Solution Of Triangles Theory Notes Math.pdf 4b211o

Solutions of Triangles

Theory Notes - Solutions of Triangles 1.

ELEMENTS OFATRIANGLE In a triangle ABC the angles are denoted by capital letters A, B and C and the length of the sides opposite to these angles are denoted by small letters a, b and c. Semi perimeter of the triangle is given by s = a  b  c and its area is denoted by  . 2 SINE RULE In a triangle ABC,

a b c   = 2R (where R is circum radius) sin A sin B sin C

Note : Area of triangle =

1 1 1 bc sinA = ac sinB = ab sinC. 2 2 2

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Drill Exercise - 1

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2.

2.

If in a ABC,

3.

ABCD is a trapezium such that AB and CD are parallel and CB is perpendicular to them. If ADB = 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

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1.

a 2 sin( B  C) b 2 sin(C  A ) c 2 sin( A  B) In any triangle ABC, prove that + + =0 sin B  sin C sin C  sin A sin A  sin B

St

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sin A sin(A  B) = , prove that a2, b2, c2 are in A. P.. sin C sin(B  C)

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

1  cos  . 7  cos 

5.

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

3.

COSINE RULE

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In a triangle ABC, (i) cos A =

b2  c2  a 2 2bc

(ii) cos B =

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c2  a 2  b2 2ca

Solutions of Triangles

(iii) cos C =

a 2  b2  c2 2ab

Drill Exercise - 2 cos A cos B cos C a 2  b 2  c 2 + + = . a b c 2abc

1.

In any ABC, prove that

2.

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

3.

In a triangle ABC, C = 60º, then prove that

4.

If in a triangle ABC,

1 (m  1)(m  3) , where m (1, 3). 2 m Prove that there are two values of the third side one of which is m times the other. 1 1 3 + = ac bc a bc

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cos A  2 cos C sin B = prove that the triangle is either isosceles or right cos A  2 cos B sin C

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

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5.

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angled.

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Illustration 1:

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Find the angles of the triangle whose sides are 3 + Solution:

=

66 3 12 2

3,b=2 3,c= 

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Let a = 3 +

1 3 2 2

4.



= cos 1050

Applying Sine formula : b a



6

3,2

6.



b 2  c 2  a 2 12  6  9  3  6 3  cos A = 2bc 12 2



A = 1050

a b  , we get sin A sin B

 

1 2 3 sin 1050 =  B = 450 3 3 2



sin B = sin A 



A = 1050, B = 450, C = 300

PROJECTION FORMULAE (i) a = b cos C + c cos B

(ii) b = c cos A + a cos C

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

2 = 2b.

Solution:

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3 and

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(iii) c = a cos B + b cos C

Solutions of Triangles

5.

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. NAPIER’S ANALOGY (TANGENT RULE) A  BC bc cot  2  2  bc

B CA ca cot  2  2  ca

(i) tan

(ii) tan 

C  A  B a  b cot  2  2  ab

(iii) tan 

6.

HALF ANGLE FORMULAE

(a)

(i) sin

A  2

(iii) sin

C  2

(i) cos

A = 2

(i) tan

1.

ca

s  a s  b  ab

ss  b  ca

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ss  a  bc

B = 2

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(ii) cos

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s s  c C  2 ab

A = 2

 s  b  s  c  s s  a 

C

s  a s  b  ss  c 

(iii) tan 2 

 s  c  s  a 

B  2

(ii) tan

B = 2

s  c s  a  ss  b 

ud

(c)

bc

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(iii) cos

(ii) sin

St

(b)

s  b s  c

Drill Exercise - 3

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

(ii) sin

A 2

(iii) cos

A 2

(iv) tan

A 2

2.

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

3.

In any ABC, prove that

4.

If are the lengths of the altitudes of a ABC, prove that

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sin B c  a cos B = sin C b  a cos C

1 1 1 cot A  cot B  cot C where  is the area of ABC. 2 + 2 + 2 =    

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Solutions of Triangles

5.

In a ABC, prove that cot

7.

AREA OF TRIANGLE

A B C abc A  cot . + cot + cot = 2 2 2  bca  2

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

7.1

(i) sin A = (iii) sin C =

8.

2 2Δ s  s  a  s  b  s  c   bc bc

(ii) sinB=

2 2Δ s  s  a  s  b  s  c   ab ab

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

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 DAC =  . Then

(ii)

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(m + n) cot  = m cot  – n cot  (m + n) cot  = n cot B – m cot C

(i)

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CENTROID AND MEDIANS OF A TRIANGLE The line ing 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.

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9.

2 2Δ s  s  a  s  b  s  c   ca ca

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St

Illustration 3: Find the lengths of the medians and the angles made by the medians with the sides of a triangle ABC. Solution : a2 AD = AC + CD – 2AC. CD cos C = b + – ab cosC, 4 2

and

2

2

2

c2 = b2 + a2 – 2ab cos C.

Hence 2AD2 – c2 = b2 – so that AD =

a2 , 2 A

1 1 2 2b 2  2c 2  a 2 = b  c 2  2bc cos A 2 2

  F E

Similarly, BE =

G

1 1 2c 2  2a 2  b 2 , and CF = 2a 2  2b 2  c 2 2 2

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B

 D

C

Solutions of Triangles

If BAD   and CAD  , we have sin  DC a   sinC AD 2AD 

sin  =

a sin C a sin C  2AD 2b 2  2c 2  a 2

Similarly sin  =

a sin B 2 b 2  2c 2  a 2

Drill Exercise - 4 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

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1.

1 (cot C ~ cot B) 2 Prove that the distance between the middle point of BC and the foot of the perpendicular from A is b2 ~ c2 . 2a

10.

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St

5.

ud

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

CIRCUM CIRCLE The circle which es 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. A

E

F

O A

A

B a/2

D

a/2

C

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Solutions of Triangles

Radius of the circumcircle is given by the following formulae 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, R2 and R3 are the circumradii

R=

a

b

c

abc

of the triangles OBC, OCA and OAB respectively, then prove that R  R  R  3 . R 1 2 3 Solution: Clearly, in the  OBC,  BOC = 2A, OB = OC = R, BC = a. 

2R1 =

a {using sine rule in  BOC) sin 2A

Similarly, 2R2 =

A

b c and 2R 3  sin 2B sin 2C

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a b c  R  R  R = 2(sin2A + sin2B + sin 2C) 1 2 3

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= 2.4 sin A sin B sin C,

O

B

C

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a b c abc . .  3 . =8 2R 2R 2R R

a

b

c

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

Page 6 of 35

St

Solution :

ud

that x  y  z  4xyz .

a

Let M be the circumcenter. MD  BC. So BD = DC = and  BMD = A. 2 BD In  BDM, = tan A or MD b

a 2 = tan A, i.e., a = tan A, 2x x

c

A

F

Similarly, 2 y = tan B, = tan C 2z

E

z M x

A



a b c tan A + tan B + tan C = 2x  2y  2z

and

tan A. tan B. tan C = 2x . 2 y . 2z

a

b

B

c

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

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y

D

C

Solutions of Triangles

a b c abc    x y z 4xyz



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 ing the feet of these perpendiculars is called the pedal triangle i.e.  DEF is the pedal triangle of  ABC. A

F E P B

900 – C C

D

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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)

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Similarly

c cos B cos C = 2R cos B cos C sin C

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= AB cos B cot C =

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

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Again

A

AP = AE sec DAC = c cos A cosec C c cos A = 2 R cos A sin C

F

St

=

E P

so, BP = 2R cos B and = 2R cos C

B

90 – C C

D

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

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Let O be the circumcenter and P be the orthocenter of the  ABC If OF be perpendicular to AB, we have

OAF  900  C Also

PAL  900  C



OAP  C – B

A F P O

B

Also OA = R and PA = 2R cosA OP2 = OA2 + PA2 – 2OA. PA cos OAP 

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D

K

L

C

Solutions of Triangles

= 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 

12.

OP = R 1 8 cos A cos B cos C BISECTORS OF THE ANGLES If AD bisects the angle A and divide the base into portions x and y, we have, by Geometry, x AB c   y AC b



x y xy a    c b bc bc

ac ab and y = bc bc Also let  be the length of AD we have  ABD +  ACD =  ABC

x=

A



1 A 1 A 1 c sin  b sin  bc sin A, 2 2 2 2 2

i.e.,





bc sin A 2bc A  cos b  c sin A b  c 2 2

B

x

y

C

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Drill Exercise - 5

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

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

2.

In any ABC, prove that a cosA + b cosB + c cosC = 4R sinA sinB sinC.

3.

If p1, p2 p3 are respectively the perpendiculars from the vertices of a triangle to the opposite sides,

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1.

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a 2 b 2c 2 prove that p1p2p3 = . 8R 3 4.

In a ABC, 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 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.

Page 8 of 35

A

E

r I

900– B/2

r

B/2

B

D

F r

C/2

C

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Solutions of Triangles

Radius of the Incircle is given by the following formulae r=

 A B C A B C = (s – a) tan = (s – b) tan = (s – c) tan = 4R sin sin sin . s 2 2 2 2 2 2

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. OAF  900  C. 

A

OAI  IAF  OAF



F



A A ABC CB  =  900  C   C  2 2 2 2

E

O

Also,

B

R

2

 1  16 sin 2

B C B C B C B 2C sin – 8sin sin  cos cos  sin sin  2 2 2 2 2 2 2 2

ud

OI 2

B 2 C B C CB sin – 8R2 sin sin cos 2 2 2 2 2

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= R2 + 16R2 sin2 

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OI 2  OA 2  AI 2  2OA. AI cos OAI

te



IE r B C   4R sin sin A A 2 2 sin sin 2 2

C

ps

AI =

I

1 – 8 sin

B C B C B C sin  cos cos  sin sin  2 2 2 2 2 2

=

1 – 8 sin

B C A sin sin 2 2 2



OI  R 1  8 sin

St

=

. . . (i)

B C A sin sin . 2 2 2

Illustration 9: If the distances of the vertices of a triangle ABC from the points of s of the incircle with sides be  ,  and  , then prove that

r2 

  

Solution: Let the incircle touches the side AB at P, where AP =  . Let I be the incentre. From the right-angled  IPA, 

Page 9 of 35

r A A B C  tan ;    r cot similarly,  = r cot and   r cot  2 2 2 2

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Solutions of Triangles

In  ABC, we have the identity cot

A B C A B C  cot  cot  cot cot cot 2 2 2 2 2 2 A



         . . r r r r r r

 P

or

1        13  r r



 r2 =      .

E I

C

B

Illustration 10: Show that the line ing the in-centre to the circumcenter of a triangle ABC is inclined to the side BC  cos B  cos C  1  at an angle tan 1  .  sin C  sin B 

IL IM  LM IM  ON    OL BM  BN BM  NC

r  R cos A B r cot  R sin A 2

te

tan  

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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

St

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A B C sin sin  R cos A 2 2 2  A B C B 4R sin sin sin .cot  R sin A 2 2 2 2 4R sin



 14.

Page 10 of 35

A I O B

L

NM

C

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

 cos B  cos C  1    tan 1  .  sin C  sin B 

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 = R 2  2Rr .

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Solutions of Triangles

(iii)

The distance between incentre and orthocenter is 2r 2  4R 2 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 4R sin

B C C A A B sin , 4R sin sin , 4R sin sin respectively.. 2 2 2 2 2 2 A B C cot cot . 2 2 2

In a ABC, prove that ratio of the area of the incircle to that of the triangle is  : cot

4.

Prove that a cot A  b cot B  c cot C  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 I1 and r1 respectively. Radii of the excircles are given by the following formulae A

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3.

(ii)

(iii) 16.

Page 11 of 35

r2 =

r3 =

 A A B C  s tan  4R sin cos cos sa 2 2 2 2

ud

r1 =

B

D1

C

F1

 B A B C  s tan  4R cos sin cos sb 2 2 2 2

St

(i)

E1 L

 C A B C  s tan  4R cos cos sin . sc 2 2 2 2

I1 M

EXCENTRAL TRIANGLE The triangle formed by ing the three excentres I1, I2 and I3 of 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

A B , 4 R cos and 2 2

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Solutions of Triangles

4 R cos (iv)

 A  B  C C and its angles are – , – and – . 2 2 2 2 2 2 2

II1 = 4 R sin

A B C : II2 = 4 R sin : II3 = 4 R sin 2 2 2

Illustration 11 : If the exradii r1, r2 and r3 of a  ABC are in HP, show that its sides a, b and c are in A.P.. Solution: We know that r1 =

   , r2  , r3  sa sb sc

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

1.

ps

Drill Exercise - 7

.in



Prove the following :

te

r  r1  r2  r3 1 abc 1 (A  B) C  (r1  r2  r3  r )   (a  b) sec sec 4rs 4 4 cos C 4 2 2 2 2 2 2 (r  r )(r2  r )(r3  r ) r s  1 1  1 1  1 1  a b          1  2c sin( A  B) 4  r r1  r r2  r r3  4r 2

ud

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R

A

 (b  c) tan 2   4 sin A 4 cos A

St

 a cos A

2.



r  sin A

2 sin A

Prove the following : A A A  s(s  a ) tan  cos bc(s  b)(s  c) 2 2 2 1 sin A sin B 2  Rr (sin A  sin B  sin C)  (b 2 sin 2C  c 2 sin 2B)  (a  b 2 ) 4 2 sin( A  B)

  rr1r2 r3  rr1 cot



a 2  b2  c2 (abc) 2 / 3  (sin 2A  sin 2B  sin 2C)1 / 3 5/3 4(cot A  cot B  cot C) 2

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, R2 and R3 be the diameter of the excircles of a ABC (opposite to the vertices A, B and C

Page 12 of 35

respectively), then prove that

R  R2  R3 a b c    1 . R1 R 2 R 3 abc

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Solutions of Triangles

5.

Prove that r 2  r12  r2 2  r3 2  16R 2  (a 2  b 2  c 2 ) .

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.

Page 13 of 35

If two sides b and c and the included angle A are given, then tan

using

BC bc A BC  cot , we get . 2 bc 2 2

Also

BC 2

.in

(ii)

A , so that B and C can be evaluated. 2

= 90 0 –

b sin A . sin B

te

The third side is given by a =

ps

(i)

b2  c2  a 2 If the sides a, b and c are given, then cos A = . B and C can be obtained in the 2bc similar way.

yS

c If two sides b and c and the angle B (opposite to side b) are given, then sin C = sin B, A = b b sin A give the remaining elements. If b < c sin B, there is no triangle sin B 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).

ud

1800 – (B + C) and a =

St

(iii)

A

A

c

c b

B

(Fig 1)

c sinB

D

B

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b c sinB

D (Fig 2)

Solutions of Triangles A

A b

c b

b D

B

C2

b

c sinB

c

c sinB

C2

C1

B

C1

(Fig 4)

(Fig 3)

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.

b 2  c 2  a 2 100  36  196 1     A  1200 cos A = 2bc 120 2

ps



.in

Solution: Let a = 14, b = 10, c = 6  The largest angle is opposite the largest side.

te

Illustration 13:

yS

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





2

ud

2 a 2  b2  c2 1 1 3  4  c cos C = .  2ab 2 2 1  3 .2

Also,

c2 = 6 

St





c=



6

sin A sin B sin C   a b c



3 sin A sin B   2  2 1 3 6



A = 1800 – (450 + 600) = 750

sin B =

1  B = 450 2

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 sin Solution: Given b.c = k2

Page 14 of 35

A . 2

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Solutions of Triangles

b2  c2  a 2 Now cosA = 2bc

A 2

 k2  2 or 2k cosA = b +    a b   4 2 2 or b – (a + 2k cosA). b2 + k4 = 0 Since b2 is real, (a2 + 2k2) (a2 + 2k2 cosA – 2k2)  0 2

2



 2 2 2 A  2 2 2 A  a  2k .2 cos   a  2k .2sin 0 2  2 



 2 2 2 A  2 2 2 A  a  4k cos   a  4k sin 0 2 2 



a 2  4k 2 sin 2



A  A   a  2k sin   a  2k sin   0 2 2 



a   2k sin

b

c

C

B a

.in

A A  0 [since a2 + 4k2 cos2 is always positive] 2 2

ps

A A or a  2k sin (since 2ksin(A/2) is real) 2 2

yS

A . 2

ud

Hence a  2 k sin

te

A But a must be positive.  a  – 2k sin is rejected 2

DRILL EXERCISE - 8

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

St

1.

areas is 4.

1 2 2 b sin A. 2

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

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

18.

Page 15 of 35

3 1 : 3 1

a 2  b 2 sin 2 A

INSCRIBED & CIRCUMSCRIBED POLYGONS

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Solutions of Triangles

(Important Formulae) I.

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

II.

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

III.

Side of Inscribed polygone = 2r sin

IV.

Side of Circumscribed polygone = 2r tan

1 2 2 nr sin 2 n

1 2  nr tan 2 n

 . n

 . n

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

1  2   BOL     2 n  n If a be a side of the polygon, we have

.in

2 n

ps

BOC 

te



 a   r  cot . A n 2 n Now the area of the regular polygon = n times the area of the

ud

R

R B

L

C

St

1 1 a  na 2  OBC  n  OL.BC   n . cot .a  cot . 2 2 n 4 n 2 

D

O

yS

a = BC = 2BL = 2RsinBOL = 2Rsin

DRILL EXERCISE - 9

1.

If a, b, c, d are the sides of a quadrilateral described about a circle then prove that A C ad sin 2  bc sin 2 . 2 2

2.

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

3.

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

4.

With reference to a given circle, A1 & B1 are the areas of the inscribed and circumscribed regular

Page 16 of 35

polygons of n sides, A 2 & B 2 are corresponding quantities for regular polygons of 2n sides. Prove that A 2 is a geometric mean between A1 & B1 and B2 is a hormonic mean between A 2 & B1 . www.StudySteps.in

Solutions of Triangles

Answer Key Drill Exercise - 1 3.

25 3 43 3 Drill Exercise - 2

5.

313 338

Drill Exercise - 3 (i) 84

(ii)

1 5

(iv)

1 2

ps

Drill Exercise - 6 a = 13, b = 15, c = 14

te

1.

2 5

(iii)

.in

1.

yS

Drill Exercise - 8

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

St

ud

1.

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 = a2 – c2 (b) 3a2 = b2 – 3c2 (c) b2 = a2 – c2 (d) a2 + b2 = 5c2

Page 17 of 35

Solution: From the right angled  CAD, we have

A 90 c

b 2b a 2  b 2  c 2   a/2 a 2ab 2 2 2 2 a + b – c = 4b  a2 – c2 = 3b2.

cos C =

A–

0

b

/2

B a/2

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D

a/2

C

Solutions of Triangles

Example 2: There exists a triangle ABC satisfying sin A sin B sin C   2 3 7

(a) tanA + tanB + tanC = 0

(b)

(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 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 1  a 2  b2  c2 If (a + b) = c + ab, then = = cos C = , which is possible 2 3 2ab Hence (c) is the correct answer. 2

2

yS

te

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

ps

.in

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

c b

  and hence C = . 2 2

St

We get A + B =

tan A  tan B 1  tan A tan B

ud

Since tan (A + B) =

A

B

a

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 2

2

2

sin A + sin B + sin C =

a2 c2



b2 c2

1 

a 2  b2 c2

 1  1  1  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

Page 18 of 35

Solution: If p1, p2, p3, are altitude from A, B, C respectively,

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C

Solutions of Triangles

1

1

1

2

2

2

then  = ap1 = bp2 = 3  p1 = , p2 = , p3 = 2 2 2 a b c By the law of sines a b c   = k (say) sin A sin B sin C

2Δ 2Δ 2 , p2 = , p3 = k sin C k sin A k sin B Now, sin A, sin B, sin C are in A.P.   p1 =

p1, p2, p3 are in H.P.

(a)

25 9

(b)

25 3

(c)

25 18

(d)

10 3

.in

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

ps

Solution: Let O be the point of intersection of the medians of triangle ABC. Then the area of  ABC is three 2

10

yS

te

times that of  AOC. Now, in  AOC, AO = AD = . Therefore, applying the sine rule to 3 3  AOC, we get

1

ud

OC AO 10 sin  / 8   OC  . sin  / 8 sin  / 4  3 sin  / 4 

St

area of  AOC = . AO.OC. sin  AOC 2 =

=

1 10 10 sin  / 8    . . . .sin   2 3 3 sin  / 4 2 8

B

50 sin  / 8cos / 8 50 25 .   9 sin  / 4  18 9

25 25   area of  ABC = 3. 9 3

E

D O A

/8

/4

C

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

Page 19 of 35

Solution:

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Solutions of Triangles

We have tan

cot A / 2 cot B / 2   1  0 A  B C AB  = cot = tan  90  = cot A / 2  cot B / 2 2  2 2 

6 37 . 1 222  100 122 2 5 20    = 6 37 120  185 305 5  5 20

Also tan 

A C tan = 2 2

5 2 sb .  6 5 s



s  b s  c  s  a s  b  ss  a  ss  c  3s  b   s

 2s  3b

 a + b + c = 3b  a + c = 2b

Which shows that a, b and c are in A.P.

ps

.in

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:

63 tan2

St

Now



te

AB AB 1  1  tan  2 2 63

ud



AB 1  tan 2 31 2   32 1  tan 2 A  B 2

yS

AB 2 cos (A – B) = 2 AB 1  tan 2 1  tan 2

tan

AB ab C 1 54 C  cot   cot 2 ab 2 2 63 5  4

tan

C = 2

Also, cos C =

63 9 1  tan 2 C / 2 

1  tan C / 2  2



1  63 / 81 18 1   1  63 / 81 144 8

c2 = a2 + b2 – 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

Page 20 of 35

(a)

5 4

(b)

4 5

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Solutions of Triangles

(c)

7 4

(d)

4 7

Solution: From the given relation, we have s tan

A B C = 2s tan = 3s tan 2 2 2

tan(A / 2) tan(B / 2) tan(c / 2)    k (say) 6 3 2

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

A B B C C A tan  tan tan  tan tan  1 2 2 2 2 2 2

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

2 tanB / 2



12k 1  36k 2

6k

1  tan B / 2 1  9k 2

2



1

.in

and sin B =

1  tan 2  A / 2 



4 5

ps



2 tan  A / 2  sin A =

k = 1/6

a:b=5:4

Example 9:

Page 21 of 35

yS



a sin A 5   b sin B 4

ud



te

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

St

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

a2 is equal to 8

(a) m 22  m 32  2m12

(b) m12  m 22  2m 32

(c) m 22  m 32  2m12

(d) none of these

Solution: In  ABC, AD2 = m12 =

A

c2  b2 a 2  2 4

a 2 2   2 2 In  ABD, AE = m2 = AD  c   2  2 4

2

B

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E

D

F

C

Solutions of Triangles

a 2 2   AD  b 2   AF2 = m32 = 2 4  m22 + m32 = AD2 +



2

b2  c2 a 2 a2 a2 a2   m12  m12    2m12  2 8 4 8 8

m 2 2  m32  2m12 

a2 8

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

(b) sin

A B C : sec : sec 2 2 2

(d) none of these

ps

Solution: Here BD : DC = c : b But BD + DC = a;

te

c .a bc

yS

 BD =

A B C : sin : sin 2 2 2

.in

(c) sec

A B C : cosec : cosec 2 2 2

B

St

ca sin B 2 A .  cos ec  AD = b  c sin A b  c 2 2

Also,

AI AB c bc    ID BD ca / b  c  a



AI =



A

I

ud

BD AD  In  ABD, sin A sin B 2

D

bc  A  B  C .AD  cos ec Similarly BI = cos ec , CI  cos ec abc s 2 s 2 s 2

IA : IB : IC  cos ec

A B C : cos ec : cos ec 2 2 2

Example 11:

Page 22 of 35

In a  ABC, the value of (a)

R r

C

a cos A  b cos B  c cos C is equal to abc

(b)

R 2r

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Solutions of Triangles

r R Solution:

(c)

(d)

2r R

a cos A  b cos B  ccos C 2R sin A cos A  2R sin Bcos B  2R sin Ccos C  a bc 2s

= But

4R abc abc R R . 3 .sin 2A  sin 2B  sin 2C  = .4 sin A sin B sin C = 2s 2s 2s 8R 4sR 2 R=

abc  ,r= 4 s

So, the value =

4R r   4. .R 2 R r

  cos 5 10

(b)

te

2  cos ec 5 10 Solution:

(d) none of these

yS

(c)

1 1 . r . r. sin 720 = r2 cos 180 2 2

2r 2 2

360 0 = 720 5

St

 ar (  AOB) =

ud

In the  OAB, OA = OB = r and  AOB =

 A1 : A2 =

2  sec 5 10

ps

(a)

.in

Example 12: The area of a circle is A1and the area of a regular pentagon inscribed in the circle is A2. Then A1 : A2 is

0

5r cos18



D

E

C

O r

2  sec 5 10

A

B

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

Page 23 of 35

(a) 2 13

(b)

5 13 12

5 13 3

(d)

3 13 2

(c)

Solution:

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Solutions of Triangles

AG =

A

2 2 A A1, BG = BB1 3 3

 AG =

1 3

and BG =

1 2a 2  2c 2  b 2 3

 AG =

1 1 2 a , BG  b  4c 2 3 3

 AG =

5 1 2 , BG  16  36  13 3 3 3

B1

2b 2  2c2  a 2

G

B

A1

C

as a2 = b2 + c2

1 Also, AB = c = 3 and  GAB   ABC  2 3 If ‘R1’ be the circumradius of triangle GAB then

AG BG AB  5 . 2 4 GAB

3 3

13.3.

1 4 .2

.in

R1 =

ps

5 13 units. 12 Example 14: A variable triangle ABC 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)

St

ud

yS

te

=

  = 2 s(s – a)  2



2 s

2



2(s  a )  1 (radius of incircle of triangle ABC) s

a = constant. s

Now  =

1 aH , where ‘Ha’ is the distance of ‘A’ from BC. 2 a

 1 aH a 2s  = 1  Ha = = constant s 2 s a  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

Page 24 of 35

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Solutions of Triangles

(a)

D

B

3  r  d

(b)   r 2  d 2

(c)   r 3  d 3

(d)

O

2  r  d

C A

Solution:  AC = d, OA = OB = r , CD = BD =  ,  COA = 3  AC2 = OA2 + OC2 – 2AOOC. cos

 d2 = 2r2 – 2r2 .

 3

1 = r2 2

D

B

2   Also,  BOD =  COD = 3.2 3

O C

 BD    r 3d 3 3 OB r Hence the correct answer is (c)

A

ps

.in

 tan

te

SOLVED SUBJECTIVE EXAMPLES

Page 25 of 35

St

ud

yS

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  =

n  12  n  22  n 2 2n  1n  2  n 2   n  1   n  2  2

and cos 2  =

2n  n  1

 2

n  1n  5 n  5 n 2  6n  5 = 2n  1n  2   2n  2  2n  1n  2  n 2  2n  3 n  1n  3 n  3 = = 2n n  1  2n 2n n  1 2

 n5  n 3  2 But cos 2  = 2 cos  – 1; so  1 2n  2n  2   2

or

n  3 n  52   1 or (n – 3)(n+2)2 = n{(n + 5)2 – 2 (n + 2)2} 2 2n 2n  2 

or or or

(n – 3)(n2 + 4n + 4) = n (– n2 + 2n + 17) n3 + n2 – 8n – 12 = – n3 + 2n2 + 17n (n – 4)(2n2 + 7n + 3) = 0 n = 4 or 2n2 + 7n + 3 = 0. 

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Solutions of Triangles

Roots of 2n2 + 7n + 3 = 0 are i.e., – 

 7  49  24 4

1 and – 3 which are not natural numbers. 2

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

B C , tan are rational. 2 2

(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.

Page 26 of 35

Now, (i) 

a, b, c,  , s are rational.

yS

numbers are rational. Thus (i) 

te

B C and tan are rational because sum, difference, product and quotient of nonzero rational 2 2

(ii).

B C , tan be rational 2 2

St

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

ud

So tan

ps

.in

B s  c s  a  C s  a s  b  abc  , tan  and s = 2  2  2

Now, tan

B 2 = rational, because tan B is rational. Now, sin B = B 2 1  tan 2 2 2 tan

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

B C s  c s  a  s  a s  b   s  a   s  b  s  c  s  a a .   1 . Now, tan . tan = = 2 2   s  s  a  s  b  s  c  s s 2



But

(ii)

 

s is rational b + c is rational, because a is rational.

a b c   sin A sin B sin C



a bc rational   sin A sin B  sin C rational

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Solutions of Triangles 

a is rational. But a is rational. So sinA is rational sin A

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

a b c   sin A sin B sin C



b

a sin B = rational sin A

and

c=

a sin C = rational sin A



Thus (iii)

1  = 2 bc sin A = rational.



(i).

Example 3: Solution : AB 2 4 A B 5 1  tan 2 2

But tan 

tan

AB 1  2 3

or tan2

AB 1  2 9

( A > B).

St

or

AB 2  54 2 54

2 tan 2

ud

By componendo and dividendo,

te



ps

1  tan 2

yS

4 Here, cos (A – B) = , 5

4 , find the area of the triangle. 5

.in

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

AB ab C  cot  2 ab 2

The area of the triangle =

1 63 C  cot 3 63 2

or cot

C  = 1;  C = . 2 2

1 1  ab sin C  .6.3. sin  9 sq. units. 2 2 2

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 a2b2 sin2 C = a2p2 + b2q2 – 2abpqcos C. Solution : Let  ACE =  . Clearly, from the figure, we get

Page 27 of 35

p q  sin ,  sin   C  AC BC

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p

Solutions of Triangles



p q  sin ,  sin . cos C  cos .sin C b a



q p  cos C  cos .sin C a b

or

q 2 p2 2pq p2  p2  2 2  cos C  cos C  1   1   cos C a 2 b2 ab b2  b2  q2

or

a

2



p2 b

2



2

q p  2 2   cos C   cos .sin C = a b  

or

In a  ABC, prove that cos A. cos C =





2 c2  a 2 , where AD is the median through A and 3ca

Page 28 of 35

b2  c2  a 2 2bc

AC

b

. . . (i)

2b

te

  From the  CAD, cos C = CD a / 2 a

ps

From the  ABC, cos A =

.in

AD  AC. Solution:

sin(A  90 )



AB sin ADB

ud

0

a/2 c a c    cos A sin 900  C or 2 cos A cos C



cos A=



from (i),

or

b2 + c2 – a2 = – 2b2



St



a cos C a 2b b  .  ,  2c  2c a c

from (ii)

Now, cosA. cosC =

B



D

. . . (iii)

b 2  c 2  a 2 2b b 2  c 2  a 2 .  2bc a ca





=

3b 2  3 c 2  a 2 a 2  c2  3 c2  a 2  , 3ca 3ca

=

2 c2  a 2 . 3ca



A

b2  c2  a 2  b  2bc c

c2 – a2 = – 3b2



. . . (ii)

yS

BD

or

or



2pq cos C  sin 2 C or a2p2 + b2q2 – 2abpqcosC = a2 b2sin2C. ab

Example 5:

From the  ABD,



 p2  1   1  cos 2 C  b2   

from (iii)



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C

Solutions of Triangles

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

Similarly P, D, B and F lie on a circle and therefore  PDF =  PBF = 900 – A, Hence  FDE = 1800 – 2A Similarly  DEF = 1800 – 2B  EFD = 1800 – 2C Also, from the triangle AEF we have A

EF AE AB cos A c cos A c cos A     sin A sin AFE cos PFE cos PAE sin C 

F E

c sin A cos A = a cosA EF = sin C

P 90 – C

B

C

D

.in

similarly DF = b cosB and DE = c cosC

yS

  1 1  1 1  1        4 1  2   t1 t 2  t 2 t 3   t2 

te

ps

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:

Page 29 of 35

ud

Solution: Let the points P and Q divide the side BC in three equal parts such that BP = PQ = QC = x A Also let,  BAP =  ,  PAQ =  ,  QAC = 

St



 

and  AQC =  From question,

tan  = t1, tan  = t2, tan  = t3. 

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

B

P

(2x + x) cot  = 2x cot (  +  ) – x cot  from  APC, we get

. . . (i)

(x + x) cot  = x cot  – x cot  dividing (i) by (ii), we get

. . . (ii)

2 cot      cot  3 = or cot   cot  2 or or

3 cot  – cot  =

Q

4cot . cot   1 cot   cot 

3 cot2  – cot  cot  + 3 cot  . cot  – cot  . cot  = 4 cot  . cot  – 4 4 + 4 cot2  = cot2  + cot  . cot  + cot  . cot  + cot  .cot 

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C

Solutions of Triangles

or

4(1 + cot2  ) = (cot  + cot  )(cot  + cot  )

or

 1   1 1  1 1  4 1  2    t  t  t  t  3  t 2   1 2  2

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

a b c   = 2 (tan A + tanB + tan C).   

yS

St

ud

CD tan B = DE

Adding (i) and (ii) we get, tan B + tanC = Similarly tan C + tan A =

Hence,

. . . (i)

te

BD DE From the right angled triangle CDE,

tan C =

A

ps

 AEB =  ACB = C and  AEC =  ABC = B From the right angled triangle BDE,

.in

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,

D C

B

. . . (ii)

C

B

E

a 

b C and tan A + tan B =  

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 a b c abc that    . x y z xyz Solution: Let H be the orthocentre. Then 0  BHC = 180 –  HBC –  HCB = 1800 – (900 – C) – (900 – B) = B + C =  – A.

Page 30 of 35

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Solutions of Triangles 

ar(  BHC) = =

1 BH. CH sin  BHC 2

1 1 yz sin (  – A) = yzsinA. 2 2

A x

1 Similarly, ar(  CHA) = zx sin B 2

H y

1 ar(  AHB) = xy sin C 2 

ar(  ABC) = =

B

1 xyz 2

 sin A sin B sin C      y z   x

1

1 a

b

c

.in

abc abc , i.e.,  = 4 4R

. . . (i)

ps

Also, we know that R =

te

 a b c abc 1 = xyz  x  y  z   4R 4R  

a b c abc    x y z xyz .

yS

(i) gives,

C

1 1 1 yz sin A + zx sin B + xy sin C 2 2 2

= 2 xyz. 2R  x  y  z   



z

ud

Example 10:

St

Prove that in a  ABC, R  2r.. Solution: We have r = 4R sin A/2 sin B/2 sin C/2



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

1 Also we know that sin A/2 sin B/2 sin C/2  , 8



r 1   4R 8

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.

Page 31 of 35

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Solutions of Triangles

Now, r1 + r2 + r3 – r =  1

Δ Δ Δ Δ    sa sb sc s 1 

1  1

=   s  a  s    s  b  s  c      

a



a

=   ss  a   s  b s  c     = a = a

=

s  s  a  s  b  s  c 

2s 2  sa  b  c   bc 2





a abc abc   . 2s 2  2s 2  bc   4R  R     4  

r1 + r2 + r3 = r + 4R.

.in



s 2  s  b  c   bc  s 2  as

St

ud

yS

te

ps

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/2 Now, cot A/2 – sinA, cotB/2 – sin B, cotC/2 – sin C are in A.P. as cotA/2, cotB/2, cotC/2 are in A.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 a  ABC, prove that 8r R {cos2A/2 + cos2 B/2 + cos2 C/2} = 2bc + 2ca + 2ab – a2 – b2 – c2. Solution:

Page 32 of 35







    abc  abc 2 2 cos 2 A / 2   cos A / 2 = L.H.S. = 8  .  s  s   4 

=

abc  1  cos A  s

abc  b 2  c 2  a 2  = s  1   2bc  

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

Solutions of Triangles

abc  2bc  b 2  c 2  a 2  = s   2bc  

abc  b  c 2  a 2  = s   2bc  



=

abc  a  b  c b  c  a   ,   s 2bc 

=

abc 2s  b  c  a     2bc  = s

where a + b + c = 2s

 a b  c  a    ab  bc  a 2 

= 2bc  2ca  2ab  a 2  b 2  c 2 8rR{cos2A/2 + cos2 B/2 + cos2C/2} = 2bc + 2ab + 2ca – a2 – b2 – c2 .

t12

1



t2

2



1 t3

2



abc abc

te

1

ps

.in

Example 14: If ‘t1’, ‘t2’ and ‘t3’ are the lengths of the tangents drawn from centre of ex-circle to the circum circle of the  ABC, then prove that

ud

yS

Solution: Let S and I1 be respectively the centres of the circumcircle and the excircle touching BC. It can be shown that In  SI1 P , SI12 = R2 + t12

St

SI1  R 2  2Rr1

R2 + 2Rr1 = R2 + t12 ,

1

t12

A

1  2Rr1

R P

S

C

B

Similarly 1  1 , 1  1 t 22 2Rr2 t 32 2Rr3

I1

1 1 1 1 1 1 1   2 2      2 t1 t2 t3 2R  r1 r2 r3  = =

1 2R

s  a s  b s  c  1 s s        = 2R   2R

abc abc

proved

Example 15: If a, b and A are given in a triangle and c1, c2 are the possible values of the third side, prove that

Page 33 of 35

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Solutions of Triangles

c12  c 22  2c1 c 2 cos A = 4a2 cos2 A

Solution:

Page 34 of 35

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

We have



 and

cosA =

c1  c 2  2b cos A   c1c 2  b 2  a 2 

. . . (i)

c12  c 22  2c1c 2 cos 2A

  

(c1 + c2)2 – 2c1c2 – 2c1c2 cos 2A [using (i)] 2 (c1 + c2) – 2c1c2 (1 + cos 2A) 4b2 cos2A – 2(b2 – a2). 2cos2A = 4a2 cos2A



c12  c 22  2c1c 2 cos A  4a 2 cos 2 A

St

ud

yS

te

ps

.in



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Page 35 of 35

St

ud

yS

te

ps

.in

Solutions of Triangles

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