solution of triangles theory notes math.pdf

(1)

1

1.. EELLEEMMEENNTTS S OOFFA TA TRRIIAANNGGLLEE

In a triangle ABC the angles are denoted by capital letters A, B and C

In a triangle ABC the angles are denoted by capital letters A, B and C and the length of and the length of the sidesthe sides opposite to these angles are denoted by small letters a, b and c.

opposite to these angles are denoted by small letters a, b and c. Semi perimeter of the triSemi perimeter of the triangle isangle is given by s =

given by s = aa bb cc 2 2



 

_{and its area is denoted by}_{and its area is denoted by}

_

_..

2

2.. SSIINNE E RRUULLEE In a tri

In a triangle ABC,angle ABC, aa bb cc

ssiinn AA



ssiinn BB



ssiinn CC = 2R (where R is = 2R (where R is circum radius)circum radius) Note :

Note : Area of triaArea of trianglengle



== 2 2 1 1 bc sinA = bc sinA = 2 2 1 1 ac sinB = ac sinB = 2 2 1 1 ab sinC. ab sinC. Drill Exercise - 1 Drill Exercise - 1 1.

1. In any triangle ABCIn any triangle ABC, prove that, prove that

C C sin sin B B sin sin )) C C B B sin( sin( aa22





+ + A A sin sin C C sin sin )) A A C C sin( sin( b b22





+ + B B sin sin A A sin sin )) B B A A sin( sin( cc22





= 0 = 0 2. 2. If in aIf in a



ABC,ABC, C C sin sin A A sin sin = = )) C C B B sin( sin( )) B B A A sin( sin(



, prove that a

, prove that a22_{, b}_{, b}22_{, c}_{, c}22_are_{are in A. P.}_{in A. P...}

3.

3. ABCD is a trapezABCD is a trapezium such that AB and CD are parallel and CB isium such that AB and CD are parallel and CB is perpendicular to them. If perpendicular to them. If



ADB = 60º, BCADB = 60º, BC = 4 and CD = 3, then fi= 4 and CD = 3, then find the length of sind the length of side AB.de AB. 4.

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

by



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









cos cos 7 7 cos cos 1 1 .. 5.

5. Through the angular points of a triThrough the angular points of a triangle are drawn straight lines which make the same angleangle are drawn straight lines which make the same angle



with the with the opposite sides of the triangle; prove that area of

opposite sides of the triangle; prove that area of the triangle formed by them is to the triangle formed by them is to the area of thethe area of the original triangle as 4 cos

original triangle as 4 cos22



_:_{: 1.}_1.

3

3.. CCOOSSIINNE E RRUULLEE In a tri

In a triangle ABC,angle ABC, (i)

(i) cos A =cos A =

bc bc 2 2 aa cc b b22



22



22 (ii) (ii) cos B = cos B = ca ca 2 2 b b aa cc22



22



22

Theory

(2)

(iii)

(iii) cos cos C C ==

ab ab 2 2 cc b b aa22



22



22 Drill Exercise - 2 Drill Exercise - 2 1.

1. In anyIn any



ABC, prove thatABC, prove that aa A A cos cos + + b b B B cos cos + + cc C C cos cos = = abc abc 2 2 cc b b aa22



22



22 .. 2.

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

m m )) 3 3 m m )( )( 1 1 m m (( 2 2 1 1





, where m , where m



(1, 3).(1, 3). Prove that there are two values of the thi

Prove that there are two values of the third side one of which is m times the other.rd side one of which is m times the other. 3.

3. In a In a triangle ABC,triangle ABC,



C = 60º, then prove thatC = 60º, then prove that

cc aa 1 1



++ bb cc 1 1



= = aa bb cc 3 3



4.

4. If in a tIf in a triangle riangle ABABC,C,

B B cos cos 2 2 A A cos cos C C cos cos 2 2 A A cos cos



= = C C sin sin B B sin sin

prove that the triangle

prove that the triangle is either is either isosceles or rightisosceles or right angled.

angled. 5.

5. A ring, 10 cm, in dA ring, 10 cm, in diameteriameter, is suspended from a point , is suspended from a point 12 cm, above it12 cm, above its centre by 6 equal stringss centre by 6 equal strings attached to its circu

attached to its circumference at equal intervals. Find the cosine of themference at equal intervals. Find the cosine of the angle between consecutiveangle between consecutive strings.

strings.

Illustration Illustration 1: 1:

Find the angles of t

Find the angles of the triangle whoshe triangle whose sides are e sides are 3 +3 + 33 ,, 22 33 andand 66 ..

Solution: Solution:

Let a = 3 +

Let a = 3 + ₃₃, b, b = 2= 2 ₃₃, c =, c = ₆₆



cos cos A A ==

2 2 12 12 3 3 6 6 3 3 9 9 6 6 12 12 bc bc 2 2 aa cc b b22 22 22













= = 2 2 2 2 3 3 1 1 2 2 12 12 3 3 6 6 6 6







= cos 105 = cos 10500

_

_{A = 105}_{A = 105}00

Applying Sine formula : Applying Sine formula :

B B sin sin b b A A sin sin aa

_

, we get , we get



sin B =sin B = bbssiinn AA 2 32 3 ssiin 1n 10

  

05500 aa



33



33 = = 1 1 2 2



B = 45 B = 45 0 0



A = 105A = 10500_{, B = 45}_{, B = 45}00_{, C = 30}_{, C = 30}00 4 4.. PPRROOJJEECCTTIIOON N FFOORRMMUULLAAEE (i)

(i) a = b cos C + c cos Ba = b cos C + c cos B (ii)(ii) b = c cos A + a cos Cb = c cos A + a cos C (iii) (iii) c = a cos B + b cos Cc = a cos B + b cos C

If A = 45

If A = 4500_{, B = 75}_{, B = 75}00_{, prove that a + c}_{, prove that a + c}

2

2 = 2b. = 2b.

Solution: Solution:

(3)

(iii)

(iii) cos cos C C ==

ab ab 2 2 cc b b aa22



22



22 Drill Exercise - 2 Drill Exercise - 2 1.

1. In anyIn any



ABC, prove thatABC, prove that aa A A cos cos + + b b B B cos cos + + cc C C cos cos = = abc abc 2 2 cc b b aa22



22



22 .. 2.

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

m m )) 3 3 m m )( )( 1 1 m m (( 2 2 1 1





, where m , where m



(1, 3).(1, 3). Prove that there are two values of the thi

Prove that there are two values of the third side one of which is m times the other.rd side one of which is m times the other. 3.

3. In a In a triangle ABC,triangle ABC,



C = 60º, then prove thatC = 60º, then prove that

cc aa 1 1



++ bb cc 1 1



= = aa bb cc 3 3



4.

4. If in a tIf in a triangle riangle ABABC,C,

B B cos cos 2 2 A A cos cos C C cos cos 2 2 A A cos cos



= = C C sin sin B B sin sin

prove that the triangle

prove that the triangle is either is either isosceles or rightisosceles or right angled.

angled. 5.

5. A ring, 10 cm, in dA ring, 10 cm, in diameteriameter, is suspended from a point , is suspended from a point 12 cm, above it12 cm, above its centre by 6 equal stringss centre by 6 equal strings attached to its circu

attached to its circumference at equal intervals. Find the cosine of themference at equal intervals. Find the cosine of the angle between consecutiveangle between consecutive strings.

strings.

Find the angles of t

Find the angles of the triangle whoshe triangle whose sides are e sides are 3 +3 + 33 ,, 22 33 andand 66 ..

Solution: Solution:

Let a = 3 +

Let a = 3 + ₃₃, b, b = 2= 2 ₃₃, c =, c = ₆₆



cos cos A A ==

2 2 12 12 3 3 6 6 3 3 9 9 6 6 12 12 bc bc 2 2 aa cc b b22 22 22













= = 2 2 2 2 3 3 1 1 2 2 12 12 3 3 6 6 6 6







= cos 105 = cos 10500

_

_{A = 105}_{A = 105}00

Applying Sine formula : Applying Sine formula :

B B sin sin b b A A sin sin aa

_

, we get , we get



sin B =sin B = bbssiinn AA 2 32 3 ssiin 1n 10

  

05500 aa



33



33 = = 1 1 2 2



B = 45 B = 45 0 0



A = 105A = 10500_{, B = 45}_{, B = 45}00_{, C = 30}_{, C = 30}00 4 4.. PPRROOJJEECCTTIIOON N FFOORRMMUULLAAEE (i)

(i) a = b cos C + c cos Ba = b cos C + c cos B (ii)(ii) b = c cos A + a cos Cb = c cos A + a cos C (iii) (iii) c = a cos B + b cos Cc = a cos B + b cos C

If A = 45

If A = 4500_{, B = 75}_{, B = 75}00_{, prove that a + c}_{, prove that a + c}

2

2 = 2b. = 2b.

Solution: Solution:

(4)

As A = 45

As A = 4500_{, B = 75}_{, B = 75}00_we_{we have}_{have C =}_{C = 60}₆₀00



2b = 2 2b = 2 (a cos (a cos C + c C + c cos cos A) A) = 2(a c= 2(a cos 60os 6000_{+ c cos 45}_{+ c cos 45}00₎₎

= a

= a + c+ c ₂₂ = L.H.S. = L.H.S. 5.

5. NAPIER’S ANALOGY (TANGENT RULE)NAPIER’S ANALOGY (TANGENT RULE)

(i) (i) 2 2 A A cot cot cc b b cc b b 2 2 C C B B tan tan



















 

(ii) (ii) tantan

2 2 B B cot cot aa cc aa cc 2 2 A A C C



















 

(iii) (iii) tantan

2 2 C C cot cot b b aa b b aa 2 2 B B A A



















 

6 6.. HHAALLF F AANNGGLLE E FFOORRMMUULLAAEE ((aa)) ((ii)) sisinn



 

 

bc bc cc ss b b ss 2 2 A A





_(ii)_{(ii) ssin}_inBB



 

s c s as c s a





 

2 2 caca







(iii)

(iii) sinsin



 

 

ab ab b b ss aa ss 2 2 C C





((bb)) ((ii)) coscos

2 2 A A = =





b bcc aa ss ss



(ii) (ii) coscos

2 2 B B = =





ca ca b b ss ss



(iii)

(iii) coscos CC ss ss cc





2

2 abab





((cc)) ((ii)) tantan

2 2 A A = =



ss b

_{ }

b ss cc

   

_



ss ss aa



(ii)(ii) tantan BB₂₂ ==









ss bb



ss aa ss cc ss



(iii)

(iii) tantan



_



_



cc ss ss b b ss aa ss 2 2 C C





Drill Exercise - 3 Drill Exercise - 3 1.

1. In In aa



ABC, if a = 13, b = 14, and c = 15, find tABC, if a = 13, b = 14, and c = 15, find the followingshe followings (i)

(i)



(ii)(ii) sinsin 2 2 A A

(iii) (iii) coscos

2 2 A A

(iv) (iv) tantan

2 2 A A

2.

2. The sides of a triangle are xThe sides of a triangle are x22_{+ x + 1, 2x + 1 and x}_{+ x + 1, 2x + 1 and x}22_–1;_{–1; prove}_{prove that}_{that the}_{the greatest}_{greatest angle}_{angle is}_{is 120º.}_120º.

3.

3. In anyIn any



ABC, prove thatABC, prove that

C C sin sin B B sin sin = = C C cos cos aa b b B B cos cos aa cc



4.

4. If If

  

are the lengths of the altitudes of aare the lengths of the altitudes of a



ABC, prove thatABC, prove that 2 2 1 1



+ + 22 1 1



+ + 22 1 1



= =





cotcotBB cotcotCC A

A cot cot

where

(5)

5. In a



ABC, prove that cot 2 A + cot 2 B + cot 2 C =

















a c b c b a cot 2 A . 7. AREA OF TRIANGLE Area of triangle = s s a s b s c(



)(



)(



) 7.1 (i) sin A = 2 s s a

  

s b s

  

c 2Δ bc

   

bc (ii) sinB=

  

  

2 2Δ s s a s b s c ca

   

ca (iii) sin C = 2 s s a

     

s b s c 2Δ ab

   

ab 8. m-n THEOREM

Let 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 TRIANGLE

The line joining any vertex of a triangle to the mid point of the opposite side of the triangle is called the median of the triangle. The three medians of a triangle are concurrent and the point of concurrency of the medians of any triangle is called the centroid of the triangle. The centroid divides the median in 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_{= AC}2_{+ CD}2_{– 2AC. CD cos C = b}2₊

4 a2

– ab cosC, and c2_{= b}2_{+ a}2_{– 2ab cos C.}

Hence 2AD2_{– c}2_{= b}2_– 2 a2 , so that AD = 2b2 2c2 a2 2 1





₌ b c 2bccosA 2 1 ₂ ₂



Similarly, D A F B E C G    BE = 2c2 2a2 b2 2 1





_{, and CF =} ₂_a2 ₂_b2 _c2 2 1





(6)

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



= 2 2 2 ₂_c _a b 2 B sin a





Drill Exercise - 4

1. If the medians of a



ABC make angles

  

with each other, prove that cot



+ 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 equal sides 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 the

triangle 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 is

2 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

a 2 c ~ b2 2 . 10. CIRCUM CIRCLE

The circle which passes through the angular points of a

_

ABC, is called its circumcircle. The centre of 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 E O A _A a/2 a/2

(7)

Radius of the circumcircle is given by the following formulae

R = a b c abc

2 sin A



2 sin B



2 sin C



4



Illustration 4:

If in a



ABC, O is the circumcenter and R is the circumradius and R₁, R₂ and R₃ are the circumradii of the triangles OBC, OCA and OAB respectively, then prove that 3

3 2 1 R abc R c R b R a





_. Solution:

Clearly, in the



OBC,



BOC = 2A, OB = OC = R, BC = a.



2R₁=

A 2 sin

a

{using sine rule in



BOC)

Similarly, 2R₂= C 2 sin c R 2 and B 2 sin b 3



A O B _C



3 2 1 R c R b R a

_

= 2(sin2A + sin2B + sin 2C) = 2.4 sin A sin B sin C,

= 8 ₃ R abc R 2 c . R 2 b . R 2 a



_. Illustration 5:

If the distances of the sides of a



ABC from its circumcenter be x, y and z respectively, then prove that _xa



b_y



_zc



₄abc_xyz.

Solution :

Let M be the circumcenter. MD

_

BC. So BD = DC =

2 a and



BMD = A.A. In



BDM, MD BD = tan A or x 2 a = tan A, i.e., x 2 a = tan A,A, Similarly, ₂b_y = tan B, z 2 c = tan C y A M B _D _C z A x E F



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

2x



2y



2z

and tan A. tan B. tan C = ₂a_x.₂b_y.₂c_z

(8)



_xa



_yb



_zc



₄abc_xyz

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 the point of cuncurrency of the altitudes of the triangle is called the orthocenter of the triangle. The triangle 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 = C sin c

cos B cos C = 2R cos B cos C Similarly

PE = 2R cosA cosC and PF = 2R cosA cosB Again

AP = AE sec DAC = c cos A cosec C = C sin c cos A = 2 R cos AA D A F B E C P 90 – C so, 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



ABC If OF be perpendicular to AB, we have

0 OAF 90 C







Also



PAL



900



C A B _K C D F L O P





OAP



C – B

Also OA = R and PA = 2R cosA

(9)

= R2_{+ 4R}2_cos2_{A – 4R}2_{cosA cos (C – B)}

= R2_{– 4R}2_{cosA [cos(B + C) + cos (C – B)] = R}2_{– 8R}2_{cos A cos B cosC}



OP = R ₁

_

₈_cos_A_cos_B_cos_C 12. BISECTORS OF THE ANGLES

If AD bisects the angle A and divide the base into portions x and y, we have, by Geometry,

b c AC AB y x



_

c b a c b y x b y c x











x = ac b c



and y = ab b c



Also let



be the length of AD

we have

_

ABD +

_

ACD =

_

ABC



bcsinA, 2 1 2 A sin b 2 1 2 A sin c 2 1









y A B _C D x  i.e., 2 A cos c b bc 2 2 A sin A sin c b bc











Drill Exercise - 5

1. Show that the distances of the orthocentre from the sides of a triangle ABC are A cos C cos R 2 & A cos C cos R 2 , C cos B cos R 2 .

2. In any



ABC, prove that a cosA + b cosB + c cosC = 4R sinA sinB sin C.

3. If p₁, p₂p₃ are respectively the perpendiculars from the vertices of a triangle to the opposite sides, prove that p₁p₂p₃ = ₃ 2 2 2 R 8 c b a .

4. In a



ABC, if 8R2_{= a}2_{+ b}2_{+ c}2_{, 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 opposite sides : prove that the diameter of the circumcircles of the triangle AEF, BDF, and CDE are respectively a 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. INCIRCLE

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



ABC.

D A B F C E I C/2 B/2 r r r 90 – B/20

(10)

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 A sin 2 B sin 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.

. C 90 OAF



0





OAF IAF OAI











O A F B E I C =





2 B C 2 C B A C 2 A C 90 2 A ₀















Also, AI = 2 C sin 2 B sin R 4 2 A sin r 2 A sin IE



OAI cos AI . OA 2 AI OA OI2



2



2





= R2_{+ 16R}2_sin2 2 B sin2 2 C – 8R 2_sin 2 B sin 2 C cos 2 B C



2 C sin 2 B sin 16 1 R OI 2 2 2 2







_{– 8sin}













_

2 C sin 2 B sin 2 C cos 2 B cos 2 C sin 2 B = 1 – 8 sin













_

2 C sin 2 B sin 2 C cos 2 B cos 2 C sin 2 B = 1 – 8 sin 2 A sin 2 C sin 2 B . . . (i) 2 A sin 2 C sin 2 B sin 8 1 R OI







. 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















2 r Solution:

Let the incircle touches the side AB at P, where AP =



. Let I be the incentre.



From the right-angled



IPA,

2 A cot r ; 2 A tan r











similarly,



= r cot 2 C cot r and 2 B





(11)

In



ABC, we have the identity cot 2 C cot 2 B cot 2 A cot 2 C cot 2 B cot 2 A







r . r . r r r r



















A P B E C I  or

















₃



r 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 tan 1 cos B cos C 1

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





tan IL IM LM IM ON r R cos A B OL BM BN BM NC r cot R sin A 2



 





_

A B C

4Rsin sin sin R cos A

2 2 2

A B C B

4Rsin sin sin .cot R sin A

2 2 2 2







B O L I C A NM cosA cos B cos C 1 cos A cos B cos C 1

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

















tan 1 cos B cos C 1 sin C sin B











 

_

_







.

14. THE DISTANCES BETWEEN THE SPECIAL POINTS

(i) The distance between circumcenter and orthocenter is = R. ₁

_

₈_cos_A_cos_B_cos_C (ii) The distance between circumcenter and incentre is = _R2

_

₂_Rr.

(12)

(iii) The distance between incentre and orthocenter is ₂_r2

_

₄_R2_cos_A_cos_B_cos_C_.

Drill Exercise - 6

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

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

2 B sin 2 A sin R 4 , 2 A sin 2 C sin R 4 , 2 C sin 2 B sin R 4 _{respectively..}

3. In a



ABC, prove that ratio of the area of the incircle to that of the triangle is



: cot 2 A cot 2 B cot 2 C . 4. Prove that acotA



bcotB



ccotC



2(R



r).

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

_

_cos_C

_

₁.

15. ESCRIBED CIRCLES

The circle which touches the side BC and the two sides AB and AC produced is called the escribed circle opposite the angle A. Its centre and radius will be denoted by I₁and r₁ respectively.

Radii of the excircles are given by the following formulae (i) r₁ = 2 C cos 2 B cos 2 A sin R 4 2 A tan s a s







(ii) r₂ = 2 C cos 2 B sin 2 A cos R 4 2 B tan s b s







B A M E1 C D1 F1 I1 L (iii) r₃ = 2 C sin 2 B cos 2 A cos R 4 2 C tan s c s







. 16. EXCENTRAL TRIANGLE

The triangle formed by joining the three excentres I₁, I₂ and I₃of



ABC is called the excentral or excentric triangle. Not that " (i) Incentre I of



ABC is the orthocenter of the excentral



I₁I₂I₃.

(ii)



ABC is the pedal triangle of the



I₁I₂I₃. (iii) The sides of the excentral triangle are 4 R cos

2 A , 4 R cos 2 B and

(13)

4 R cos 2 C

and its angles are 2



– 2 A , 2



– 2 B and 2



– 2 C . (iv) II₁ = 4 R sin 2 A : II₂ = 4 R sin 2 B : II₃ = 4 R sin 2 C Illustration 11 :

If the exradii r₁, r₂ and r₃of a



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

Solution: We know that r₁ = c s r , b s r , a s 2 3



















r₁, r₂, r₃ are in HP



s

_



a, s



_

b, s

_



c_{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 :









_



_









































































A sin 2 A sin r A cos 4 2 A tan ) c b ( A sin 4 A cos a r 1 r 1 r 1 r 1 r 1 r 1 4 s r r 4 ) r r )( r r )( r r ( ) B A sin( c 2 b a 2 C sec 2 ) B A ( sec ) b a ( 4 1 C cos 4 r r r r ) r r r r ( 4 1 rs 4 abc R 3 2 1 2 2 2 3 2 1 2 2 3 2 1 3 2 1

2. Prove the following :

3 / 1 3 / 5 3 / 2 2 2 2 2 2 2 2 1 3 2 1 ) C 2 sin B 2 sin A 2 (sin 2 ) abc ( ) C cot B cot A (cot 4 c b a ) b a ( ) B A sin( 2 B sin A sin ) B 2 sin c C 2 sin b ( 4 1 ) C sin B sin A (sin Rr ) c s )( b s ( bc 2 A cos 2 A tan ) a s ( s 2 A cot rr r r rr































3. Show that the radii of the three escribed circles of a triangle are the roots of the equation, x3_{– x}2_{(4R + r) + xs}2_{– rs}2_{= 0}

4. If R₁, R₂ and R₃ be the diameter of the excircles of a



ABC (opposite to the vertices A, B and C respectively), then prove that

c b a R R R R c R b R a ₁ ₂ ₃ 3 2 1







.

(14)

5. Prove that r2



r₁2



r₂2



r₃2



16R2



(a2



b2



c2).

17. SOLUTION OF TRIANGLES

When any three of the six elements (except all the three angles) of a triangle are given, the triangle is known completely. This process is called the solution of triangles.

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

2 2 2

b 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 tan B C b ccotA 2 b c 2







, we get B C 2



. Also B C 2



= 900 _– A

2 , so that B and C can be evaluated. The third side is given by a = bsinA

sinB .

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

b sin B, A = 1800_{– (B + C) and a =} bsinA

sin B give 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 tw o values of angle C (fig 3). If c < b and B is an acute angle, then there is only one triangle (f ig 4).

A B _D c b _{c sinB} A B D c b c sinB (Fig 2) (Fig 1)

(15)

A B D c b c sinB C1 C2 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-angled with 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 0 b c a 100 36 196 1 A 120 2bc 120 2

 

_

 

_

_{  }

Illustration 13:

If in a triangle ABC, a = (1 + _{3 ) cm, b = 2 cm and C = 60}0_{, then find the other two angles and the}

third side Solution: cos C = 2 2 2 a b c 2ab

 

.











2 ₂ 1 3 4 c 1 2 _{2 1} _{3 .2}



 







c2_{= 6}



_{c =} 6 Also, sin A sin B sin C

a



b



c



3 sin A sin B ₂ 2 1



3



6



sin B = 1 2



B = 45 0



A = 1800_{– (45}0_{+ 60}0_{) = 75}0 Illustration 14 :

Given the base of a triangle, the opposite angle A, and the product k 2 _{of the other two sides, show}

that it is not possible for a to be less than 2k sin A 2 .

Solution:

(16)

Now cosA = 2 2 2 b c a 2bc

 

or 2k 2_{cosA = b}2₊ 2 2 2 k a b

 



 

 

A B C a c b or b4_{– (a}2_{+ 2k}2_{cosA). b}2_{+ k}4_{= 0}

Since b2_{is real, (a}2_{+ 2k}2_{) (a}2_{+ 2k}2_{cosA – 2k}2₎

_

₀



a2 2k .2 cos2 2 A a2 2k .2 sin2 2 A 0 2 2



_



_



_















a2 4k cos2 2 A a2 4k sin2 2 A 0 2 2



_





_



_



















a2 4k sin2 2 A 0 2





[since a2_{+ 4k}2_cos2 A

2 is always positive]



a 2k sin A a 2k sin A 0 2 2



_

 

_



_



 





 





a 2k sinA or a 2k sinA 2 2

 



_{(since 2ksin(A/2) is real)}

But a must be positive.



a

_

– 2k sin A

2 is rejected Hence a 2 k sin A

2



_.

DRILL EXERCISE - 8

1. A right triangle has c = 64,



A = 61º and



C = 90º. Find the remaining parts. 2. _{Solve the triangle in which b = 100, c = 100 2 and B = 30º.}

3. In a



ABC if a, b and A are given, then prove that two triangles are formed such that the sum of their areas is

2 1

b2_sin2_A.

4. The lengths of two sides of a triangle are 12 cm and 12 2 cm respectively, and the angle opposite the shorter side is 30º; prove that there are two triangle satisfying these conditions, find their angles and show that their areas are in the ratio ₃

_

₁ : ₃

_

₁

5. In a



ABC, if a, b and A are given, then there are two triangles with third sides c₁ and c₂ such that c₁ – c₂ = 2 _a2

_

_b2_sin2_A

(17)

(Important Formulae)

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

n 2 sin nr 2 1 2



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

n tan nr 2

1 2



III. Side of Inscribed polygone =

n sin r

2



_. IV. Side of Circumscribed polygone =

n tan r

2



_.

Illustration 15 :

Find the radii of the inscribed and the circumscribed circles of a regular polygon of n sides with each side 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 incircle and the circumcircle of the polygon

2 BOC n









BOL 1 2 2 n n











_

_







If a be a side of the polygon, we have a = BC = 2BL = 2RsinBOL = 2Rsin n





r a cot 2 n





.

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

OBC



n 1OL.BC n . cot .a1 a na2 cot

2 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 C sin bc 2 A sin ad 2



2 _.

2. Two regular polygons of n & 2n sides have the same perimeter, show that their areas are in the ratio n cos 1 : n cos 2







_.

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







_.

4. With reference to a given circle, A₁ & B₁are the areas of the inscribed and circumscribed regular polygons of n sides, A₂ & B₂ are corresponding quantities for regular polygons of 2n sides. Prove that A is a geometric mean between₂ A₁ & B₁ and B is a hormonic mean between₂ A₂ & B₁.

(18)

Answer Key

Drill Exercise - 1 3. 3 3 4 3 25



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₍ ₆



₂₎ ,



A = 15º ,



C = 135º 4. 45º and 105º; 135º and 15º

SOLVED OBJECTIVE EXAMPLES

Example 1:

If D is the mid point of the side BC of a triangle ABC and AD is perpendicular to AC, then (a) 3b2_{= a}2_{– c}2 _{(b) 3a}2_{= b}2_{– 3c}2

(c) b2_{= a}2_{– c}2 _{(d) a}2_{+ b}2_{= 5c}2 Solution:

From the right angled



CAD, we have

a / 2 A c B b C D 900 a / 2 A – / 2 cos C = ab 2 c b a a b 2 2 / a b 2



2



2





a2_{+ b}2_{– c}2_{= 4b}2

_

_a2_{– c}2_{= 3b}2_.

(19)

Example 2:

There exists a triangle ABC satisfying

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

7 C sin 3 B sin 2 A sin

_

(c) (a + b)2_{= c}2₊_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 of tan A, tan B, tan C can be zero, (a) is not possible

If (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 _{= c}2_{+ ab, then} 2 2 2 a 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_{– c}2_{x + ab = 0, then}

(a) tan A = a/b (b) tan B = b/a

(c) cos C = 0 (d) sin2_{A + sin}2_{B + sin}2_{C = 2} Solution:

From the given equation, we get

tan A + tan B = c2_{/ ab and tan A tan B = 1.}

Since tan (A + B) = B tan A tan 1 B tan A tan





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 + sin}2_{B + sin}2_{C =} ₁ ₁ ₁ ₂

c b a 1 c b c a 2 2 2 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:

(20)

then

_

= 2 1 ap₁ = 2 1 bp₂ = 2 1 cp₃



p₁ = a 2



, p₂ = b 2



, p₃ = c 2



By the law of sines

a b c

sin A



sin B



sin C = k (say)



p₁ = 2Δ

ksinA, p2 =

2Δ

ksinB, p3 = _k_sin_C

2



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



p₁, p₂, p₃ are in H.P.

Example 5:

In a triangle ABC, medians AD and CE are drawn. If AD = 5,

_

DAC =



/ 8 and



ACE =



/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 2

AD =

3 10

. Therefore, applying the sine rule to



AOC, we get















/ 4



sin 8 / sin . 3 10 OC 4 / sin AO 8 / sin OC













area of

_

AOC =

2 1

. AO.OC. sin

_

AOC

=



_



_















_



8 2 sin . 4 / sin 8 / sin . 3 10 . 3 10 . 2 1 D B O E A /8 /4 _C

= 50₉ .sin

   



_sin / 8

_

_

cos_/₄

_



/ 8



₁₈50



25₉



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

(21)

We have tan 2 C = tan













_



2 B A 900 _{= cot} 2 B A



= cot_cot

   

_{   }

A_A / _/2₂cot

_

_cotB / _B2_/



₂1

= 6 37 . 1 222 100 122 2 5 20 6 37 _{120 185} ₃₀₅ ₅ 5 20



_





Also tan 2 A tan 2 C =

  

s



_s

_

b_s

_

s_a



_

c

  

s



_s

_

a_s

_

s_c



_

b



s b s 5 2 . 6 5





_

₃

_

_s

_

_b

_

_

_s

_

₂_s

_

₃_b



a + b + c = 3b



a + c = 2b Which 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 thir d side c is equal to

(a) 6 (b) 8 (c) 4 (d) none of these Solution: cos (A – B) = 2 B A tan 1 2 B A tan 1 32 31 2 B A tan 1 2 B A tan 1 2 2 2 2



















63 tan2 63 1 2 B A tan 1 2 B A











Now tan 2 C cot 4 5 4 5 63 1 2 C cot b a b a 2 B A



















tan 2 C = 9 63 Also, cos C =









8 1 144 18 81 / 63 1 81 / 63 1 2 / C tan 1 2 / C tan 1 2 2













c2_{= a}2_{+ b}2_{– 2ab cos C = 25 + 16 – 2.5.4. (1/8) = 36}

_

_{c = 6}

Hence (a) is the correct answer.

Example 8:

In a triangle ABC, if r₁= 2r₂ = 3r₃, then a : b is equal to (a) 5

4 (b)

4 5

(22)

(c) 7

4 (d)

4 7

Solution:

From the given relation, we have s tan 2 A = 2s tan 2 B = 3s tan 2 C k 2 ) 2 / c tan( 3 ) 2 / B tan( 6 ) 2 / A tan(

_

(say) Also, since A/2 + B/2 + C/2 = 900_{, we get}

tan 1 2 A tan 2 C tan 2 C tan 2 B tan 2 B tan 2 A







6k. 3k + 3k. 2k + 2k. 6k = 1



36k 2_{= 1}

_

_{k = 1/6}



sin A =









2 2 2 tan A / 2 12k 1 1 tan



A / 2



1 36k





and sin B =



_



_

5 4 k 9 1 k 6 2 / B tan 1 2 / B tan 2 2 2



_





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



4 5 B sin A sin 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 = m₂, AF = m₃, then

8 a2 is equal to (a) m2₂



m₃2



2m₁2 (b) m₁2



m2₂



2m2₃ (c) m2₂



m₃2



2m₁2 (d) none of these Solution: In



ABC, AD2 _{= m} 1 2₌ 4 a 2 b c2 2 2





A B C F D E In



ABD, AE2_{= m} 2 2₌ 2 2 2 a AD c 2 2 4

 

 



_

_{ }

(23)

AF2_{= m} 3 2₌ 2 2 2 a AD b 2 2 4

 

 



_

_{ }



m₂2_{+ m} 3 2_{= AD}2₊ 8 a m 2 8 a 4 a m m 8 a 2 c b ₂ 2 1 2 2 2 1 2 1 2 2 2















2 2 2 2 2 3 1 a m m 2m 8









Example 10:

If I is the incentre of a triangle ABC, then the ratio IA : IB : IC is equal to (a) cosec 2 A : cosec 2 B : cosec 2 C (b) sin 2 A : sin 2 B : sin 2 C (c) sec 2 A : sec 2 B : sec 2 C (d) none of these Solution: Here BD : DC = c : b But BD + DC = a;



BD = .a c b c



In



ABD, sinB AD 2 A sin BD



A B C D I



AD = 2 A ec cos c b 2 2 A sin B sin . c b ca









Also,

_

_

a c b c b / ca c BD AB ID AI

_







AI = 2 A ec cos s AD . c b a c b







Similarly BI = cos ecB, CI cos ecC

s 2 s 2





A B C

IA : IB : IC cos ec : cos ec : cos ec

2 2 2





Example 11:

In a



ABC, the value of

c b a C cos c B cos b A cos a



is equal to (a) r R (b) r 2 R

(24)

(c) R r (d) R r 2 Solution:

a cos A b cos B c cos C 2R sin A cos A 2R sin Bcos B 2R sin Ccos C

a b c 2s





 

= .



sin2A sin2B sin2C



s 2

R



= .4sinA sinB sinC

s 2 R = ₃ ₂ 4R abc abc . 2s 8R



4sR But R =



4 abc , r = s



So, the value =

R r R . r . 4 R 4 2





Example 12:

The area of a circle is A₁and the area of a regular pentagon inscribed in the circle is A₂. Then A₁: A₂is (a) 10 cos 5



(b) 10 sec 5 2



(c) 10 ec cos 5 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 1 r2_{cos 18}0 O C D E A B r



A₁ : A₂ = 10 sec 5 2 18 cos r 5 r 2 0 2 2

_





Example 13:

In a triangle ABC a = 5, b = 4 and c = 3. ‘G’ is the centroid of the triangle. Circumradius of triangle GAB is equal to (a) ₂ ₁₃ (b) 13 12 5 (c) 13 3 5 (d) 13 2 3 Solution:

(25)

AG = 3 2 A AA₁, BG = 3 2 BB₁



AG = 3 1 2 2 2 2b



2c



a A B A 1 B1 C G and BG = 2a2 2c2 b2 3 1







AG = b2 4c2 3 1 BG , a 3 1





as a2_{= b}2_{+ c}2



AG = 13 3 2 36 16 3 1 BG , 3 5







Also, AB = c = 3 and GAB ABC

1

2 3

   

If ‘R ₁’ be the circumradius of triangle GAB then R₁ =

  

AG₄ BG AB



5₃.₃2 13.3.₄1_.₂ GAB





= 12 13 5 units. Example 14:

A variable triangleABC is circumscribed about a fixed circle of unit radius. Side BC always touches the circle at D and has fixed direction. If B and C vary in such a way that (BD). (CD) = 2 then locus of 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)}



₁ s ) a s ( 2 s2 2









(radius of incircle of triangle ABC)



s a = constant. Now



= 2 1

aH_a, where ‘H_a’ is the distance of ‘A’ from BC.



1 aHa s 2 s





_{= 1}

_

_H a = _a s 2 = 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,,}

(26)

A 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_{= OA}2_{+ OC}2_{– 2AOOC. cos} 3





d2_{= 2r}2_{– 2r}2_. 1 2 = r 2 A B D C O

Also,

_

BOD =

_

COD = 2

3.2 3

 

_



tan BD r 3 d 3 3 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 twice the 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 oposite to the side n is



. Now cos

_

=

   

  

2

  

n 1 n 2 5 n 6 n 2 n 1 n 2 n 2 n 1 n 2 2 2 2











=

  

_{    }

2 n 2 5 n 2 n 1 n 2 5 n 1 n







and cos 2



=

   





2 2 2 n n 1 n 2 2n n 1

   



= 2n



n 1



3 n 2 n2





=

  

n₂



_n1

_

_nn

_



₁

_

3



n₂



_n3

But cos 2



= 2 cos2

_

_{– 1; so}



n 2



1 2 5 n 2 n 2 3 n 2





















or







n 2



1 2 5 n n 2 3 n 2 2









or (n – 3)(n+2)2_{= n{(n + 5)}2_{– 2 (n + 2)}2_} or (n – 3)(n2_{+ 4n + 4) = n (– n}2_{+ 2n + 17)} or n3_{+ n}2_{– 8n – 12 = – n}3_{+ 2n}2_{+ 17n} or (n – 4)(2n2_{+ 7n + 3) = 0}

_

_{n = 4 or 2n}2_{+ 7n + 3 = 0.}

(27)

Roots of 2n2_{+ 7n + 3 = 0 are} 4 24 49 7





i.e., – 2 1

and – 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, tan 2 B , tan 2 C are 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







 

_





  

s



a

_

s



b 2 C tan , a s c s 2 B and s = 2 c b a



Now, (i)



a, b, c,

_

, s are rational. So tan 2 B and tan 2 C

are rational because sum, difference, product and quotient of nonzero rational numbers are rational.

Thus (i)



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

2 B , tan 2 C be rational Now, sin B = 2 B tan 1 2 B tan 2 2



= rational, because tan 2 B is rational. sin C = 2 C tan 1 2 C tan 2 2



= rational, because tan 2

C is rational. Now, tan 2 B . tan 2 C =

     









s a s b . a s c s =

     

2 s a s b s c s a a 1 s s a s b s c s s



_



 



.



(ii)



s is rational



b + c is rational, because a is rational. But C sin c B sin b A sin a



_

rational rational C sin B sin c b A sin a







(28)



A sin

a

is 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.  C sin c B sin b A sin a

_



A sin B sin a b



_{= rational} and c = A sin C sin a = rational





= 2 1 bc 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 2 B A tan 1 2 B A tan 1 2 2









By componendo and dividendo,

4 5 4 5 2 2 B A tan 2 2









or tan2 9 1 2 B A



_

or tan 3 1 2 B A





( A > B). But tan 2 C cot b a b a 2 B A









2 C cot 3 6 3 6 3 1









_{or cot} 2 C = 1;



C = 2



.



The area of the triangle = 9 2 sin . 3 . 6 . 2 1 C sin ab 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 the vertex C, then prove that

a2_b2_sin2_{C = a}2_p2_{+ b}2_q2_{– 2abpqcos C.} Solution :

Let

_

ACE =



. Clearly, from the figure, we get

p



C



sin BC q , sin AC p











(29)



sin .cosC cos .sinC a q , sin b p

_

_

_

_

_

_



cosC cos .sinC

b p a q







_or q p 2 2 2

cos C cos .sin C a b



_



_

_









=





C cos 1 b p 1 2 2 2

















or 2 2 2 2 2 2 2 2 2 2 q p 2pq p p

cos C cos C 1 1 cos C

a b ab b b











  

_

_





or cosC sin C ab pq 2 b p a q 2 2 2 2 2







_{or a}2_p2_{+ b}2_q2_{– 2abpqcosC = a}2_b2_sin2_C. Example 5:

In a

_

ABC, prove that cos A. cos C =

ca 3

a c 2 2



2

, where AD is the median through A and AD

_

AC.

Solution:

From the

_

ABC, cos A =

bc 2 a c b2



2



2 . . . (i)

From the



CAD, cos C =

a b 2 2 / a b CD AC



_{. . . (ii)}

From the

_

ABD,

0 BD AB sin ADB sin(A 90 )







or

_

_

C 90 sin c A cos 2 / a 0

_





or a c 2 cos A



cos C





cos A= , c b a b 2 . c 2 a c 2 C cos a











from (ii)



from (i), c b bc 2 a c b2 2 2









or b2_{+ c}2_{– a}2_{= – 2b}2 A B C D or c2_{– a}2_{= – 3b}2 _{. . . (iii)}

Now, cosA. cosC =

ca a c b a b 2 . bc 2 a c b2 2 2 2



2



2







=



ca 3 a c 3 c a ca 3 a c 3 b 3 2 2 2 2



2



2



2







, from (iii) = ca 3 a c 2 2



2 .

(30)

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_{– A}_A

Similarly P, D, B and F lie on a circle and therefore



PDF =



PBF = 900_{– A, Hence}

_

_{FDE = 180}0_{– 2A}

Similarly



DEF = 1800_{– 2B}



EFD = 1800_{– 2C}

Also, from the triangle AEF we have

EF AE AB cos A c cos A c cos A sin A



sin



AFE



cos PFE





cos PAE





sin C

 EF = c sinAcosA sinC = 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 t₁, t₂ , t₃ be the tangents of the angles subtended by these parts at the opposite vertex, prove that:

2 1 2 2 3 2 1 1 1 1 1 4 1 t t t t _t

































_

_











Solution:

Let the points P and Q divide the side BC in three equal parts such tha t BP = PQ = QC = x Also let,



BAP =



,



PAQ =



,

_

QAC =



and

_

AQC =

_

From question,

tan



= t₁, tan



= t₂, tan



= t₃.

B _C A Q P  _   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 =





_



_







_



cot cot cot cot 2 or 3 cot



– cot



=



_



_



_





cot cot 1 cot . cot 4

or 3 cot2



_{– cot}



_cot



_{+ 3 cot}



_{. cot}



_{– cot}



_{. cot}



_{= 4 cot}



_{. cot}



_{– 4}

(31)

or 4(1 + cot2



_{) = (cot}



_{+ cot}



_)(cot



_{+ cot}



₎ or 4













































3 2 2 1 2 2 t 1 t 1 t 1 t 1 t 1 1 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











c b a

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

Solution :

Let AD be perpendicular from A on BC. When AD is produced, it meets the circumscribing circle at E.

From question, DE =



.

Since, angle in the same segment are equal,



AEB =



ACB = C and



AEC =



ABC = B From the right angled triangle BDE,

tan C = DE BD

. . . (i) From the right angled triangle CDE,

tan B = CD DE . . . (ii) B A C B C 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,

_

a



_

b



_

c = 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 that a b c abc

x

  

y z xyz .

Solution:

Let H be the orthocentre. Then



BHC = 1800_–



HBC –

_

HCB = 1800_{– (90}0_{– C) – (90}0_{– B)}

(32)



ar(

_

BHC) = 2 1 BH. CH sin

_

BHC = 2 1 yz sin (



– A) = 2 1 yzsinA. A B _C H x z y

Similarly, ar(

_

CHA) =

2 1 zx sin B ar(

_

AHB) = 2 1 xy sin C



ar(

_

ABC) = 2 1 yz sin A + 2 1 zx sin B + 2 1 xy sin C = 2 1 xyz















z C sin y B sin x A sin =















z c y b x a R 2 1 . xyz 2 1 . . . (i)

Also, we know that R =



4 abc , i.e.,

_

= R 4 abc



(i) gives, R 4 abc = R 4 1 xyz















z c y b x a



_xa



_yb



_zc



_xyzabc. Example 10:

Prove that in a



ABC, R

_

2r..

Solution:

We have

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



R 4

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 R 4 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 r₁, r₂, r₃ inradius be r and circumradius be R. Then we have to prove that r₁ + r₂ + r₃ = r + 4R.