Solutions o f Triangles Triangles
Theory Notes - Solution Solutions s of Triangle riangles s 1.
ELEMENTS OF A TR TRIANGLE
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 tri angle is given by s = a b c and its area is denoted by . 2 2.
SINE RULE a
In a triangle tri angle ABC,
sin A
Note : Area of triangle triangle =
1 2
b sin B
bc sinA =
c sin C 1 2
= 2R (where R is circum radius)
ac sinB =
1 2
ab sinC.
Drill Exercise - 1
a 2 sin( B C)
b 2 sin( C A )
c 2 sin( A B)
1.
In any triangle ABC, ABC, prove that
2.
If in a ABC,
3.
ABCD is a trapezium trapez ium such that AB and CD are parallel and CB is perpendicular to them. If fi nd the length of side si de AB. ADB = 60º, BC = 4 and CD = 3, then find
4.
If the sides of a triangle triangle are in arithmetic progression, and if its greatest angle a ngle exceeds the least angle
sin A sin C
=
sin B sin C
sin(A B) sin(B C)
+
sin C sin A
+
sin A sin B
= 0
, prove that a2, b2, c2 are in A. P..
by , show that the sides are in the t he ration 1 – x : 1 : 1 + x, where x =
1 cos 7 cos
.
5.
Through the angular points of a triangle tri angle 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
In a triangle tri angle ABC, (i) cos A =
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b2
c2 a 2 2bc
(ii) cos B =
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c2
a 2 b2 2ca
Solutions of Triangles
(iii) cos C =
a2
b2 c2 2ab
Drill Exercise - 2
cos A
cos B
cos C
a 2 b2
1.
In any ABC, prove that
2.
Let ABC be a triangle such s uch that 2b 2 b = (m + 1)a 1 )a and cos A =
3.
In a triangle ABC, C = 60º, then prove that
4.
If in a triangle t riangle ABC, ABC,
a
+
b
+
c
=
c2
2abc
.
(m 1)(m 3)
1
, where m (1, 3). 2 m Prove that there are two values of the third thi rd side one of which is m times the other.
cos A 2 cos C cos A 2 cos B
=
1 ac
sin B sin C
+
1 bc
=
3 abc
prove that the triangle triangle is is either isosceles or right
angled. 5.
A ring, 10 cm, in diameter d iameter,, is suspended from a point 12 cm, above its it s centre by 6 equal strings attached to its circumference circumference at equal intervals. Find the cosine of the angle between consecutive strings.
Illustration 1: 1: Find the angles of the the triangle whose whose sides are 3 +
3 and
3,2
6.
Solution: Let a = 3 +
=
66 3 12 2
3,b=2 3,c=
1 3 2 2
4.
sin A
a sin A
2 3
b sin B
sin B =
A = 1050, B = 45 0, C = 30 0
3 3
2bc
12 6 9 3 6 3 12 2
A = 1050
1 2
B = 450
PROJECTION FORMULAE (i) a = b cos C + c cos B
(ii) b = c cos A + a cos C
Illustration 2: 2: If A = 450, B = 75 0, prove that a + c
2 = 2b.
Solution:
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c2 a 2
, we get
sin 10 1050 =
a
cos A =
= cos 1050
Applying Sine formula : b
6
b2
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(iii) c = a cos B + b cos C
Solutions o f Triangles Triangles
As A = 450, B = 75 0 we have C = 60 0
5.
2b = 2 (a cos C + c cos A) = 2(a cos cos 600 + c cos 45 0)
= a + c 2 = L.H.S. NAPIER’S ANALOGY (TANGENT RULE)
B C b c cot A 2 2 b c
C A c a cot B 2 2 c a
(i) tan
(ii) tan
A B a b cot C 2 2 a b
(iii) tan
6.
HALF ANGLE FORMULAE
(a)
(i) sin sin
A 2 C
(iii) sin
(b)
A
C
(iii) cos
(c)
2
(i) tan
2
(iii) tan
C 2
s s a
(ii) cos
bc
ca
B 2
=
s s b ca
s s c
2
s c s a
ab
ab
s b s c = s s a
A
B
s a s b
=
2
(ii) sin
bc
2
(i) cos
s b s c
s c s a (ii) tan = s s b 2 B
s a s b s s c Drill Exercise - 3
1.
In a ABC, if a = 13, b = 14, and c = 15, find the t he followings (i)
(ii) sin
A
(iii) cos
2
A 2
(iv) tan
A 2
2.
The sides of a triangle are x 2 + x + 1, 2x + 1 and x 2 –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 1
2
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1
+
2
1
+
2
=
sin B sin C
=
c a cos B b a cos C
cot A cot B cot C
ABC. where is the area of
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Solutions of Triangles
5.
In a ABC, prove that cot
7.
AREA OF TRIANGLE
(i) sin A =
2
8.
+ cot
2 ab
+ cot
2
s s a s b s c
bc
(iii) sin C =
2
B
C 2
a b c A cot . = 2 b c a
s(s a )(s b)( s c)
Area of triangle =
7.1
A
s s a s b s c
2Δ bc
(ii) sinB=
2 ca
s s a s b s c
2Δ ca
2Δ 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
DAC = . Then
9.
(i)
(m + n) cot = m cot – n cot
(ii)
(m + n) cot = n cot B – m cot C
CENTROID AND MEDIANS OF A TRIANGLE
The line joining any vertex of a triangle to the mid point of the opposite side of the triangle is called the median of the triangle. The three medians of a triangle are concurrent and the point of concurrency of the medians of any triangle is called the centroid of the triangle. The centroid divides the median in the ratio 2 : 1.
Illustration 3: Find the lengths of the medians and the angles made by the medians with the sides of a triangle ABC.
Solution : 2
2
2
2
AD = AC + CD – 2AC. CD cos C = b +
a2 4
– ab cosC,
c2 = b2 + a2 – 2ab cos C.
and
2
2
2
Hence 2AD – c = b –
so that AD =
1 2
2b 2
a2 2
,
2c 2 a 2 =
1 2
b2
A
c 2 2bc cos A
F E
Similarly, BE =
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1 2
G
2c
2
2a 2 b 2 , and CF =
1 2
2a
2
2b 2 c 2
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B
D
C
Solutions o f Triangles
If BAD and CAD , we have sin sinC
sin =
DC AD
a 2AD
a sin C 2AD
a sin C 2b 2 2c 2 a 2
Similarly sin =
a sin B 2b 2
2c 2 a 2 Drill Exercise - 4
1.
If the medians of a ABC make angles with each other, prove that cot + cot + cot + cotA + cotB + cotC = 0
2.
In an isosceles right angled triangle a straight line is drawn from the middle point of one of the equal sides to the opposite angle. Show that it divides the angle into parts whose cotangents are 2 and 3.
3.
D, E and F are the middle points of the sides of the triangle ABC; prove that the centroid of the triangle DEF is the same as that of ABC, and that its orthocentre is the circumcentre of ABC.
4.
Prove that the median through A divides it into angles whose cotangents are 2 cot A + cot C and 2 cotA + cotB, and makes with the base an angle whose cotangent is
5.
2
(cot C ~ cot B)
Prove that the distance between the middle point of BC and the foot of the perpendicular from A is b2 ~ c2 2a
10.
1
.
CIRCUM CIRCLE
The circle which passes through the angular points of a ABC, is called its circumcircle. The centre of this circle i.e., the point of concurrency of the perpendicular bisectors of the sides of the ABC, is called the circumcenter. A
E
A B
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a/2
F
O
A D
a/2
C
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Solutions of Triangles
Radius of the circumcircle is given by the following formulae a
R=
2 sin A
b
2 sin B
c 2 sin C
abc 4
Illustration 4: If in a ABC, O is the circumcenter and R is the circumradius and R 1, R2 and R3 are the circumradii a
of the triangles OBC, OCA and OAB respectively, then prove that R 1
b R2
c R3
abc R3
.
Solution:
Clearly, in the OBC, BOC = 2A, OB = OC = R, BC = a.
2R1 =
a sin 2A
Similarly, 2R2 = a
R 1
b R2
{using sine rule in BOC)
b sin 2B
c R3
and 2R 3
c
A
sin 2C
= 2(sin2A + sin2B + sin 2C)
O
= 2.4 sin A sin B sin C, B
=8
a
b
c
. . 2R 2R 2R
abc R3
C
.
Illustration 5: If the distances of the sides of a ABC from its circumcenter be x, y and z respectively, then prove a
b
c
abc
that x y z 4xyz .
Solution : a
Let M be the circumcenter. MD BC. So BD = DC = and BMD = A. 2 a
In BDM,
BD MD b
= tan A or
a 2 = tan A, i.e., = tan A, 2x x
c
A
F
Similarly, 2 y = tan B, = tan C 2z
E
z M A
and
a
b
c
tan A + tan B + tan C = 2x 2y 2z a
.
b
.
B
c
tan A. tan B. tan C = 2x 2 y 2z
But in a triangle ABC, tan A + tan B + tan C = tan A. tan B. tan C
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y
x D
C
Solutions o f Triangles
a
11.
x
b
c
abc
y
z
4xyz
ORTHOCENTER AND PEDAL TRIANGLE OF A TRIANGLE.
In a triangle the altitudes drawn from the three vertices to the opposite sides are concurrent and the point of cuncurrency of the altitudes of the triangle is called the orthocenter of the triangle. The triangle formed by joining the feet of these perpendiculars is called the pedal triangle i.e.
DEF is the pedal triangle of ABC. A
F E P 0
B
90 – C C
D
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 (90 0 – C) = AB cos B cot C =
c sin C
cos B cos C = 2R cos B cos C
Similarly PE = 2R cosA cosC and PF = 2R cosA cosB Again
A
AP = AE sec DAC = c cos A cosec C =
c sin C
F
cos A = 2 R cos A
E P
so, BP = 2R cos B and CP = 2R cos C
B
90 – C C
D
Illustration 7: Find the distance between the circumcenter and the orthocenter of a triangle ABC Solution : Let O be the circumcenter and P be the orthocenter of the ABC A
If OF be perpendicular to AB, we have
OAF 900 C Also
PAL 900 C
OAP C – B
F P O
B
Also OA = R and PA = 2R cosA
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OP2 = OA2 + PA2 – 2OA. PA cos OAP
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D
K
L
C
Solutions of Triangles
= R2 + 4R2 cos2 A – 4R 2 cosA cos (C – B) = R2 – 4R 2 cosA [cos(B + C) + cos (C – B)] = R 2 – 8R 2 cos A cos B cosC
12.
1 8 cos A cos B cos C
OP = R
BISECTORS OF THE ANGLES
If AD bisects the angle A and divide the base into portions x and y, we have, by Geometry, x y
x=
AB AC
ac bc
c
x
b
c
and y =
y b
xy
bc
a bc
ab bc
Also let be the length of AD
A
we have ABD + ACD = ABC
i.e.,
1 2
c sin
A 2
1 2
b sin
bc sin A A bc sin 2
A 2
2 bc bc
1 2
cos
bc sin A,
A
B
x
2
D
y
C
Drill Exercise - 5
1.
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 sin C.
3.
If p1, p2 p3 are respectively the perpendiculars from the vertices of a triangle to the opposite sides, 2 2 2
prove that p1p2p3 =
a b c 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. A
E
r I
F r
0
90 – B/2
B
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r
B/2
D
C/2
C
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Solutions o f Triangles
Radius of the Incircle is given by the following formulae r=
s
A
= (s – a) tan
2
= (s – b) tan
B
= (s – c) tan
2
C 2
= 4R sin
A 2
sin
B 2
sin
C 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. OAI IAF OAF
A
F
=
A 2
900 C
A 2
ABC
C
2
CB
E
O
2
Also,
B
IE A sin 2
AI =
OI 2
r
A
sin
2
2
OA 2 AI2 2OA. AI cos OAI
OI 2 R
C
C
2
B
= R2 + 16R2 sin2
B
4R sin sin
I
1 16 sin 2
2
B
sin
1 – 8 sin
=
1 – 8 sin
OI R 1 8 sin
B 2
sin
2
sin 2
C 2
C
B
2
2
– 8R 2 sin
– 8sin
B 2
sin
sin
C 2
cos
CB 2
C B C B C cos cos sin sin 2 2 2 2 2
C B C B C cos cos sin sin 2 2 2 2 2
=
2
B
2
sin2
C 2
sin
B 2
A
. . . (i)
2
sin
C 2
sin
A 2
.
Illustration 9: If the distances of the vertices of a triangle ABC from the points of contacts of the incircle with sides be , and
, then prove that
r2
Solution: Let the incircle touches the side AB at P, where AP = . Let I be the incentre. From the right-angled IPA,
r
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tan
A 2
;
r cot
A 2
similarly, = r cot
B 2
and r cot
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C 2
Solutions of Triangles
In ABC, we have the identity cot
A 2
B
cot
2
cot
C 2
A
cot
2
cot
B 2
cot
C 2 A
. . r
r
r
r r r
P
or
1 r
1 r
3
E
I
C
B
r2 = .
Illustration 10: Show that the line joining the in-centre to the circumcenter of a triangle ABC is inclined to the side BC
cos B cos C 1 . sin C sin B
at an angle tan 1
Solution: Let I be the in-centre of O be the circumcenter of the triangle ABC. Let OL be parallel to BC. Let
IOL . IM = r OC = R, NOC A
tan
IL OL
4Rsin
4Rsin
14.
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A 2
IM LM BM BN
A 2
sin
sin B 2
B 2
sin
sin C 2
IM ON BM NC
C 2
r R cos A B r cot R sin A 2
A
R cos A
.cot
B 2
I O
R sin A
cosA cos B cos C 1 cos A sin A sin C sin B sin A
tan 1
B
L
NM
C
cos B cos C 1 sin C sin B
cos B cos C 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 =
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R2
2Rr .
Solutions o f 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 2
sin
C 2
, 4R sin
C 2
sin
A 2
, 4R sin
A 2
sin
B 2
respectively..
3.
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
A 2
cot
B 2
cot
C 2
.
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 (i) (ii)
r1 = r2 =
A
A A B C s tan 4R sin cos cos sa 2 2 2 2
B F1
B A B C s tan 4R cos sin cos sb 2 2 2 2
16.
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r3 =
C E1
L
(iii)
D1
I1
C A B C s tan 4R cos cos sin . sc 2 2 2 2
M
EXCENTRAL TRIANGLE The triangle formed by joining 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 2
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, 4 R cos
B 2
and
Solutions of Triangles
4 R cos (iv)
C 2
and its angles are
II1 = 4 R sin
A 2
2
–
B
: II2 = 4 R sin
2
B C , – and – . 2 2 2 2 2
A
: II3 = 4 R sin
C 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 =
, sa
r2
, sb
r3
sc
r1, r2, r3 are in HP
sa sb sc , , are in AP
s – a, s – b, s – c are in AP a, b, c are in AP
Drill Exercise - 7
1.
Prove the following :
R
abc 4rs
1 4
r r1 r2 r3
( r1 r2 r3 r )
a 2 b2 2c sin(A B)
( r1 r )(r2
4 cos C
r )(r3 r )
4r 2
1
4
(a b) sec
(A B) 2
sec
C 2
1 1 1 1 1 r r1 r r2 r r3
r 2s 2 1 4
A ( b c ) tan a cos A 2 r sin A 4 sin A 4 cos A 2 sin A 2.
Prove the following :
rr1r2 r3
rr1 cot
A 2
s(s a ) tan
Rr (sin A sin B sin C)
1 4
A 2
cos
A 2
bc(s b)(s c)
( b 2 sin 2C c 2 sin 2B)
sin A sin B 2 sin( A B)
(a 2
b2 )
b2 c2 (abc) 2 / 3 5 / 3 (sin 2A sin 2B sin 2C)1 / 3 4(cot A cot B cot C) 2 a2
3.
Show that the radii of the three escribed circles of a triangle are the roots of the equation, x3 – x2 (4R + r) + xs 2 – 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 respectively), then prove that
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a R1
b R2
c R3
R1 R 2
R3 . abc
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Solutions o f Triangles
5.
Prove that r 2 r12
r2 2 r32 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. (i)
If the sides a, b and c are given, then cos A =
b 2 c2 a 2 2bc
. B and C can be obtained in the
similar way. (ii)
If two sides b and c and the included angle A are given, then tan
BC
Also
2
bc bc
BC 2
cot
=
A
, we get
2
90 0
–
The third side is given by a =
(iii)
A 2
BC 2
,
bsinA sinB
so
.
that
B
and
C can
be
evaluated.
.
If two sides b and c and the angle B (opposite to side b) are given, then sin C = 1800 – (B + C) and a =
using
bsinA
c b
sin B, 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 tw o values of angle C (fig 3). If c < b and B is an acute angle, then there is only one triangle (f ig 4). A
A
c
c b
B
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(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
b
C2
C2
c sinB
c
c sinB
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 120 0. Solution: Let a = 14, b = 10, c = 6 The largest angle is opposite the largest side.
b2 c2 a 2
cos A =
2bc
100 36 196 120
1
A 1200 2
Illustration 13: If in a triangle ABC, a = (1 +
0 3 ) cm, b = 2 cm and C = 60 , then find the other two angles and the
third side
Solution:
cos C =
Also,
a
2
2
b c
2
.
2ab
c2 = 6 sin A a
2
c=
sin B b
1
1 3
2
4 c2
2 1 3 .2
6
sin C c 3
sin A 1 3
sin B 2
2 6
sin B =
1 2
B = 450
A = 1800 – (450 + 600) = 750
Illustration 14 : Given the base of a triangle, the opposite angle A, and the product k 2 of the other two sides, show that it is not possible for a to be less than 2k sin
A 2
.
Solution: Given b.c = k 2
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Solutions o f Triangles
Now cosA =
b
2
c2 a 2 2bc
A 2
or
k 2 2 b a
2k2 cosA = b2 + 4
2
2
2
b
c
4
or b – (a + 2k cosA). b + k = 0 2 Since b is real, (a2 + 2k 2) (a2 + 2k 2 cosA – 2k 2) 0
a 2 2k 2 .2 cos 2
a 2 4k 2 cos 2
a2
a 2k sin A a 2k sin A 0 2 2
a 2k sin
2
A 2
A 2
a 2 4k 2 sin 2 2
A 2
C
B
A
0 2
a
A 2
0
0 [since a2 + 4k 2 cos2
or a 2k sin
But a must be positive. Hence a 2 k sin
a 2 2k 2 .2 sin 2 2
A
A
4k 2 sin 2
A
A 2
is always positive]
(since 2ksin(A/2) is real)
a – 2k sin
A 2
is rejected
. DRILL EXERCISE - 8
1.
A right triangle has c = 64, A = 61º and C = 90º. Find the remaining parts.
2.
Solve the triangle in which b = 100, c = 100 2 and B = 30º.
3.
In a ABC if a, b and Aare given, then prove that two triangles are formed such that the sum of their areas is
4.
1 2
b2 sin2A.
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.
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3 1
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.
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 = 2 r tan
n
1 2
1 2
nr 2 sin
nr 2 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
BOC
2 n
1 2 BOL 2 n n
If a be a side of the polygon, we have a = BC = 2BL = 2RsinBOL = 2Rsin
r
a
cot
.
n 2 n Now the area of the regular polygon = n times the area of the
1 1 a na 2 OBC n OL.BC n . cot .a cot . 4 n 2 2 2 n
D
O
A
R
R B
L
C
DRILL EXERCISE - 9
1.
If a, b, c, d are the sides of a quadrilateral described about a circle then prove that 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
3.
n
: 1 cos
n
.
If 2a be the side of a regular polygon of n sides, R & r be the circumradius & inradius, prove that R r a cot
4.
2n
.
With reference to a given circle, A1 & B1 are the areas of the inscribed and circumscribed regular polygons of n sides, A 2 & B2 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 .
Page 16 of 35
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Solutions o f Triangles
Answer Key Drill Exercise - 1 3.
25 3 43 3 Drill Exercise - 2
5.
313 338 Drill Exercise - 3
1.
(i) 84
(ii)
1
(iii)
5
2
(iv)
5
1 2
Drill Exercise - 6 1.
a = 13, b = 15, c = 14
Drill Exercise - 8 1.
A = 29º , a = 64 cos 29º, b = 64 sin 29º
2.
a = 50 ( 6 2 ) , A = 15º , C = 135º
4.
45º and 105º; 135º and 15º
SOLVED OBJECTIVE EXAMPLES Example 1: If D is the mid point of the side BC of a triangle ABC and AD is perpendicular to AC, then (a) 3b2 = a2 – c 2 (b) 3a2 = b2 – 3c 2 (c) b2 = a2 – c2 (d) a2 + b2 = 5c2 Solution: A
From the right angled CAD, we have
0
90
cos C = 2
2
b a / 2 2
2b
a + b – c = 4b
a 2
a2
c
b2 c2
b
/ 2
2ab
a2 – c2 = 3b2. B a/2
Page 17 of 35
A –
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D
a/2
C
Solutions of Triangles
Example 2: There exists a triangle ABC satisfying sin A
sin B
(a) tanA + tanB + tanC = 0
(b)
(c) (a + b)2 = c2 + ab
(d) none of these
2
3
sin C 7
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 2
2
If (a + b) = c + ab, then
a 2 b 2 c2
2ab Hence (c) is the correct answer.
=
1 2
= cos C =
3
, which is possible
Example 3: If the tangents of the angles A and B of a triangle ABC satisfy the equation abx2 – c2x + ab = 0, then (a) tan A = a/b (b) tan B = b/a (c) cos C = 0 (d) sin2 A + sin2 B + sin2 C = 2 Solution: From the given equation, we get tan A + tan B = c2 / ab and tan A tan B = 1. Since tan (A + B) = We get A + B =
2
A
tan A tan B 1 tan A tan B
and hence C =
c b
2
. 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 c
2
b2 c
2
1
a2
b2 c
2
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 Solution: If p1, p2, p3, are altitude from A, B, C respectively,
Page 18 of 35
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C
Solutions o f Triangles
1
1
2
1
2
2
then = ap1 = bp2 = cp p1 = , p2 = , p3 = 2 2 2 3 a b c By the law of sines a sin A
b sin B
p1 =
c sin C
2Δ ksinA
= k (say)
, p2 =
2Δ ksinB
, p3 =
Now, sin A, sin B, sin C are in A.P.
2 k sin C
p1, p2, p3 are in H.P.
Example 5: In a triangle ABC, medians AD and CE are drawn. If AD = 5, DAC = / 8 and ACE = /4, then the area of the triangle ABC is equal to (a)
(c)
25
(b)
9 25
(d)
18
25 3 10 3
Solution: Let O be the point of intersection of the medians of triangle ABC. Then the area of ABC is three times that of AOC. Now, in AOC, AO =
2 3
AD =
10 3
. Therefore, applying the sine rule to
AOC, we get OC sin / 8
AO sin / 4
area of AOC = =
=
1 2
10 sin / 8 . 3 sin / 4
OC
. AO.OC. sin AOC
1 10 10 sin / 8 . . . .sin 2 3 3 sin / 4 2 8
50 sin / 8 cos / 8 . 9 sin / 4
50 18
area of ABC = 3.
25 9
B
E
25
D
9
O
25 3
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 Solution:
Page 19 of 35
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Solutions of Triangles
We have tan
C 2
0 = tan 90
A B
A B
2
2
6 37 . 1 222 100 5 20 = 6 37 120 185 5
Also tan
5 2 . 6 5
A 2
= cot
122 305
=
cot A / 2 cot B / 2 1 cot A / 2 cot B / 2
2 5
20
tan
C 2
sb 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.
Example 7: If in a triangle ABC, a = 5, b = 4 and cos (A – B) = 31/32, then the thir d side c is equal to (a) 6 (b) 8 (c) 4 (d) none of these Solution: 1 tan 2
cos (A – B) =
1 tan 2
63 tan2
AB 2
AB 2
31
AB
32
2
1
tan
1 tan 2 1 tan 2
AB 2
AB 2 AB 2
1 63
Now tan
tan
Also,
AB 2 C 2
=
cos C =
ab ab
cot
C 2
1 63
54 5 4
cot
C 2
63 9 1 tan 2 C / 2 1 tan 2 C / 2
1 63 / 81 1 63 / 81
18 144
1 8
c2 = a2 + b2 – 2ab cos C = 25 + 16 – 2.5.4. (1/8) = 36 Hence (a) is the correct answer.
Example 8: In a triangle ABC, if r 1 = 2r2 = 3r3, then a : b is equal to (a)
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5 4
(b)
4 5
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c = 6
Solutions o f Triangles
(c)
7
(d)
4
4 7
Solution: From the given relation, we have s tan
A 2
= 2s tan
tan(A / 2) 6
B
= 3s tan
2
tan(B / 2) 3
C 2
tan(c / 2)
k (say)
2
Also, since A/2 + B/2 + C/2 = 900, we get tan
A 2
tan
B 2
tan
B 2
tan
C 2
tan
C
tan
2
A 2
1
6k. 3k + 3k. 2k + 2k. 6k = 1 36k 2 = 1
2 tan A / 2 sin A =
and sin B =
1 tan 2 A / 2 2 tan B / 2
1 tan 2 B / 2
12k 1 36k 2
6k
1 9k 2
k = 1/6
1
4 5
Hence, by the law of sines, sin A/a = sin B/b, we have
a b
sin A sin B
5 4
a:b=5:4 Example 9: Let AD be a median of the
ABC. If AE and AF are medians of the triangles ABD and ADC
respectively and AD = m1, AE = m2, AF = m3, then
a2 8
is equal to
(a) m 22 m 32 2m12
(b) m12 m 22 2m 32
(c) m 22 m 32 2m12
(d) none of these
Solution: In
A
ABC, AD = m = 2
2 1
c2 b 2 2
2 2 In ABD, AE2 = m22 = AD c 2
Page 21 of 35
a2 4
a 2
2
B
4
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E
D
F
C
Solutions of Triangles
2
AD
2 3
AF = m =
2 3
m2
2
2
m3 2m
2
2
b2
2
b 2 c2
m + m = AD + 2 2
2
a 2
2 1
2
4
a2
8
m12
m12
a2 4
a2 8
2m12
a2 8
a2 8
Example 10: If I is the incentre of a triangle ABC, then the ratio IA : IB : IC is equal to A
(a) cosec
(c) sec
2
A
: sec
2
B
: cosec B 2
2
: cosec
: sec
C
(b) sin
2
C
A 2
: sin
B 2
: sin
C 2
(d) none of these
2
Solution: Here BD : DC = c : b But BD + DC = a; c
BD = b c .a BD In ABD, sin A 2 ca
AI
ID
AI =
sin B I
sin B A bc sin 2
AD =
Also,
AB BD
A
AD
.
c
2 bc
ca / b c
bc abc
IA : IB : IC
.AD
B
s
cos ec
cos ec
2
A 2
bc a
cos ec A
A 2
Similarly BI =
: cos ec
B 2
: cos ec
s
cos ec
C 2
Example 11: In a ABC, the value of
(a)
Page 22 of 35
R r
C
D
a cos A b cos B c cos C abc
(b)
is equal to
R 2r
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B 2
, CI
s
cos ec
C 2
Solutions o f Triangles
(c)
r
(d)
R Solution: a cos A b cos B c cos C a bc R =
2s
But
abc 4
2s R
r
2s
.4 sin A sin B sin C
=
4R abc . 3 2s 8R
abc 4sR
2
s
4 R 4.
=
,r=
So, the value =
R
2R sin A cos A 2R sin Bcos B 2R sin Ccos C
. sin 2A sin 2B sin 2C
R=
2r
.R 2
r R
Example 12: The area of a circle is A1and the area of a regular pentagon inscribed in the circle is A 2. Then A1 : A2 is (a)
(c)
5
cos
2
10
cos ec
5 Solution:
In the
(b)
2 5
sec
10
(d) none of these
10
OAB, OA = OB = r and AOB =
ar ( AOB) =
1 2
. r . r. sin 720 =
1 2
360 0 5
= 720
r2 cos 180
D
E
C
O r
A1 : A2 =
2r 2 2
0
5r cos18
2 5
sec
A
10
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 (a) 2 13 (c)
5 3
13
(b)
(d)
5 12 3 2
13
13
Solution:
Page 23 of 35
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Solutions of Triangles
AG =
2 3
2
A A1, BG =
AG =
and BG =
AG =
AG =
1
3 1 3 5 3
BB1 B1
2b2 2c2 a 2
3 1
3
A
G
2a 2 2c 2 b 2 a , BG , BG
1
3
Also, AB = c = 3 and
A 1
C
b 2 4c 2 as a2 = b2 + c2
3
1
B
16 36
2 3
13
1
GAB ABC 2
3 If ‘R 1’ be the circumradius of triangle GAB then
R1 = =
AG BG AB 4 GAB
5 2 1 . 13.3. 3 3 4 .2
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 Awill 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)
a s
2
= 2 s(s – a)
s
2
2(s a ) s
1 (radius of incircle of triangle ABC)
= constant.
Now =
2
1 2
aHa, where ‘Ha’ is the distance of ‘A’ from BC.
1 aH a
= 1 Ha =
2s
= 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 o f Triangles
D
B
(a) (c)
3 r d
(b)
r
(d)
3d 3
r
2
d
2
O
2 r d
C A
Solution: , COA =
AC = d, OA = OB = r , CD = BD =
3
AC2 = OA2 + OC2 – 2AOOC. cos 3
2
2
2
d = 2r – 2r .
1 2
= r
Also, BOD = COD =
BD
2
3.2
O
3
C A
r 3d 3 3 OB r Hence the correct answer is (c)
tan
D
B
2
SOLVED SUBJECTIVE EXAMPLES Example 1: The lengths of sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Determine the sides of the triangle. Solution: Let the lengths of the sides be n, n + 1, n + 2, where n N. From the question, the largest angle opposite to the side n + 2 is 2 while the smallest angle oposite to the side n is . n 5 n 1 n 5 n 1 2 n 2 2 n 2 n 2 6n 5 Now cos = = 2 n 1 n 2 2 n 2 2 n 1 n 2 2 n 1 n 2
and cos 2 =
n2
2
n 1 n 2 2n n 1
But cos 2 = 2 cos2 – 1; so
or
Page 25 of 35
n 3 2n
n 3 2n
2
=
2n 3 n 1 n 3 n 3 = 2n n 1 2n 2n n 1
n2
2
n 5 2 1 2 n 2
n 5 2 1 or (n – 3)(n+2)2 = n{(n + 5)2 – 2 (n + 2) 2} 2 2 n 2
or
(n – 3)(n2 + 4n + 4) = n (– n 2 + 2n + 17)
or
n3 + n2 – 8n – 12 = – n 3 + 2n2 + 17n
or
(n – 4)(2n2 + 7n + 3) = 0
n = 4 or 2n 2 + 7n + 3 = 0.
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Solutions of Triangles
Roots of 2n2 + 7n + 3 = 0 are
7 49 24 4
1
i.e., – 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. B
, tan
C
(ii)
a, tan
(iii)
a, sin A, sin B, sin C are rational.
2
2
Prove that (i) Solution :
are rational.
(ii)
(iii)
(i)
Let (i) be true, i.e., a, b, c and be rational numbers. B
Now, tan
2
s c s a
Now, (i) B
So tan
2
, tan
C
2
s a s b
and s =
abc 2
a, b, c, , s are rational.
and tan
C
are rational because sum, difference, product and quotient of nonzero rational
2
numbers are rational.
Thus (i)
(ii).
Let (ii) be true, i.e., a, tan 2 tan
Now, sin B = 1 tan
2 tan
sin C =
But
Page 26 of 35
2
C
, tan
2
be rational
B 2 2
B
= rational, because tan
B 2
is rational.
2
C
2 = rational, because tan C is rational. C 2 1 tan 2 2 B
Now, tan
B
2
. tan
sin A
2
=
(ii) a
C
b sin B
s c s a s a s b .
2
=
s a s b s c s a a 1 . s s a s b s c s s
s is rational b + c is rational, because a is rational. c sin C
a sin A
bc sin B sin C
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rational rational
Solutions o f Triangles
a
sin A
is rational. But a is rational. So sinA is rational
Thus (ii)
(iii)
Let (iii) be true, i.e., a, sin A, sin B, sin C be rational.
a sin A
b
and
c=
b
sin B
a sin B
sin C
= rational
sin A
a sin C
= rational
sin A
Thus (iii)
c
=
1
bc sin A = rational.
2
(i).
Example 3: If in a triangle ABC, a = 6, b = 3 and cos (A – B) =
4 5
, find the area of the triangle.
Solution : 4
Here, cos (A – B) =
5
AB
1 tan 2
,
2
AB
1 tan 2
By componendo and dividendo,
2 tan
2
But tan
tan
AB 2
AB 2
1
ab
5
AB 2
54 54
or tan2
AB 2
1 9
( A > B).
3
ab
4
2
2
or
cot
C 2
1
The area of the triangle =
3 1 2
63 63
ab sin C
cot
1 2
C 2
.6.3. sin
or cot
2
C 2
= 1; C =
2
.
9 sq. units.
Example 4: If p, q are perpendiculars from the angular points A and B of the ABC drawn to any line through the vertex C, then prove that a2b2 sin2 C = a2p2 + b2q2 – 2abpqcos C.
Solution : Let ACE = . Clearly, from the figure, we get p AC
Page 27 of 35
sin ,
q BC
sin C
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p
Solutions of Triangles
p
b q
a
q
sin , sin . cos C cos .sin C a
q2
or
a2 q2
or
a
2
p b
2
cos C cos . sin C
p2 b2 p2 b
2
cos C 2
2pq
ab
2pq ab
2
q p cos C cos2 .sin 2 C = 1 p 2 1 cos 2 C b a b
or
cos C 1
p2 b2
p2 2 1 2 cos C b
cos C sin 2 C or a2p2 + b2q2 – 2abpqcosC = a 2 b2sin2C.
Example 5: 2 c2
In a ABC, prove that cos A. cos C =
a2
3ca
, where AD is the median through A and
AD AC.
Solution: b2
From the ABC, cos A =
c2 a 2
. . . (i)
2bc AC
b
2b
From the CAD, cos C = CD a / 2 a From the ABD, a / 2
or
cos A
cos A=
BD sin(A 900 )
sin 90
a cos C
2c b2
0
C 2c
c2 a 2
or
b2 + c2 – a 2 = – 2b 2
2bc
2b
sin ADB
2 cos A
a
b c
,
c cos C
from (ii)
c B
3b 2
3 c2 a 2 3ca
2 c2
a2
3ca
c 2 a 2 2b 2bc
a2
D
. . . (iii)
b2
.
a
c2 3 c2 a 2 3ca
A
b
c2 – a2 = – 3b2
Now, cosA. cosC =
=
.
AB
a
or
a
from (i),
or
Page 28 of 35
c
=
. . . (ii)
b2
c2 a 2 ca
,
from (iii)
.
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C
Solutions o f 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 EF sin A
AE sin AFE
EF =
AB cos A cos PFE
c cos A cos PAE
A
c cos A sin C
F E
c sinC
sinAcosA = a cosA
P 90 – C
B
C
D
similarly DF = b cosB and DE = c cosC
Example 7: The base of a triangle is divided into three equal parts. If t 1, t2 , t3 be the tangents of the angles subtended by these parts at the opposite vertex, prove that:
1 1 1 1 1 t t t t 4 1 2 1 2 2 3 t 2 Solution: Let the points P and Q divide the side BC in three equal parts such tha t BP = PQ = QC = x A Also let,
BAP = , PAQ = , QAC = and
AQC =
From question, tan
= t1, tan = t2, tan = t3.
Applying,
B
m : n rule in triangle ABC, we get (2x + x) cot = 2x cot ( + ) – x cot
Q
P
C
. . . (i)
from APC, we get (x + x) cot = x cot – x cot
. . . (ii)
dividing (i) by (ii), we get 3 2
or or
Page 29 of 35
=
2 cot cot cot cot
or
3 cot – cot
=
4 cot . cot 1
3 cot2 – cot cot + 3 cot . cot – cot . cot = 4 cot
cot cot
. cot – 4 4 + 4 cot2 = cot2 + cot . cot + cot . cot + cot .cot
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Solutions of Triangles
or
4(1 + cot2 ) = (cot + cot )(cot + cot )
or
1 1 1 1 1 4 1 2 t 2 t1 t 2 t 2 t 3
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).
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, A
AEB = ACB = C and AEC = ABC = B From the right angled triangle BDE, tan C =
BD
. . . (i)
DE
From the right angled triangle CDE, tan B =
D C
B
CD
. . . (ii)
DE
C
B
E
Adding (i) and (ii) we get, tan B + tanC =
Similarly tan C + tan A =
Hence,
a
b
b
a
C
and tan A + tan B =
c
= 2 (tan A + tan B + tan C)
Example 9: If x, y, z are the distance of the vertices of the ABC respectively from the orthocentre then prove that
a x
b
c
abc
y
z
xyz
.
Solution: Let H be the orthocentre. Then BHC = 1800 – HBC – HCB = 1800 – (900 – C) – (90 0 – B) = B + C = – A.
Page 30 of 35
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Solutions o f Triangles
ar( BHC) = =
1 2
1
BH. CH sin BHC
2
yz sin ( – A) =
1 2
A
yzsinA.
x
Similarly, ar( CHA) = ar( AHB) =
ar( ABC) =
=
1 2 1 2
1
zx sin B xy sin C
B
1 2
zx sin B +
1 2
sin A sin B sin C y z
1
xyz x 2 1
Also, we know that R =
(i) gives,
abc 4R
C
xy sin C
1
a
b
c
= 2 xyz. 2R x y z
z
y
yz sin A +
2
H
=
abc
. . . (i)
abc
, i.e., = 4 4R 1
a
b
c
xyz x y z 4R
a x
b
c
abc
y
z
xyz
.
Example 10: Prove that in a ABC, R 2r..
Solution: We have r = 4R sin A /2 sin B/2 sin C/2
r 4R
= sin A/2 sin B/2 sin C/2
Also we know that sin A/2 sin B/2 sin C/2
r 4R
1 8
1 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 r 1 + r2 + r3 = r + 4R.
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Solutions of Triangles
Now, r1 + r2 + r3 –
Δ r =
sa
Δ sb
Δ sc
Δ s
1 1 1 1 s a s s b s c
=
a a s s a s b s c
=
=
a
= a
=
a
s 2 s b c bc s 2 as s s a s b s c 2s 2
. 2s 2
s a b c bc 2 2s 2 bc
abc 4R R 4
abc
r1 + r2 + r3 = r + 4R.
Example 12: If a, b, c are in A.P., prove that cos A cot A/2, cosB. cot B/2, cosCcot C/2 are in A.P. Solution: a, b, c are in A.P. cotA/2, cot B/2, cotC/2 are in A.P. Now, cosA cotA /2, cosB cotB/2, cosC cotC/2 are (1 – 2 sin2A/2) cotA/2, (1 – 2sin 2 B/2) cotB/2, (1 – 2 sin 2 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 – a 2 – b2 – c2.
Solution:
. abc cos 2 A / 2 = s 4
L.H.S. = 8
=
=
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abc s
abc s
abc s
2 cos
2
1 cos A b 2 c 2 a 2 1 2bc
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A / 2
Solutions o f Triangles
abc
=
s
abc =
=
=
s abc s
2bc b 2 c 2 a 2 2bc b c 2 a 2 2bc
a b c b c a
abc 2s s
,
2bc
b c a
=
2bc
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 + 2a b + 2ca – a2 – b2 – c2 .
Example 14: If ‘t1’, ‘t2’ and ‘t3’ ar e the lengths of the tangents drawn from centre of ex-circle to the circum circle of the ABC, then prove that 1 2
1
t1
t2
2
1 t3
abc
2
abc
Solution: Let S and I1 be respectively the centres of the circumcircle and the excircle touching BC. It can be shown that SI1 R 2 2Rr1
In
SI1 P , SI12 = R2 + t12 1
R2 + 2Rr1 = R2 + t12 ,
2
t1
A
1 2Rr1
R P
S
C
B
Similarly
1 t 12
1 t 22
1 t 22
1
t 32
1
,
1
2Rr2 t 32
1 2Rr3
I1
1 1 1 2R r1 r2 r3 1
=
=
s a s b s c = 2R 1
a bc abc
1 s 2R
s 2R
proved
Example 15: If a, b and A are given in a triangle and c 1, c2 are the possible values of the third side, prove that
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Solutions of Triangles
c12
c 22 2 c1 c 2 cos A = 4a2 cos2 A
Solution: We have
cosA =
2bc
c2 – 2bc cos A + b 2 – a2 = 0, which is quadratic in ‘c’
c1 c 2
and
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b 2 c2 a 2
2b cos A c1c2 b 2 a 2
c12
(c1 + c2)2 – 2c1c2 – 2c1c2 cos 2A (c1 + c2)2 – 2c1c2 (1 + cos 2A)
c12
. . . (i)
c 22 2c1c 2 cos 2A
4b2 cos2A – 2(b2 – a2). 2cos2A
[using (i)] = 4a2 cos2A
c 22 2c1c2 cos A 4a 2 cos2 A
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