n2 + 7n + 3 = 0.


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Roots of 2n2 + 7n + 3 = 0 are4

24497

i.e., –2

1and – 3 which are not natural numbers.

n = 4 and hence sides are 4, 5, 6.

Example 2:Consider the following statements concerning a ABC:(i) The sides a, b, c and the area are rational.

(ii) a, tan2

B, tan

2

Care rational.

(iii) a, sin A, sin B, sin C are rational.Prove that (i) (ii) (iii) (i)

Solution :Let (i) be true, i.e., a, b, c and be rational numbers.

Now, tan

bsas

2

Ctan,

ascs

2

Band s =

2

cba

Now, (i) a, b, c, , s are rational.

So tan2

Band tan

2

Care rational because sum, difference, product and quotient of nonzero rational

numbers are rational.Thus (i) (ii).

Let (ii) be true, i.e., a, tan2

B, tan

2

Cbe rational

Now, sin B =

2B

tan1

2B

tan2

2= rational, because tan

2

Bis rational.

sin C =

2C

tan1

2C

tan2

2 = rational, because tan

2

C is rational.

Now, tan2

B. tan

2

C=

bsas

.ascs

=

2s a s b s c s a a

1s s a s b s c s s

.

(ii) s is rational

b + c is rational, because a is rational.

ButCsin

c

Bsin

b

Asin

a rational

rational

CsinBsin

cb

Asin

a


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Asin

ais rational. But a is rational. So sinA is rational

Thus (ii) (iii)Let (iii) be true, i.e., a, sin A, sin B, sin C be rational.

Csin

c

Bsin

b

Asin

a

Asin

Bsinab = rational

and c =Asin

Csina= rational =

2

1bc sin A = rational.

Thus (iii) (i).

Example 3:

If in a triangle ABC, a = 6, b = 3 and cos (A – B) =5

4, find the area of the triangle.

Solution :

Here, cos (A – B) =5

4,

5

4

2BA

tan1

2BA

tan1

2

2

By componendo and dividendo,45

45

22

BAtan2 2

or tan2

9

1

2

BA

or tan3

1

2

BA

( A > B).

But tan2

Ccot

ba

ba

2

BA

2

Ccot

36

36

3

1

or cot2

C = 1; C =

2

.

The area of the triangle = 92

sin.3.6.2

1Csinab

2

1

sq. units.

Example 4:If p, q are perpendiculars from the angular points A and B of the ABC drawn to any line through thevertex C, then prove thata2b2 sin2 C = a2p2 + b2q2 – 2abpqcos C.

Solution :Let ACE = . Clearly, from the figure, we get

p

CsinBC

q,sin

AC

p


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Csin.cosCcos.sina

q,sin

b

p

Csin.cosCcosb

p

a

q or

22 2q p

cos C cos .sin Ca b

= Ccos1b

p1 2

2

2

or2 2 2 2

2 22 2 2 2

q p 2pq p pcos C cosC 1 1 cos C

a b ab b b

or CsinCcosab

pq2

b

p

a

q 22

2

2

2

or a2p2 + b2q2 – 2abpqcosC = a2 b2sin2C.

Example 5:

In a ABC, prove that cos A. cos C =

ca3

ac2 22 , where AD is the median through A and

AD AC.Solution:

From the ABC, cos A =bc2

acb 222 . . . (i)

From the CAD, cos C =a

b2

2/a

b

CD

AC . . . (ii)

From the ABD,0

BD AB

sin ADBsin(A 90 )

or C90sin

c

Acos

2/a0

or

a c

2cos A cos C

cos A= ,c

b

a

b2.

c2

a

c2

Ccosa

from (ii)

from (i),c

b

bc2

acb 222

or b2 + c2 – a2 = – 2b2

A

B CD

or c2 – a2 = – 3b2 . . . (iii)

Now, cosA. cosC =ca

acb

a

b2.

bc2

acb 222222

=

ca3

ac3ca

ca3

ac3b3 2222222

, from (iii)

=

ca3

ac2 22 .


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Example 6:Find the sides and angles of the pedal triangle.

Solution:Since the angle PDC and PEC are right angles, the points P, E, C and D lie on a circle,

PDE = PCE = 900 – AASimilarly P, D, B and F lie on a circle and therefore

PDF = PBF = 900 – A, Hence FDE = 1800 – 2ASimilarly DEF = 1800 – 2BEFD = 1800 – 2CAlso, from the triangle AEF we have

EF AE ABcos A ccos A ccos A

sin A sin AFE cos PFE cos PAE sin C

EF =c

sin A cos Asin C

= a cosA

D

A

F

B

E

C

P

90 – C

similarly DF = b cosB and DE = c cosC

Example 7:The base of a triangle is divided into three equal parts. If t

1, t

2 , t

3 be the tangents of the angles

subtended by these parts at the opposite vertex, prove that:

21 2 2 3 2

1 1 1 1 14 1

t t t t t

Solution:Let the points P and Q divide the side BC in three equal parts such that BP = PQ = QC = xAlso let,

BAP = , PAQ = , QAC =

and AQC = From question,

tan = t1, tan = t

2, tan = t

3.

B C

A

QP

Applying,m : n rule in triangle ABC, we get

(2x + x) cot = 2x cot ( + ) – x cot . . . (i)

from APC, we get

(x + x) cot = x cot – x cot . . . (ii)

dividing (i) by (ii), we get

2

3 =

cotcot

cotcot2or 3 cot – cot =

cotcot

1cot.cot4

or 3 cot2 – cot cot + 3 cot . cot – cot . cot = 4 cot . cot – 4

or 4 + 4 cot2= cot2 + cot . cot + cot . cot + cot .cot


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or 4(1 + cot2 ) = (cot + cot )(cot + cot )

or 4

322122 t

1

t

1

t

1

t

1

t

11

Hence the result.Example 8:

Perpendiculars are drawn from the angular points A, B and C of an acute angled ABC on the

oposite sides and produced to meet the circumscribing circle. If these produced parts be , and respectively, show that

cba = 2 (tan A + tanB + tan C).

Solution :Let AD be perpendicular from A on BC. When AD is produced, it meets the circumscribing circle atE.From question, DE = .Since, angle in the same segment are equal,

AEB = ACB = C and AEC = ABC = BFrom the right angled triangle BDE,

tan C =DE

BD. . . (i)

From the right angled triangle CDE,

tan B =CD

DE. . . (ii)

B

A

C

BC

E

D

Adding (i) and (ii) we get, tan B + tanC =a

Similarly tan C + tan A =b

and tan A + tan B =C

Hence,

cba = 2 (tan A + tan B + tan C)

Example 9:If x, y, z are the distance of the vertices of the ABC respectively from the orthocentre then prove

thata b c abc

x y z xyz .

Solution:Let H be the orthocentre. Then

BHC = 1800 – HBC – HCB = 1800 – (900 – C) – (900 – B)= B + C = – A.


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ar( BHC) =2

1BH. CH sin BHC

=2

1yz sin ( – A) =

2

1yzsinA. A

B C

H

x

zy

Similarly, ar( CHA) =2

1zx sin B

ar( AHB) =2

1xy sin C

ar( ABC) =2

1yz sin A +

2

1zx sin B +

2

1xy sin C

=2

1xyz

z

Csin

y

Bsin

x

Asin

=

z

c

y

b

x

a

R2

1.xyz

2

1. . . (i)

Also, we know that R =4

abc, i.e., =

R4

abc

(i) gives,R4

abc =

R4

1xyz

z

c

y

b

x

a xyz

abc

z

c

y

b

x

a .

Example 10:Prove that in a ABC, R 2r..

Solution:We have

r = 4R sin A/2 sin B/2 sin C/2

R4

r = sin A/2 sin B/2 sin C/2

Also we know that sin A/2 sin B/2 sin C/2 8

1,

8

1

R4

r R 2r..

Example 11:Prove that in a triangle the sum of exradii exceeds the inradius by twice the diameter of the circumcircle.

Solution:Let the exradii be r

1, r

2, r

3 inradius be r and circumradius be R.

Then we have to prove that r1 + r

2 + r

3 = r + 4R.


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Now, r1 + r

2 + r

3 – r =

Δ Δ Δ Δs a s b s c s

=

cs

1

bs

1

s

1

as

1

=

csbs

a

ass

a

=

2 2s s b c bc s asa

s s a s b s c

=

2

2 bccbass2a

=

4

abcRR4

abcbcs2s2.

a 22

r1 + r

2 + r

3 = r + 4R.

Example 12:If a, b, c are in A.P., prove that cos A cot A/2, cosB. cot B/2, cosCcot C/2 are in A.P.

Solution:a, b, c are in A.P.

cotA/2, cot B/2, cotC/2 are in A.P.Now, cosA cotA/2, cosB cotB/2, cosC cotC/2 are

(1 – 2 sin2A/2) cotA/2, (1 – 2sin2 B/2) cotB/2, (1 – 2 sin2 C/2). cot C/2Now, cot A/2 – sinA, cotB/2 – sin B, cotC/2 – sin C are in A.P. as cotA/2, cotB/2, cotC/2 are inA.P. and sinA, sin B, sin C are in A.P.So, cos A cot A/2, cosB. cot B/2, cosCcot C/2 are in A.P.

Example 13:If r and R are radii of the incircle and circumcircle of aABC, prove that8r R {cos2A/2 + cos2 B/2 + cos2 C/2} = 2bc + 2ca + 2ab – a2 – b2 – c2.

Solution:

L.H.S. = 8

2/Acos

4

abc.

s2

= 2/Acos2s

abc 2

= Acos1s

abc

=

bc2

acb1

s

abc 222


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=

bc2

acbbc2

s

abc 222

=

bc2

acb

s

abc 22

=

bc2

acbcba

s

abc, where a + b + c = 2s

=

bc2

acb

s

s2abc= 2abcabacba

= 222 cbaab2ca2bc2 8rR{cos2A/2 + cos2 B/2 + cos2C/2}

= 2bc + 2ab + 2ca – a2 – b2 – c2 .

Example 14:If ‘t

1’, ‘t

2’ and ‘t

3’ are the lengths of the tangents drawn from centre of ex-circle to the circum circle of

the ABC, then prove that

cba

abc

t

1

t

1

t

12

32

22

1

Solution:Let S and I

1 be respectively the centres of the circumcircle and the excircle touching BC. It can be

shown that

12

1 Rr2RSI In SI1 P , SI

12 = R2 + t

12

R2 + 2Rr1 = R2 + t

12 ,

12

1 Rr2

1

t

1

SR P

I1

CB

A

Similarly2 2

2 32 3

1 1 1 1,

2Rr 2Rrt t

2 2 21 2 3 1 2 3

1 1 1 1 1 1 1

t t t 2R r r r

=

csbsas

R2

1 =

1 s s

2R 2R

=abc

cba proved

Example 15:If a, b and A are given in a triangle and c

1, c

2 are the possible values of the third side, prove that


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2 21 2 1 2c c 2c c cos A = 4a2 cos2 AA

Solution:

We have cosA =2 2 2b c a

2bc

c2 – 2bc cos A + b2 – a2 = 0, which is quadratic in ‘c’

1 2

2 21 2

c c 2b cos A

and c c b a

. . . (i)

2 21 2 1 2c c 2c c cos 2A

(c1 + c

2)2 – 2c

1c

2 – 2c

1c

2 cos 2A [using (i)]

(c1 + c

2)2 – 2c

1c

2 (1 + cos 2A)

4b2 cos2A – 2(b2 – a2). 2cos2A = 4a2 cos2A

2 2 2 21 2 1 2c c 2c c cos A 4a cos A


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