Page # 20 SOLUTION OF TRIANGLES

EXERCISE – IV HINTS & SOLUTIONS

Sol.1 Rr (sinA + sin B + sin C) =  Aliter :

a b c  2 A B
L.H.S. = r     = rS =  L.H.S. = 1 +  cos  sin2  + cos2 C
2 2 2  2 2 2

C A B C
Sol.2 2R cosA = 2R + r – r1 = 1 + sin cos   + 1 – sin2
2  2  2
A sin B sin C  cos B cos C 
R.H.S. = 2R + 4R sin 
2  2 2 2 2  C cos A  B   sin C 
= 2 + sin  
2   2  2
A B  C 
= 2R – 4R sin cos  
2  2  C cos A  B   cos A  B 
= 2 + sin  
2   2   2 
A B  C 
= 2R – 4R sin cos  
2  2  C A B
= 2 + sin . 2 sin sin
2 2 2
 2 A
= 2R 1  2 sin  = 2RcosA = L.H.S. A r
 2 = 2 + 2sin =2+ = R.H.S.
2 2R

A B C s2 Sol.5
Sol.3 cot + cot + cot =
2 2 2 
A B C
tan tan tan
L.H.S.= 2 + 2 + 2
s(s  a) s(s  b) s(s  c) (s  b)(s  c) (s  a)(s  c) (s  a)(s  b)
L.H.S. = + +
  

s 3r 3rs
= (s – a + s – b + s – c) = =
 (s  a)(s  b)(s  c) s(s  a)(s  b)(s  c)

3 3
s s2 = =
= (3s – 2s) = 2

 

Sol.6 r1 = r + r2 + r3
A B C r
Sol.4 cos2 + cos2 + cos2 =2+    
2 2 2 2R  = + +
sa s sb sc
1  cos A 1  cos B 1  cos C
L.H.S. = + + 1 1 1 1
2 2 2  – = +
sa s sb sc
3 1
= + (cosA + cosB + cos C) a a
2 2  =
s(s  a) (s  b)(s  c)
3 1 A s2 – as = s2 – (b + c) s + bc
= + (1 + 4 sin )
2 2 2 bc = s (b + c – a)
2bc = (a + b + c) (b + c – a)
r
=2+ = R.H.S. 2bc = (b + c)2 – a2
2R

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 SOLUTION OF TRIANGLES Page # 21

2bc = b2 + c2 + 2bc – a2 Sol.9 B = 3C

a2 = b2 + c2  A + B + C = 180º
ABC is right angled triangle A + 3C + C = 180º{ C  first quadrant}
A + 4C = 180º
by sine rule
Sol.7 2(2R)2 = a2 + b2 + c2
8R2 = a2 + b2 + c2 a b

c
=
2 = sin2 A+ sin2B + sin2 C sin A sin3C sin C

2 = sin2A +1– cos2B + sin2C b sin C = c sin 3C

1 = sin2A – cos (B + C) cos (B – C) sin C [b – 3C – 4c sin2C] = 0
b – 3c + 4 c sin2C = 0 {sin C  0}
cos2A = cos A cos (B – C)
cosA [cos (B – C) – cos A] = 0 3c  b
sin2 C =
 cos A [cos (B – C) + cos (B + C)] = 0 4c
2 cos A cos B cos C = 0
3c  b cb
A = 90º or B = 90º or C = 90º  cos2 C = 1 – =
4c 4c

Sol.8  = ABD + ACD

bc
 cos C = {cos C > 0}
4c
A
A  A + 4C = 180°  = 90º – 2C
2
A
c b sin = cos 2C = 2cos2 C –1
2

b  c 
ra =2  –1= bc
ra  4 c  2c
B D C
a
3c
Sol.10 BD =
4
In BCD sine rule
1 1
 = c ra + b ra
2 2
b
A
ra 2 2
 = (b + c) ra =
2 bc D
b
2 2 c 2
rb = , rc =
ca ab

1 1 1 B a C
R.H.S. = r + r + r
a b c

(b  c)  (c  a)  (a  b) 2(a  b  c) b 3c
c 2b
= = 2 = 4  =
2 2 sinC 3
1 sinC

2s 2
= = c b
 r {ABC sine rule =
sin C sin B

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 Page # 22 SOLUTION OF TRIANGLES

0 = cot  – 3 cot 2

b 2b 3
=  sin B = cos  = 3 cot 2
sinB 3 2 tan 2 = 3 tan 
B = 60º or 120º
2 tan 
But B  60º  B = 120º { B > 90º}  = 3 tan 
1  tan2 

Sol.11 In ADB sine rule tan  [2 – 3 + 3 tan2] = 0

tan [3tan2 –1] = 0
D q
C
 1
 tan  0 or tan2 =
3
p
2

q
2
P 1
tan = ±   = 30º, 150º
3
180º(  ) 
A B 5 
{  = not possible}  =
6 6
AB p2  q2
=
sin  sin(180º  ) Sol.13 AC = d2, BD = d1
R2 = 25 R1 = 12.5
In ABC, sine rule
sin  p2  q2 sin  p2  q2
AB = =
sin(  ) sin  cos   cos  sin 

A
p2  q2 sin 
 q   p   
=   sin     cos 
 2 2 
p q   2 2 
  p q  90º
B O D

(p2  q2 ) sin 
=
p cos   q sin 

C
BD 1
Sol.12 =
CD 3
Use M–N theorem d1
= 2.(12.5)
sin 2
d1 = 25 sin2
A
& In ACD, by sine rule

   d2
= 2.(25)
sin(180  2)
d2 = 50 sin2
d2 = 2d1
90º 90º2 Sine Rule, in AOB
B 9 D 9 9 C
4 4 2  d1   d2 
   
2   2  d1 d2
=  =
sin  sin(90º) sin  cos 
(1 + 3) cot 90º = 1.cot – 3 cot 2

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 SOLUTION OF TRIANGLES Page # 23

d1 1 sin2 A  sin2 B  sin2 C

tan = d  tan = =
2 2 2 sin2 C

1 101 sin2 C  sin2 C

2  =
2 2 sin2 C
 sin2 = 2
1
1  101  1 100
2 = = = 50
2 2

4
sin2 = B  C 
5 A
Sol.15 BIC = 180º –   = 90º +
 2  2
4
d1 = 25 × A
5
d1 = 20 & d2 = 40 A A
1 1 2 2
Area of ABCD = d . d2 = 20 . 40
2 1 2
= 400 sq. unit I
180º
Sol.14 a2 + b2 = 101 c2
B/2 C/2
or sin2A + sin2B = 101 sin2C
B a C
Now

cos C B
cot C sinC BI = r cosec
= cos A cosB 2
cot A  cot B 
sin A sinB C
CI = r cosec
2
cos C sin A sinB In BIC, Sine Rule

sinC sinC(A  B)
C B
a r cos ec r cos ec
cos C(2 sin A sin B) = 2 = 2 = 2x
=  A B C
2 sin2 C sin 90º  sin sin
 2 2 2
cos C[cos(A  B)  cos(A  B)]
=
2 sin2 C a r r
 = = = 2x
A B C B C
cos sin sin sin sin
cos C[cos(A  B)  cos C] 2 2 2 2 2
=
2 sin2 C
2a r
 =
 cos(A  B) cos(A  B)  cos2 C A A A B C
= 2 sin cos sin sin sin
2 sin2 C 2 2 2 2 2

2a
 (cos2 A  sin2 B)  1  sin2 C  = 4R  a = 2R sinA
= sin A
2 sin2 C
a 2R sin A
sin2 B  (1  cos2 A)  sin2 C) 2x =  2x =
A A
= cos cos
2 sin2 C 2 2

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 Page # 24 SOLUTION OF TRIANGLES

A a b
 x = 2R sin Sol.17 (i) =
2 cos A cos B

B C
III'y y = 2R sin & z = 2R sin sin A sin B
2 2  =
cos A cos B
L.H.S. = 4R3 – R(x2 + y2 + z2) – xyz
 tanA = tan B  A=B
A B C
= 4R3 – 4R3 (sin2 + sin2 + sin2 ) now 2 sinA cos B = 2 sinB cos A
2 2 2
= sin (A + B) – sin (A – B)
A B C  2 sinA cos B = sin C A=B
– 8R3 sin sin sin
2 2 2 (ii) 2 sinA cosB = sinC
sin (A + B) + sin (A – B) = sinC
A B C
= 4R3 [cos2 – sin2 – sin2 ] sin (A – B) = 0  A = B
2 2 2
A B
A B C  tan tan =1
– 8R3 sin sin sin 2 2
2 2 2

 A B A B A A A C C A
2 C  tan2 + tan tan + tan tan = 1
= 4R3 cos  2  cos  2   sin 2  2 sin 2  2 2 2 2 2
     

A A C
 C A B  A  B  A  tan2 + 2 tan tan –1=0
2 2 2
=4R3 sin 2 cos  2   cos  2    2 sin 2 
      
A A C
(iii) tan2 + 2 tan tan –1=0
 A B C A 2 2 2
= 4R3 2 sin sin sin  2 sin  = 0= RHS
 2 2 2 2
2
 A C C
  tan  tan  – tan2 –1=0
 2 2  2
cos A  2 cos C sin B
Sol.16 =
cos A  2 cos B sin C
 sinC cos A + 2 sinC cos C
A C
= sin B cos A + 2 sin B cos B sin2  
  2  = sec2 C
 cosA (sin B – sin C) + sin2B – sin 2C = 0 A C 2
cos2 cos2
2 2
B  C  B  C 
 cosA 2cos   sin  
 2   2 

+ 2 cos (B + C) sin (B – C) = 0 B
cos2
 2 = 1  cos2 B = cos2 A
B  C   B  C   B  C  A 2 2
2cosA sin   cos   2 cos  = 0 cos2
 2    2   2  2

B  C   B = A
cos A = 0 or sin   =0
 2   tan B = tan A

 sin A sinB a b
 A  or B – C = 0  B = C  =  =
2 cos A cosB cos A cos B

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 SOLUTION OF TRIANGLES Page # 25

cos 3A + cos (B – C) = 0
1 1 a
Sol.18  = ap1 p = 3A + B – C = 180º
2 1 2 2A + (A + B) – C = 180º
2A + 180º – C – C = 180º
2A = 2C  A = C
A
ABC is isosceles

b
p2 Sol.20 Power of point
C CD × DE = AD × DB
p3
p1 AD  DB
DE =
B a C CD

1 b 1 c A
III'y p = , p =
2 2  3 2 
E b
D
1 1 1 a b c
L.H.S = p + p – p = + – C/2
1 2 3 2 2 2
C/2
B a C
abc (a  b  c)(a  b  c)
= =
2 2(a  b  c)

(a  b)2  c2 (a2  b2  c2 )  2ab

= =
2(a  b  c) 2(a  b  c)
Now

2ab cos C  2ab ab(1  cos C) CD CD (CD)2  AD b


= = = = =
2(a  b  c) (a  b  c) DE  AD  DB  AD  DB  DB a
 
 CD 
C
2ab cos2
= 2 = R.H.S.
(a  b  c) 2
 2ab  c
  cos2 4a2b2 cos2
c
ab 2 2
= =
Sol.19 loga2 = logb2 + logc2 – log (2bc cos A)  bc   ac  abc 2
  
loga2 = logb2c2 – log (b2 + c2 – a2) ab ab
 log a2 (b2 + c2 – a2) = log b2c2
 a2b2 + a2c2 – a4 = b2c2 4ab c
 b2 (a2 – c2) – a2 (a2 – c2) = 0 = 2 cos2
c 2
 (a2 – c2) (b2 – a2) = 0
a = c or b = a CE CD  DE CD
= = +1
ABC is isosceles DE DE DE
Aliter :
4ab C
= cos2 +1
bc c2 2
2a2 bc cos A = b2c2  cosA =
2a2
4ab s(s  c) 2s(2s  2c)
= +1= +1
sin B sin C c2
ab c2
 cos A =
2 sin2 A
 2 sin2A sinA = 2 sinB sin C (a  b)2  c2  c2 (a  b)2
= =
 cosA – cos3A = cos (B – C) – cos (B + C) c2 c2

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 Page # 26 SOLUTION OF TRIANGLES


Sol.21 OM = r cos (i) A2
2 = A1B1 is true
n
2 1 1
(ii) B = A + B is also True
rue
2 1 1

O
R 2
Sol.22 r = & A = 90º = B + C
  r 1 2 3
r
n n
A B
M r1 A B C 2 3
= 4 sin cos cos =
P N Q R 2 2 2 2


AB = 2r sin
n 1 B C 2 3
4 cos cos =
2 2 2 2
1   r cos  
A1 = n . . 2r sin  
2 n  n
B   B 
   2.2 cos cos    = 2 + 3 {B+ C = }
A1 = nr2 sin cos 2 4 2 2
n n

PN  
Now = tan    
ON n = 2 cos 4  cos B  4   = 2 + 3
  

PN = r tan
n
 
  2 + 2 cos  B   = 2 + 3
1   4 
B1 = n . . 2  r tan  . r
2  n

   3
B1 = nr2 tan cos  B   =
n  4 2
III'y
   
A2 = 2nr2 sin cos B– =
2n 2n 4 6

 5 
A2 = nr2 sin B= ,C=
n 12 12

B2 = 2nr2 tan
2n b
= tan B = 2 + 3
c

1 A
Sol.23 = sin
AP 2
O

r  r AP = cosec
A
2n 2
2n

A
AI = cosec +1+r
2

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 SOLUTION OF TRIANGLES Page # 27

A Sol.24 ABC = 18 , a=c

r1 = 1
BDF = 2
P
DF = 2 2

1
I  = 18 = ac sinB  a2 sin B = 36 ...(i)
2

R
Q 1
B r2 = 4 r2 = 9 C BDF = 2 = a cos B × c cos B sin B
2
A A a2 cos2 B sin B = 4
r cosec = cosec +1+r
2 2
4 1
cos2B = = ...(ii)
A r 1 36 9
cosec =
2 r 1
1
cos B = ±
B r4 C r9 3
III'y cosec = & cosec =
2 r4 2 r 9
1 8
sin2B = 1 – =
A A 9 9
cot2 = cosec2 – 1
2 2
2 2 sin B  0
sinB =  B  I & II quad.
2 2 4r 3 
(r  1)  (r  1)
= =
(r  1) 2 (r  1)2
BDF 2 1
 ABC = 18 = 9 then ratio between sides
A 2 r
cot =
2 (r  1)
DF 1
is =
AC 3
B 4 r C 6 r
III'y cot = , cot =
2 (r  4) 2 (r  9) 1
2 2
 =  b=6 2
b 3
A A
cot =  cot
2 2 b
by Sine Rule , = 2R
sin B
2 r 4 r 6 r 48r r
 + + =
(r  1) (r  4) (r  9) (r  1)(r  4)(r  9)
6 2
(r2 – 13r + 36) + 2(r2 – 10r + 9) b 2 2 
R = = 2 
+3 (r2 – 5r + 4) = 24r 2 sin B  3 
 
6r2 – 48r + 66 = 24r
 r2 – 8r + 11 = 4r
9
r2 – 12r + 11 = 0 R=
2
(r – 11) (r – 1) = 0
r = 11 or r = 1
r = 11 r  1

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