NURTURE COURSE

SOLUTIONS OF TRIANGLE
CONTENTS
SOLUTIONS OF TRIANGLE

THEORY & ILLUSTRATIONS ...................................... Page – 01

ELEMENTARY EXERCISE ............................................ Page – 16

EXERCISE (O-1) ............................................................. Page – 18

EXERCISE (O-2) ............................................................. Page – 20

EXERCISE (S-1) .............................................................. Page – 22

EXERCISE (S-2) .............................................................. Page – 23

EXERCISE (JA) .............................................................. Page – 24

ANSWER KEY .................................................................. Page – 26

JEE (ADVANCED) SYLLABUS :

Solutions of Triangle : Relations between sides and angles of a triangle, sine rule,
cosine rule, half-angle formula and the area of a triangle.
 Solutions of Triangle
ALLEN
SOLUTIONS OF TRIANGLE
The process of calculating the sides and angles of triangle using given information is called solution
of triangle.
In a DABC, the angles are denoted by capital letters A, B and C and the length of the sides opposite
these angle are denoted by small letter a, b and c respectively.
1. SINE FORMULAE : A
In any triangle ABC
b
c
h
a b c abc
= = =l= = 2R
sin A sin B sin C 2D B C
D a
where R is circumradius and D is area of triangle.

Illustration 1 : Angles of a triangle are in 4 : 1 : 1 ratio. The ratio between its greatest side and perimeter
is

3 3 3 1
(A) (B) (C) (D)
2+ 3 2+ 3 2- 3 2+ 3
Solution : Angles are in ratio 4 : 1 : 1.
Þ angles are 120°, 30°, 30°.
If sides opposite to these angles are a, b, c respectively, then a will be the greatest side.
a b c
Now from sine formula = =
sin120° sin 30° sin 30°
a b c
Þ = =
3 / 2 1/ 2 1/ 2
a b c
Þ = = = k (say)
3 1 1
then a = 3k , perimeter = (2 + 3)k

3k 3
\ required ratio = = Ans. (B)
(2 + 3)k 2 + 3
Illustration 2 : In triangle ABC, if b = 3, c = 4 and ÐB = p/3, then number of such triangles is -
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(A) 1 (B) 2 (C) 0 (D) infinite

sin B sin C
Solution : Using sine formulae =
b c

sin p / 3 sin C 3 sin C 2

Þ = Þ = Þ sin C = > 1 which is not possible.
3 4 6 4 3
Hence there exist no triangle with given elements. Ans. (C)
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Illustration 3 : The sides of a triangle are three consecutive natural numbers and its largest angle is twice
the smallest one. Determine the sides of the triangle. A
Solution : Let the sides be n, n + 1, n + 2 cms.
n n+1
i.e. AC = n, AB = n + 1, BC = n + 2
Smallest angle is B and largest one is A. C n+2 B
Here, ÐA = 2ÐB
Also, ÐA + ÐB + ÐC = 180°
Þ 3ÐB + ÐC = 180° Þ ÐC = 180° – 3ÐB
We have, sine law as,
sin A sin B sin C sin 2B sin B sin(180 - 3B)
= = Þ = =
n+2 n n +1 n+2 n n +1

Þ sin 2B sin B sin 3B

= =
n+2 n n +1
(i) (ii) (iii)
from (i) and (ii);
2 sin Bcos B sin B n+2
= Þ cos B = ......... (iv)
n+2 n 2n
and from (ii) and (iii);
sin B 3sin B - 4 sin 3 B sin B sin B(3 - 4 sin 2 B)
= Þ =
n n +1 n n +1
n +1
Þ = 3 - 4(1 - cos 2 B) .......... (v)
n
from (iv) and (v), we get
2
n +1 æ n+2ö n +1 æ n 2 + 4n + 4 ö
= -1 + 4 ç ÷ Þ +1 = ç ÷
n è 2n ø n è n2 ø
2n + 1 n 2 + 4n + 4
Þ = Þ 2n2 + n = n2 + 4n + 4
n n2
Þ n2 – 3n – 4 = 0 Þ (n – 4)(n + 1) = 0
n = 4 or – 1
where n ¹ –1
\ n = 4. Hence the sides are 4, 5, 6 Ans.
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Do yourself - 1 :
p
(i) If in a DABC, ÐA = and b : c = 2 : 3 , find ÐB .
6
(ii) Show that, in any DABC : a sin(B – C) + b sin(C – A) + c sin(A – B) = 0.

sin A sin(A - B)
(iii) If in a DABC, = , show that a2, b2, c2 are in A.P.
sin C sin(B - C)

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 Solutions of Triangle
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2. COSINE FORMULAE :

b2 + c2 - a 2 c2 + a 2 - b2 a 2 + b2 - c2
(a) cos A = (b) cos B = (c) cosC =
2bc 2ca 2ab
or a2 = b2 + c2 – 2bc cosA

Illustration 4 : In a triangle ABC, if B = 30° and c = 3 b, then A can be equal to -

(A) 45° (B) 60° (C) 90° (D) 120°

c2 + a 2 - b2 3 3b 2 + a 2 - b 2
Solution : We have cos B = Þ =
2ca 2 2 ´ 3b ´ a

Þ a2 – 3ab + 2b2 = 0 Þ (a – 2b) (a – b) = 0

Þ Either a = b Þ A = 30°

or a = 2b Þ a2 = 4b2 = b2 + c2 Þ A = 90°. Ans. (C)

Illustration 5 : In a triangle ABC, (a2 –b2 – c2 ) tan A + (a2 – b2 +c2) tan B is equal to -
(A) (a2 + b2 –c2) tan C (B) (a2 + b2 + c2) tan C
(C) (b2 + c2 –a2) tan C (D) none of these
Solution : Using cosine law :
The given expression is equal to –2 bc cos A tan A + 2 ac cos B tan B

= 2abc æç -
sin A sin B ö
+ ÷= 0 Ans. (D)
è a b ø

Do yourself - 2 :
(i) If a : b : c = 4 : 5 : 6, then show that ÐC = 2ÐA.
(ii) In any DABC, prove that

cos A cos B cos C a 2 + b2 + c 2

(a) + + =
a b c 2abc
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b2 c2 a2 a 4 + b4 + c4
(b) cos A + cos B + cos C =
a b c 2abc

3. PROJECTION FORMULAE :
(a) b cos C + c cos B = a (b) c cos A + a cos C = b (c) a cos B + b cos A = c

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A C 3b
Illustration 6 : In a DABC, c cos
2
+ a cos 2 = , then show a, b, c are in A.P.
2 2 2
c a 3b
Solution : Here, (1 + cos A) + (1 + cos C) =
2 2 2
Þ a + c + (c cos A + a cos C) = 3b
Þ a + c + b = 3b {using projection formula}
Þ a + c = 2b
which shows a, b, c are in A.P.

Do yourself - 3 :
p 5p
(i) In a DABC, if ÐA = , ÐB = , show that a + c 2 = 2b .
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æ C Bö
(ii) In a DABC, prove that : (a) b(a cosC – c cosA) = a2 – c2 (b) 2 ç b cos 2 + c cos2 ÷ = a + b + c
è 2 2ø

4. NAPIER'S ANALOGY (TANGENT RULE) :

æ B- C ö b -c A æC-A ö c-a B æ A-Bö a-b C

(a) tan ç ÷= cot (b) tan ç ÷= cot (c) tan ç ÷ = cot
è 2 ø b+c 2 è 2 ø c+a 2 è 2 ø a+b 2

Illustration 7 : In a DABC, the tangent of half the difference of two angles is one-third the tangent of
half the sum of the angles. Determine the ratio of the sides opposite to the angles.
æ A-Bö 1 æ A +Bö
Solution : Here, tan ç ÷ = tan ç ÷ ........ (i)
è 2 ø 3 è 2 ø
æ A-Bö a-b æCö
using Napier's analogy, tan ç ÷= .cot ç ÷ ........ (ii)
è 2 ø a+b è2ø
from (i) & (ii) ;

1 æ A +Bö a -b æCö
tan ç ÷= .cot ç ÷ Þ 1 cot æç C ö÷ = a - b .cot æç C ö÷
3 è 2 ø a+b è2ø 3 è 2 ø a+b è2ø

æ B+C ö æp Cö C
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{as A + B + C = p \tan ç ÷ = tan ç - ÷ = cot }

è 2 ø è2 2ø 2
a-b 1
Þ = or 3a – 3b = a + b
a+b 3
a 2 b 1
2a = 4b or = Þ =
b 1 a 2
Thus the ratio of the sides opposite to the angles is b : a = 1 : 2. Ans.

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Do yourself - 4 :
æ B-C ö
tan ç ÷
b-c è 2 ø
(i) In any DABC, prove that =
b+c æ B+ C ö
tan ç ÷
è 2 ø
A c-b a 2 - b2
(ii) If DABC is right angled at C, prove that : (a) tan = (b) sin(A - B) =
2 c+b a 2 + b2

5. HALF ANGLE FORMULAE :

a+b+c
s= = semi-perimeter of triangle.
2
A (s - b)(s - c) B (s - c)(s - a) C (s - a)(s - b)
(a) (i) sin = (ii) sin = (iii) sin =
2 bc 2 ca 2 ab
A s(s - a) B s(s - b) C s(s - c)
(b) (i) cos = (ii) cos = (iii) cos =
2 bc 2 ca 2 ab
A (s - b)(s - c) B (s - c)(s - a) C (s - a)(s - b)
(c) (i) tan = (ii) tan = (iii) tan =
2 s(s - a) 2 s(s - b) 2 s(s - c)
D D D
= = =
s(s - a) s(s - b) s(s - c)
(d) Area of Triangle
1 1 1 1 1 1
D = s(s - a)(s - b)(s - c) = bc sin A = ca sin B = ab sin C = ap1 = bp 2 = cp 3 ,
2 2 2 2 2 2
where p1,p2,p3 are altitudes from vertices A,B,C respectively.

Illustration 8 : If in a triangle ABC, CD is the angle bisector of the angle ACB, then CD is equal to-
a+b C 2ab C 2ab C bsin ÐDAC
(A) cos (B) sin (C) cos (D)
2ab 2 a+b 2 a+b 2 sin(B + C / 2)
Solution : DCAB = DCAD + DCDB
1 1 æCö 1 æCö
Þ absinC = b.CD.sin ç ÷ + a.CD sin ç ÷
2 2 è2ø 2 è2ø
æCö æ æCö æ C öö
Þ CD(a + b) sin ç ÷ =ab ç 2 sin ç ÷ cos ç ÷ ÷
è2ø è è2ø è 2 øø
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2ab cos(C / 2)
So CD =
(a + b)
CD b
and in DCAD, = (by sine rule)
sin ÐDAC sin ÐCDA
bsin ÐDAC
Þ CD = Ans. (C,D)
sin(B + C / 2)
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s2
Illustration 9 : If D is the area and 2s the sum of the sides of a triangle, then show D £ .
3 3
Solution : We have, 2s = a + b + c, D2 = s(s – a)(s – b)(s – c)
Now, A.M. ³ G.M.
(s - a) + (s - b) + (s - c)
³ {(s - a)(s - b)(s - c)}1 / 3
3
1/3
3s - 2s æ D 2 ö
or ³ç ÷
3 è s ø
1/3
s æ D2 ö
or ³ç ÷
3 è s ø

D2 s3 s2
or £ Þ D£ Ans.
s 27 3 3
Do yourself - 5 :
(i) Given a = 6, b = 8, c = 10. Find
A A A
(a) sinA (b)tanA (c) sin (d) cos (e) tan (f) D
2 2 2
A B C
(ii) Prove that in any DABC, (abcs) sin .sin .sin = D 2 .
2 2 2
6. m-n THEOREM : A

a
(m + n) cot q = m cot a – n cot b
b

h
(m + n) cot q = n cot B – m cot C.
q
B C
m D n

7. RADIUS OF THE CIRCUMCIRCLE 'R' :

A
Circumcentre is the point of intersection of perpendicular bisectors of
the sides and distance between circumcentre & vertex of triangle is R
c b
called circumradius 'R'. O
R R
a b c abc
R= = = = . B D a C
2 sin A 2 sin B 2 sin C 4 D
8. RADIUS OF THE INCIRCLE 'r' :
Point of intersection of internal angle bisectors is incentre and
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perpendicular distance of incentre from any side is called inradius 'r'.

A
D A B C A B C
r = = (s - a) tan = (s - b) tan = (s - c) tan = 4R sin sin sin .
s 2 2 2 2 2 2
I r
B C A C B A r
sin sin sin sin sin sin
=a 2 2 =b 2 2 =c 2 2 r
A B C B C
cos cos cos
2 2 2
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 Solutions of Triangle
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Illustration 10 : In a triangle ABC, if a : b : c = 4 : 5 : 6, then ratio between its circumradius and inradius
is-
16 16 7 11
(A) (B) (C) (D)
7 9 16 7
R abc D (abc)s R abc
Solution : = = Þ = ....(i)
r 4D s 4D2 r 4(s - a)(s - b)(s - c)

a b c
Q a:b:c=4:5:6 Þ = = = k (say)
4 5 6
Þ a = 4k, b = 5k, c = 6k
a + b + c 15k 7k 5k 3k
\ s= = ,s–a= ,s–b= ,s–c=
2 2 2 2 2
R (4k)(5k)(6k) 16
using (i) in these values = = Ans. (A)
r æ 7k ö æ 5k ö æ 3k ö 7
4 ç ÷ ç ÷ ç ÷
è 2 ø è 2 ø è 2 ø
r
Illustration 11 : If A, B, C are the angles of a triangle, prove that : cosA + cosB + cosC = 1 + .
R
æ A+Bö æ A-Bö
Solution : cosA + cosB + cosC = 2 cos ç ÷ .cos ç ÷ + cos C
è 2 ø è 2 ø
C æ A-Bö 2 C C é æ A -Bö æ C öù
= 2 sin .cos ç ÷ + 1 - 2 sin = 1 + 2 sin êcos ç ÷ - sin ç ÷ ú
2 è 2 ø 2 2ë è 2 ø è 2 øû
C é æ A-Bö æ A + B öù ì C æ A + B öü
= 1 + 2 sin ê cos ç ÷ - cos ç ÷ íQ = 90° - ç ÷ý
2ë è 2 ø è 2 ø úû î 2 è 2 øþ
C A B A B C
= 1 + 2 sin .2 sin .sin = 1 + 4 sin .sin .sin
2 2 2 2 2 2
r
= 1+ {as, r = 4R sin A/2 . sinB/2 . sinC/2}
R
r
Þ cosA + cosB + cosC = 1 + . Hence proved.
R

Do yourself - 6 :
(i) If in DABC, a = 3, b = 4 and c = 5, find
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(a) D (b) R (c) r

(ii) In a DABC, show that :
a 2 - b2 A B C D abc
(a) = 2R sin(A - B) (b) r cos cos cos = (c) a + b + c =
c 2 2 2 4R 2Rr
(iii) Let D & D' denote the areas of a D and that of its incircle. Prove that
æ A B Cö
D : D' = ç cot .cot .cot ÷ : p
è 2 2 2ø

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9. RADII OF THE EX-CIRCLES : A
Point of intersection of two external angles and one internal angle bisectors c b
is excentre and perpendicular distance of excentre from any side is called B a C
exradius. If r1 is the radius of escribed circle opposite to ÐA of DABC and
r1 r1
so on, then - I1
B C
a cos cos
D A A B C 2 2
(a) r1 = = s tan = 4R sin cos cos =
s-a 2 2 2 2 A
cos
2
A C
b cos cos
D B A B C 2 2
(b) r2 = = s tan = 4R cos sin cos =
s-b 2 2 2 2 B
cos
2
A B
D C A B C c cos 2 cos 2
(c) r3 = = s tan = 4R cos cos sin =
s-c 2 2 2 2 C
cos
2
I1, I2 and I3 are taken as ex-centre opposite to vertex A, B, C repsectively.

b-c c-a a -b
Illustration 12 : Value of the expression + + is equal to -
r1 r2 r3
(A) 1 (B) 2 (C) 3 (D) 0
(b - c) (c - a) (a - b)
Solution : + +
r1 r2 r3

æs-a ö æs-bö æs-cö

Þ (b – c) ç ÷ + (c - a) ç ÷ + (a - b). ç ÷
è D ø è D ø è D ø
(s - a)(b - c) + (s - b)(c - a) + (s - c)(a - b)
Þ
D
s(b - c + c - a + a - b) - [ab - ac + bc - ba + ac - bc] 0
= = =0
D D
b-c c-a a-b
Thus, + + =0 Ans. (D)
r1 r2 r3
Illustration 13 : If r1 = r2 + r3 + r, prove that the triangle is right angled.
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Solution : We have, r1 – r = r2 + r3
D D D D s-s+a s-c +s-b
Þ - = + Þ =
s-a s s-b s-c s(s - a) (s - b)(s - c)
a 2s - (b + c)
Þ = {as, 2s = a + b + c}
s(s - a) (s - b)(s - c)
a a
Þ = Þ s2 – (b + c) s + bc = s2 – as
s(s - a) (s - b)(s - c)
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(b + c - a)(a + b + c)
Þ s(–a + b + c) = bc Þ = bc
2
Þ (b + c)2 – (a)2 = 2bc Þ b2 + c2 + 2bc – a2 = 2bc
Þ b2 + c2 = a2
\ ÐA = 90°. Ans.

Do yourself - 7 :
(i) In an equilateral DABC, R = 2, find
(a) r (b) r1 (c) a

(ii) In a DABC, show that

1 2 2 æ 1 1 öæ 1 1 ö æ 1 1 ö
(a) r1r2 + r2r3 + r3r1 = s2 (b) r s ç - ÷ç - ÷ ç - ÷ = R
4 è r r1 øè r r2 øè r r3 ø
(c) rr1 r2 r3 = D

10. ANGLE BISECTORS & MEDIANS :

An angle bisector divides the base in the ratio of corresponding sides. A

BD c ac ab c
= Þ BD = & CD = b
CD b b+c b+c
B C
D
If ma and ba are the lengths of a median and an angle bisector from the
angle A then,
A
2bc cos
1 2
ma = 2b2 + 2c 2 - a 2 and ba =
2 b+c

3 2
Note that m a + m b + m c = (a + b + c )
2 2 2 2 2

11. ORTHOCENTRE : A

(a) Point of intersection of altitudes is orthocentre & the triangle KLM

M L
which is formed by joining the feet of the altitudes is called the P
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pedal triangle.
B K C
(b) The distances of the orthocentre from the angular points of the DABC
are 2R cosA, 2R cosB, & 2R cosC.

(c) The distance of P from sides are 2R cosB cosC, 2R cosC cosA and
2R cosA cosB.

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Do yourself - 8 :
(i) If x, y, z are the distance of the vertices of DABC respectively from the orthocentre, then
a b c abc
prove that + + =
x y z xyz
(ii) If p1, p2, p3 are respectively the perpendiculars from the vertices of a triangle to the opposite
sides, prove that
a 2 b2 c 2 1
(a) p1p2p3 = 3 (b) D = Rp1p 2 p3
8R 2
(iii) In a DABC, AD is altitude and H is the orthocentre prove that AH : DH = (tanB + tanC)
: tanA
(iv) In a DABC, the lengths of the bisectors of the angle A, B and C are x, y, z respectively.

1 A 1 B 1 C 1 1 1
Show that cos + cos + cos = + + .
x 2 y 2 z 2 a b c

12. THE DISTANCES BETWEEN THE SPECIAL POINTS :

(a) The distance between circumcentre and orthocentre is = R 1 - 8cos A cos Bcos C

(b) The distance between circumcentre and incentre is = R 2 - 2Rr

(c) The distance between incentre and orthocentre is = 2r 2 - 4R 2 cos A cos Bcos C

(d) The distances between circumcentre & excentres are

A B C
OI1 = R 1 + 8sin cos cos = R 2 + 2Rr1 & so on.
2 2 2

Illustration 14 : Prove that the distance between the circumcentre and the orthocentre of a triangle ABC
is R 1 - 8cos A cos Bcos C .

Solution : Let O and P be the circumcentre and the orthocentre respectively. If OF is the perpendicular
to AB, we have ÐOAF = 90° – ÐAOF = 90° – C. Also ÐPAL = 90° – C.

Hence, ÐOAP = A – ÐOAF – ÐPAL = A – 2(90° – C) = A + 2C – 180°

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= A + 2C – (A + B + C) = C – B. A

Also OA = R and PA = 2RcosA. F

O
L
Now in DAOP,
P

OP2 = OA2 + PA2 – 2OA. PA cosOAP B

K C

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= R2 + 4R2 cos2 A – 4R2 cosAcos(C – B)

= R2 + 4R2 cosA[cosA – cos(C – B)]

= R2 – 4R2 cosA[cos(B + C) + cos(C – B)] = R2 – 8R2 cosA cosB cosC.

Hence OP = R 1 - 8cos A cos Bcos C . Ans.

13. SOLUTION OF TRIANGLES :

The three sides a,b,c and the three angles A,B,C are called the elements of the triangle ABC. When
any three of these six elements (except all the three angles) of a triangle are given, the triangle is
known completely; that is the other three elements can be expressed in terms of the given elements
and can be evaluated. This process is called the solution of triangles.

A (s - b)(s - c)
* If the three sides a,b,c are given, angle A is obtained from tan =
2 s(s - a)

b2 + c2 - a 2
or cos A = .B and C can be obtained in the similar way.
2bc

B-C b -c A
* If two sides b and c and the included angle A are given, then tan = cot gives
2 b+c 2

B-C B+C A
. Also = 90° - , so that B and C can be evaluated. The third side is given by
2 2 2

sin A
a=b
sin B

or a2 = b2 + c2 – 2bc cos A.
* If two sides b and c and an angle opposite the one of them (say B) are given then

c b sin A
sin C = sin B, A = 180° - (B + C) and a = given the remaining elements.
b sin B
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Case I :
A
b < c sin B.
c b csinB
We draw the side c and angle B. Now it is obvious from the
figure that there is no triangle possible. B

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Case II : A

b = c sin B and B is an acute angle, there is only one triangle possible. c b csinB
and it is right-angled at C.
B D

Case III : A
b > c sin B, b < c and B is an acute angle, then there are two triangles possible c
b b c sinB
for two values of angle C.
D
B C2 C1

b b csinB
Case IV : c

C2 B C1
b > c sin B, c < b and B is an acute angle, then there is only one triangle.

Case V : C

b > c sin B, c > b and B is an obtuse angle. b

For any choice of point C, b will be greater than c which is a
c A
contradication as c > b (given). So there is no triangle possible. B

Case VI :
b > c sin B, c < b and B is an obtuse angle. C

We can see that the circle with A as centre and b as radius will cut the
line only in one point. So only one triangle is possible. B c A
b

C
Case VII :
b > c and B = 90°.
Again the circle with A as centre and b as radius will cut the line only B c A
in one point. So only one triangle is possible. b
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Case VIII :
b < c and B = 90°.
The circle with A as centre and b as radius will not cut the line in any B c
A
b
point. So no triangle is possible.

This is, sometimes, called an ambiguous case.

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Alternative Method :

a 2 + c2 - b2
By applying cosine rule, we have cosB =
2ac

( c cos B) - ( c2 - b 2 )
2
Þ a2 – (2c cos B)a + (c2 – b2) = 0 Þ a = c cosB ±

b 2 - ( c sin B)
2
Þ a = c cosB ±

This equation leads to following cases :

Case-I : If b < csinB, no such triangle is possible.

Case-II: Let b = c sinB. There are further following case :

(a) B is an obtuse angle Þ cosB is negative. There exists no such triangle.

(b) B is an acute angle Þ cosB is positive. There exists only one such triangle.

Case-III: Let b > c sin B. There are further following cases :

(a) B is an acute angle Þ cosB is positive. In this case triangle will exist if and only if

b 2 - ( c sin B) or c > b Þ Two such triangle is possible. If c < b, only one

2
c cosB >
such triangle is possible.

(b) B is an obtuse angle Þ cosB is negative. In this case triangle will exist if and only if

b 2 - ( c sin B) > |c cos B| Þ b > c. So in this case only one such triangle is possible. If
2

b < c there exists no such triangle.

This is called an ambiguous case.

a sin B a sin C
* If one side a and angles B and C are given, then A = 180° – (B + C), and b = ,c = .
sin A sin A

* If the three angles A,B,C are given, we can only find the ratios of the sides a,b,c by using sine
rule (since there are infinite similar triangles possible).
node06\B0AH-AI\Kota\JEE(Advanced)\Nurture\Maths\Sheet\SOT\Eng.p65

Illustration 15 : In the ambiguous case of the solution of triangles, prove that the circumcircles of the two
triangles are of same size.
Solution : Let us say b,c and angle B are given in the ambiguous case. Both the triangles will
b
have b and its opposite angle as B. so = 2R will be given for both the triangles.
sin B
So their circumradii and therefore their sizes will be same.

E 13
JEE-Mathematics
ALLEN
Illustration 16 : If a,b and A are given in a triangle and c1,c2 are the possible values of the third side, prove
that c12 + c22 - 2c1c2 cos2A = 4a2cos2A.

b2 + c2 - a 2
Solution : cos A =
2bc
Þ c2 – 2bc cosA + b2 – a2 = 0.
c1 + c2 = 2bcosA and c1c2 = b2 – a2.
Þ c12 + c22 – 2c1c2cos2A = (c1 + c2)2 – 2c1c2(1 + cos2A)
= 4b2 cos2A – 2(b2 – a2)2 cos2A = 4a2cos2A.

æ A - A2 ö c sin B
Illustration 17 : If b,c,B are given and b < c, prove that cos ç 1 ÷= .
è 2 ø b

Solution : ÐC2AC1 is bisected by AD. A

A1–A2
c
æ A - A2 ö AD c sin B b b
Þ In DAC2D, cos ç 1 ÷= =
è 2 ø AC 2 b D
B C2 C1
Hence proved.
Do yourself - 9 :
æ A - A 2 ö a1 - a 2
(i) If b,c,B are given and b<c, prove that sin ç 1 ÷=
è 2 ø 2b
(ii) In a DABC, b,c,B (c > b) are gives. If the third side has two values a1 and a2 such
that

4b 2 - c 2
a1 = 3a2, show that sin B = .
3c 2
14. REGULAR POLYGON :
A regular polygon has all its sides equal. It may be inscribed p
or circumscribed. r n
h
(a) Inscribed in circle of radius r :
a
p p
(i) a = 2h tan = 2r sin
n n
(ii) Perimeter (P) and area (A) of a regular polygon of n sides inscribed in a circle of radius r
p 1 2p
are given by P = 2nr sin and A = nr 2 sin
n 2 n
(b) Circumscribed about a circle of radius r :
node06\B0AH-AI\Kota\JEE(Advanced)\Nurture\Maths\Sheet\SOT\Eng.p65

p
p n
(i) a = 2r tan r
n
a
(ii) Perimeter (P) and area (A) of a regular polygon of n sides
p
circumscribed about a given circle of radius r is given by P = 2nr tan and
n
p
A = nr 2 tan
n
14 E
 Solutions of Triangle
ALLEN
Do yourself - 10 :
(i) If the perimeter of a circle and a regular polygon of n sides are equal, then
p
area of the circle tan n
prove that = .
area of polygon p
n
(ii) The ratio of the area of n-sided regular polygon, circumscribed about a circle, to the area
of the regular polygon of equal number of sides inscribed in the circle is 4 : 3. Find the
value of n.

15. SOME NOTES :

(a) (i) If a cos B = b cos A, then the triangle is isosceles.
(ii) If a cos A = b cos B, then the triangle is isosceles or right angled.
(b) In right angle triangle
(i) a2 + b2 + c2 = 8R2 (ii) cos2 A + cos2 B + cos2 C = 1
(c) In equilateral triangle

3R
(i) R = 2r (ii) r1 = r2 = r3 =
2
3a 2 a
(iii) r : R : r1 = 1 : 2 : 3 (iv) area = (v) R =
4 3
(d) (i) The circumcentre lies (1) inside an acute angled triangle (2) outside an obtuse angled
triangle & (3) mid point of the hypotenuse of right angled triangle.
(ii) The orthocentre of right angled triangle is the vertex at the right angle.
(iii) The orthocentre, centroid & circumcentre are collinear & centroid divides the line segment
joining orthocentre & circumcentre internally in the ratio 2 : 1 except in case of equilateral
triangle. In equilateral triangle, all these centres coincide
(e) Area of a cyclic quadrilateral = (s - a)(s - b)(s - c)(s - d)

a+b+c+d
where a, b, c, d are lengths of the sides of quadrilateral and s = .
2

ANSWERS FOR DO YOURSELF

node06\B0AH-AI\Kota\JEE(Advanced)\Nurture\Maths\Sheet\SOT\Eng.p65

1: (i) 90°
3 3 1 3 1
5: (i) (a) (b) (c) (d) (e) (f) 24
5 4 10 10 3
5
6: (i) (a) 6 (b) (c) 1
2
7: (i) (a) 1 (b) 3 (c) 2 3
10 : (ii) 6

E 15
JEE-Mathematics
ALLEN
ELEMENTARY EXERCISE
b 3
1. Angles A, B and C of a triangle ABC are in A.P. If = , then ÐA is equal to
c 2
p p 5p p
(A) (B) (C) (D)
6 4 12 2
2. If K is a point on the side BC of an equilateral triangle ABC and if ÐBAK = 15°, then the ratio of
AK
lengths is
AB

(A)
(
3 2 3+ 3 ) (B)
(
2 3+ 3 ) (C)
(
2 3- 3 ) (D)
(
3 2 3- 3 )
2 2 2 2

3. In a triangle ABC, ÐA = 60° and b : c = ( )

3 + 1 : 2 then (ÐB – ÐC) has the value equal to
(A) 15° (B) 30° (C) 22.5 ° (D) 45°
4. In an acute triangle ABC, ÐABC = 45°, AB = 3 and AC = 6 . The angle ÐBAC, is
(A) 60° (B) 65° (C) 75° (D) 15° or 75°
5. Let ABC be a right triangle with length of side AB = 3 and hypotenuse AC = 5.
BD AB
If D is a point on BC such that = , then AD is equal to
DC AC
4 3 3 5 4 5 5 3
(A) (B) (C) (D)
3 2 3 4
4
6. In a triangle ABC, if a = 6, b = 3 and cos(A – B) = , the area of the triangle is
5
15
(A) 8 (B) 9 (C) 12 (D)
2
c
7. In DABC, if a = 2b and A = 3B, then the value of is equal to
b
(A) 3 (B) 2 (C) 1 (D) 3
p
8. If the sides of a triangle are sin a, cos a, 1 + sin a cos a , 0 < a < , the largest angle is
2
(A) 60° (B) 90° (C) 120° (D) 150°
node06\B0AH-AI\Kota\JEE(Advanced)\Nurture\Maths\Sheet\SOT\Eng.p65

9. If the angle A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the
lengths of the sides opposite to A, B and C respectively, then the value of expression
æa c ö
E = ç sin 2C + sin 2A ÷ , is
èc a ø

1 3
(A) (B) (C) 1 (D) 3
2 2

16 E
 Solutions of Triangle
ALLEN
10. If in a triangle sin A : sin C = sin (A – B) : sin (B – C), then a2, b2, c2
(A) are in A.P. (B) are in G.P. (C) are in H.P. (D) none of these
A b+c
11. In triangle ABC, if cot = , then triangle ABC must be
2 a
[Note: All symbols used have usual meaning in DABC.]
(A) isosceles (B) equilateral (C) right angled (D) isoceles right angled
12. Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B
and C respectively. If a = 1, b = 3 and C = 60°, then sin2B is equal to
27 3 81 1
(A) (B) (C) (D)
28 28 28 3

13. The ratio of the sides of a triangle ABC is 1: 3 : 2 . Then ratio of A : B : C is

(A) 3 : 5 : 2 (B) 1 : 3 :2 (C) 3 : 2 : 1 (D) 1 : 2 : 3

14. In triangle ABC, If s = 3 + 3 + 2 , 3B – C = 30°, A + 2B = 120°, then the length of longest

side of triangle is
[Note: All symbols used have usual meaning in triangle ABC.]
(A) 2 (B) 2 2 (C) 2( 3 + 1) (D) 3 -1
In a triangle tan A : tan B : tan C = 1 : 2 : 3, then a : b2 : c2 equals
2
15.
(A) 5 : 8 : 9 (B) 5 : 8 : 12 (C) 3 : 5 : 8 (D) 5 : 8 : 10
A C
16. In DABC, if a,b,c (taken in that order) are in A.P. then cot cot =
2 2
[Note: All symbols used have usual meaning in triangle ABC. ]
(A) 1 (B) 2 (C) 3 (D) 4
tan C
17. In DABC if a = 8, b = 9, c = 10, then the value of is
sin B
32 24 21 18
(A) (B) (C) (D)
9 7 4 5
18. In triangle ABC, if D = a – (b – c) , then tan A =
2 2

[Note: All symbols used have usual meaning in triangle ABC. ]

15 1 8 8
(A) (B) (C) (D)
16 2 17 15
æ cos A ö p
19. In a triangle ABC, if the sides a, b, c are roots of x3 – 11x2 + 38x – 40 = 0. If å çè ÷ = , then
a ø q
node06\B0AH-AI\Kota\JEE(Advanced)\Nurture\Maths\Sheet\SOT\Eng.p65

find the least value of (p + q) where p,q Î N.

20. ABC is a triangle such that sin (2A + B) = sin (C – A) = – sin (B + 2C) = 1 . If A, B, C are in A.P.,
2
find A, B, C.

E 17
JEE-Mathematics
ALLEN
EXERCISE (O-1)
1. A triangle has vertices A, B and C, and the respective opposite sides have lengths a, b and c. This
triangle is inscribed in a circle of radius R. If b = c = 1 and the altitude from A to side BC has
2
length , then R equals
3

1 2 3 3
(A) (B) (C) (D)
3 3 2 2 2
2. A circle is inscribed in a right triangle ABC, right angled at C. The circle is tangent to the segment
AB at D and length of segments AD and DB are 7 and 13 respectively. Area of triangle ABC is
equal to
(A) 91 (B) 96 (C) 100 (D) 104
3. In a triangle ABC, if a = 13, b = 14 and c = 15, then angle A is equal to
(All symbols used have their usual meaning in a triangle.)

-1 4 -1 3 -1 3 -1 2
(A) sin (B) sin (C) sin (D) sin
5 5 4 3

4. In a triangle ABC, if b = ( 3 - 1) a and ÐC = 30°, then the value of (A – B) is equal to

(All symbols used have usual meaning in a triangle.)

(A) 30° (B) 45° (C) 60° (D) 75°
5. In triangle ABC, if AC = 8, BC = 7 and D lies between A and B such that AD = 2, BD = 4, then
the length CD equals

(A) 46 (B) 48 (C) 51 (D) 75

6. In a triangle ABC, if ÐC = 105°, ÐB = 45° and length of side AC = 2 units, then the length of the
side AB is equal to

(A) 2 (B) 3 (C) 2 +1 (D) 3 +1

8a 2 b 2c 2
7. In a triangle ABC, if (a + b + c) (a + b – c) (b + c – a) (c + a – b) = 2 , then the triangle is
a + b2 + c 2
node06\B0AH-AI\Kota\JEE(Advanced)\Nurture\Maths\Sheet\SOT\Eng.p65

[Note: All symbols used have usual meaning in triangle ABC.]

(A) isosceles (B) right angled (C) equilateral (D) obtuse angled
8. In triangle ABC, if 2b = a + c and A – C = 90°, then sin B equals
[Note: All symbols used have usual meaning in triangle ABC.]
7 5 7 5
(A) (B) (C) (D)
5 8 4 3

18 E
 Solutions of Triangle
ALLEN
9. In a triangle ABC, a3 + b3 + c3 = c2 (a + b + c)
(All symbol used have usual meaning in a triangle.)
Statement–1: The value of ÐC = 60°.
Statement –2: DABC must be equilateral.
(A) Statement–1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
10. The sides of a triangle are three consecutive integers. The largest angle is twice the smallest one.
The area of triangle is equal to

5 15 15
(A) 7 (B) 7 (C) 7 (D) 5 7
4 2 4
11. The sides a, b, c (taken in that order) of triangle ABC are in A.P.
a b c æ aö ægö
If cos a = , cos b = , cos g = then tan 2 ç ÷ + tan 2 ç ÷ is equal to
b+c c+a a+b è 2ø è 2ø
[Note: All symbols used have usual meaning in triangle ABC. ]
1 1 2
(A) 1 (B) (C) (D)
2 3 3
p p
12. AD and BE are the medians of a triangle ABC. If AD = 4, ÐDAB = , ÐABE = , then area of
6 3
triangle ABC equals
8 16 32 32
(A) (B) (C) (D) 3
3 3 3 9
13. In triangle ABC, if sin3 A + sin 3 B + sin 3 C = 3sin A.sin B.sin C , then triangle is
(A) obtuse angled (B) right angled (C) obtuse right angled (D) equilateral
r
14. For right angled isosceles triangle, =
R
[Note: All symbols used have usual meaning in triangle ABC. ]
p p p p
(A) tan (B) cot (C) tan (D) cot
12 12 8 8
node06\B0AH-AI\Kota\JEE(Advanced)\Nurture\Maths\Sheet\SOT\Eng.p65

1 1 3
15. In triangle ABC, If + = then angle C is equal to
a+c b+c a+b+c
[Note: All symbols used have usual meaning in triangle ABC. ]
(A) 30° (B) 45° (C) 60° (D) 90°

E 19
JEE-Mathematics
ALLEN
EXERCISE (O-2)
Multiple Correct Answer Type :
1. In a triangle ABC, let 2a2 + 4b2+ c2 = 2a(2b + c), then which of the following holds good?
[Note: All symbols used have usual meaning in a triangle.]
-7
(A) cos B = (B) sin (A– C) = 0
8
r 1
(C) = (D) sin A : sin B : sin C = 1 : 2 : 1
r1 5
2. In a triangle ABC, if a = 4, b = 8 ÐC = 60°, then which of the following relations is (are) correct?
[Note: All symbols used have usual meaning in triangle ABC.]
(A) The area of triangle ABC is 8 3
(B) The value of å sin 2
A=2

2 3
(C) Inradius of triangle ABC is
3+ 3
4
(D) The length of internal angle bisector of angle C is
3
3. In which of the following situations, it is possible to have a triangle ABC?
(All symbols used have usual meaning in a triangle.)
(A) (a + c – b) (a – c + b) = 4bc (B) b2 sin 2C + c2 sin 2B = ab
2p æ A-Cö æA+Cö
(C) a = 3, b = 5, c = 7 and C =
3
(D) cos ç ÷ = cos ç ÷
è 2 ø è 2 ø
4. In a triangle ABC, which of the following quantities denote the area of the triangle?
a 2 - b2 æ sin A sin B ö r1r2 r3
(A) (B)
2 çè sin(A - B) ÷ø år r 1 2

a 2 + b2 + c2 A B C
(C) (D) r2 cot ·cot cot
cot A + cot B + cot C 2 2 2
5. In DABC, angle A, B and C are in the ratio 1 : 2 : 3, then which of the following is (are) correct?
(All symbol used have usual meaning in a triangle.)
(A) Circumradius of DABC = c (B) a : b : c = 1 : 3 : 2

3 2
(C) Perimeter of DABC = 3 + 3 (D) Area of DABC = c
8
node06\B0AH-AI\Kota\JEE(Advanced)\Nurture\Maths\Sheet\SOT\Eng.p65

6. Let one angle of a triangle be 60°, the area of triangle is 10 3 and perimeter is 20 cm. If a > b > c
where a, b and c denote lengths of sides opposite to vertices A, B and C respectively, then which
of the following is (are) correct?
(A) Inradius of triangle is 3 (B) Length of longest side of triangle is 7

7 1
(C) Circumradius of triangle is (D) Radius of largest escribed circle is
3 12
20 E
 Solutions of Triangle
ALLEN
7. In triangle ABC, let b = 10, c = 10 2 and R = 5 2 then which of the following statement(s) is
(are) correct?
[Note: All symbols used have usual meaning in triangle ABC.]
(A) Area of triangle ABC is 50.
(B) Distance between orthocentre and circumcentre is 5 2
(C) Sum of circumradius and inradius of triangle ABC is equal to 10
5
(D) Length of internal angle bisector of ÐACB of triangle ABC is
2 2
8. In a triangle ABC, let BC = 1, AC = 2 and measure of angle C is 30°. Which of the following
statement(s) is (are) correct?
(A) 2 sin A = sin B

(B) Length of side AB equals 5 - 2 3

(C) Measure of angle A is less than 30°
(D) Circumradius of triangle ABC is equal to length of side AB
4 24
9. Given an acute triangle ABC such that sin C = , tan A = and AB = 50. Then-
5 7
(A) centroid, orthocentre and incentre of DABC are collinear

4
(B) sin B =
5

4
(C) sin B =
7
(D) area of DABC = 1200
10. In a triangle ABC, if cos A cos 2B + sin A sin 2B sin C = 1, then

r r p
(A) A,B,C are in A.P. (B) B,A,C are in A.P. (C) =2 (D) = 2 sin
R R 12

11. In DABC, angle A is 120°, BC + CA = 20 and AB + BC = 21, then

(A) AB > AC (B) AB < AC
node06\B0AH-AI\Kota\JEE(Advanced)\Nurture\Maths\Sheet\SOT\Eng.p65

(C) DABC is isosceles (D) area of DABC = 14 3

12. In a triangle ABC, ÐA = 30°, b = 6. Let CB1 and CB2 are least and greatest integral value of side a for
which two triangles can be formed. It is also given angle B1 is obtuse and angle B2 is acute angle.
(All symbols used have usual meaning in a triangle.)
(A) |CB1– CB2| = 1 (B) CB1+ CB2 = 9
3 9
(C) area of DB1CB2 = 6 + 7 (D) area of DAB2C = 6 + 3
2 2

E 21
JEE-Mathematics
ALLEN
13. If the lengths of the medians AD,BE and CF of triangle ABC are 6, 8,10 respectively, then-
(A) AD & BE are perpendicular (B) BE and CF are perpendicular
(C) area of DABC = 32 (D) area of DDEF = 8
14. Let P be an interior point of DABC.
Match the correct entries for the ratios of the Area of DPBC : Area of DPCA : Area of DPAB
depending on the position of the point P w.r.t. D ABC.
Column-I Column-II
(A) If P is centroid (G) (P) tanA : tanB : tanC
(B) If P is incentre (I) (Q) sin2A : sin2B : sin2C
(C) If P is orthocentre (H) (R) sinA : sinB : sinC
(D) If P is circumcentre (S) 1:1:1
(T) cos A : cosB : cosC
EXERCISE (S-1)
1. Given a triangle ABC with sides a = 7, b = 8 and c = 5. If the value of the expression

(å sin A )çæ å cot A ö÷ can be expressed in the form

p
where p, q Î N and
p
q
is in its lowest form
è 2ø q
find the value of (p + q).

2. If two times the square of the diameter of the circumcircle of a triangle is equal to the sum of the
squares of its sides then prove that the triangle is right angled.

3. In acute angled triangle ABC, a semicircle with radius ra is constructed with its base on BC and
tangent to the other two sides. rb and rc are defined similarly. If r is the radius of the incircle of triangle
2 1 1 1
ABC then prove that, = + + .
r ra rb rc
4. If the length of the perpendiculars from the vertices of a triangle A, B, C on the opposite sides are

1 1 1 1 1 1 1
p1, p2, p3 then prove that + + = = + + .
p1 p2 p3 r r1 r2 r3

5. With usual notations, prove that in a triangle ABC

a cot A + b cot B + c cot C = 2(R + r)
6. With usual notations, prove that in a triangle ABC
Rr (sin A + sin B + sin C) = D
node06\B0AH-AI\Kota\JEE(Advanced)\Nurture\Maths\Sheet\SOT\Eng.p65

7. With usual notations, prove that in a triangle ABC

A B C s2
cot + cot + cot =
2 2 2 D
8. With usual notations, prove that in a triangle ABC
a 2 + b2 + c 2
cot A + cot B + cot C =
4D

22 E
 Solutions of Triangle
ALLEN
æ cö
9. If a,b,c are the sides of triangle ABC satisfying log ç 1 + ÷ + log a - log b = log 2 .
a è ø
Also a(1 – x2) + 2bx + c(1 + x2) = 0 has two equal roots. Find the value of sinA + sinB + sinC.
10. With usual notations, prove that in a triangle ABC
b-c c-a a -b
+ + =0
r1 r2 r3
11. With usual notations, prove that in a triangle ABC
r1 r2 r3 3
+ + =
(s - b) (s - c) (s - c) (s - a ) (s - a ) (s - b) r
12. With usual notations, prove that in a triangle ABC
abc A B C
cos cos cos = D
s 2 2 2
13. With usual notations, prove that in a triangle ABC
1 1 1 1 a 2 + b 2 + c2
+ + + =
r2 2
r1 r2
2
r3
2
D2
14. With usual notations, prove that in a triangle ABC
2R cos A = 2R + r – r1
15. If r1 = r + r2 + r3 then prove that the triangle is a right angled triangle.

EXERCISE (S-2)
b+c c+a a +b
1. With usual notation, if in a D ABC, = = ; then prove that, cos A = cos B = cos C .
11 12 13 7 19 25
2. Given a triangle ABC with AB = 2 and AC = 1. Internal bisector of ÐBAC intersects BC at D. If
AD = BD and D is the area of triangle ABC, then find the value of 12D2.

3. For any triangle ABC , if B = 3C, show that cos C = b + c & sin A = b - c .
4c 2 2c
cot C
4. In a triangle ABC if a2 + b2 = 101c2 then find the value of .
cot A + cot B
5. The two adjacent sides of a cyclic quadrilateral are 2 & 5 and the angle between them is 60°. If the
area of the quadrilateral is 4 3 , find the remaining two sides.
6. If in a D ABC , a = 6, b = 3 and cos(A - B) = 4/5 then find its area.
node06\B0AH-AI\Kota\JEE(Advanced)\Nurture\Maths\Sheet\SOT\Eng.p65

a b
7. In a D ABC, (i) = (ii) 2 sin A cos B = sin C
cos A cos B
A A C
(iii) tan2 + 2 tan tan - 1 = 0, prove that (i) Þ (ii) Þ (iii) Þ (i).
2 2 2
8. Two sides of a triangle are of lengths 6 and 4 and the angle opposite to smaller side is 300. How
many such triangles are possible ? Find the length of their third side and area.

E 23
JEE-Mathematics
ALLEN
9. The triangle ABC (with side lengths a, b, c as usual) satisfies log a2 = log b2 + log c2 – log (2bc cosA).
What can you say about this triangle?
10. The sides of a triangle are consecutive integers n, n + 1 and n + 2 and the largest angle is twice the
smallest angle. Find n.

EXERCISE (JA)
1. Let ABC and ABC¢ be two non-congruent triangles with sides AB = 4, AC = AC¢ = 2 2 and
angle B = 30°. The absolute value of the difference between the areas of these triangles is [JEE 2009, 5]
2. (a) If the angle A,B and C of a triangle are in an arithmetic progression and if a,b and c denote the
length of the sides opposite to A,B and C respectively, then the value of the expression
a c
sin 2C + sin 2A , is -
c a

1 3
(A) (B) (C) 1 (D) 3
2 2
(b) Consider a triangle ABC and let a,b and c denote the length of the sides opposite to vertices A,B
and C respectively. Suppose a = 6, b = 10 and the area of the triangle is 15 3 . If ÐACB is
obtuse and if r denotes the radius of the incircle of the triangle, then r2 is equal to
p
(c) Let ABC be a triangle such that ÐACB = and let a,b and c denote the lengths of the sides
6
opposite to A,B and C respectively. The value(s) of x for which a = x2 + x + 1, b = x2 – 1 and
c = 2x + 1 is/are [JEE 2010, 3+3+3]

(
(A) - 2 + 3 ) (B) 1 + 3 (C) 2 + 3 (D) 4 3

7 5
3. Let PQR be a triangle of area D with a = 2, b = and c = , where a, b and c are the lengths of the
2 2
2 sin P - sin 2P
sides of the triangle opposite to the angles at P, Q and R respectively. Then
2sin P + sin 2P
equals [JEE 2012, 3M, –1M]

2 2
3 45 æ 3 ö æ 45 ö
(A) (B) (C) ç ÷ (D) ç ÷
4D 4D è 4D ø è 4D ø
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1
4. In a triangle PQR, P is the largest angle and cos P = . Further the incircle of the triangle touches
3
the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM
are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)
[JEE(Advanced) 2013, 3, (–1)]
(A) 16 (B) 18 (C) 24 (D) 22
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 Solutions of Triangle
ALLEN
5. In a triangle the sum of two sides is x and the product of the same two sides is y. If x2 – c2 = y,
where c is a third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle
is - [JEE(Advanced)-2014, 3(–1)]
3y 3y 3y 3y
(A) 2x(x + c) (B) 2c(x + c) (C) 4x(x + c) (D) 4c(x + c)

6. In a triangle XYZ, let x,y,z be the lengths of sides opposite to the angles X,Y,Z, respectively and

s-x s-y s-z 8p

2s = x + y + z. If = = and area of incircle of the triangle XYZ is , then-
4 3 2 3

(A) area of the triangle XYZ is 6 6

35
(B) the radius of circumcircle of the triangle XYZ is 6
6
X Y Z 4
(C) sin sin sin =
2 2 2 35
2æX+Yö 3
(D) sin ç ÷= [JEE(Advanced)-2016, 4(–2)]
è 2 ø 5

7. In a triangle PQR, let ÐPQR = 30° and the sides PQ and QR have lengths 10 3 and 10, respectively.
Then, which of the following statement(s) is (are) TRUE ? [JEE(Advanced)-2018, 4(–2)]
(A) ÐQPR = 45°
(B) The area of the triangle PQR is 25 3 and ÐQRP = 120°
(C) The radius of the incircle of the triangle PQR is 10 3 - 15
(D) The area of the circumcircle of the triangle PQR is 100p.
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E 25
JEE-Mathematics
ALLEN
ANSWERS
ELEMENTARY EXERCISE
1. C 2. C 3. B 4. C 5. B 6. B 7. D 8. C 9. D
10. A 11. C 12. A 13. D 14. C 15. A 16. C 17. A 18. D
19. 25 20. 45°,60°,75°
EXERCISE (O-1)
1. D 2. A 3. A 4. C 5. C 6. D 7. B 8. C
9. C 10. C 11. D 12. D 13. D 14. C 15. C

EXERCISE (O-2)
1. B,C 2. A,B 3. B,C 4. A,B,D 5. B,D 6. A,C 7. A,B,C
8. A,C,D 9. A,B,D 10. B,D 11. A,D 12. A,B,C,D 13. A,C,D

14. (A) S; (B) R; (C) P; (D) Q

EXERCISE (S-1)
12
1. 107 9.
5
EXERCISE (S-2)
2. 9 4. 50 5. 3 cms & 2 cms 6. 9 sq. unit

8. ( ) ( ) ( ) (
Two tringle 2 3 - 2 , 2 3 + 2 , 2 3 - 2 & 2 3 + 2 sq. units )
9. triangle is isosceles 10. 4
EXERCISE (JA)
1. 4 2. (a) D, (b) 3, (c) B 3. C 4. B,D 5. B 6. A,C,D
7. B,C,D
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26 E