Properties
And
Solution of Triangle
Some important formula :1. sin(𝐴 − 𝐵) × sin(𝐴 + 𝐵) = sin, 𝐴 − sin, 𝐵
Proof :- L.H.S = sin(𝐴 − 𝐵) × sin(𝐴 + 𝐵)
= (sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵) × ( sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵)
= (sin 𝐴 cos 𝐵)2 − (cos 𝐴 sin 𝐵)2
= sin, 𝐴 cos , 𝐵 − cos , 𝐴 sin, 𝐵
= sin, 𝐴(1 − sin, 𝐵 ) − sin, 𝐵 (1 − sin, 𝐴)
= sin, 𝐴 − sin, 𝐴 sin, 𝐵 − sin, 𝐵 + sin, 𝐴 sin, 𝐵
= sin, 𝐴 − sin, 𝐵 R.H.S.
2. sin(𝐵 − 𝐶) × sin(𝐵 + 𝐶) = sin, 𝐵 − sin, 𝐶
3. sin(𝐶 − 𝐴) × sin(𝐶 + 𝐴) = sin, 𝐶 − sin, 𝐴
Cosine law
Statement: In any triangle
ABC, the cosine law states
that
cosA =
cosB =
cosC =
b 2 +c 2 −a 2
2bc
c 2 +a 2 −b 2
2ca
a 2 +b 2 −c 2
2ab
A
A
Proof:
A
c
B
A
b
a
D
Fig.(i)
A
A
c
C
B
c
b
a
C
Fig.(ii)
D
B
b
a
Fig.(iii)
C=D
Let ABC be any triangle. Draw AD perpendicular to BC or BC
produced (if necessary). Then there may arise the following three
cases:
Case I : When ∡C is an acute in fig (i) , then from the right angled
A
triangles ABD and ADC,
AB2 = AD2 + BD2……………..(1)
AC2 = AD2 + DC2 …………… (2)
b
c
Subtracting (2) from (1), we get
AB2 – AC2 = BD2 – DC2
B
= (BC – DC)2 – DC2
2
2
= BC -2BC.DC
∴
AB2
=
AC2 +BC2
-2BC.DC
C
From right angled triangle ACD, cosC =
2
= BC -2BC.DC + DC – DC
2
a D
This implies DC = AC cosC
∴ AB2 = AC2 +BC2 – 2BC.AC cosC
i.e. c2 = b2 + a2 - 2ab.cosC
DC
AC
Case II: When ∡C is an obtuse angled in fig (ii), then from right angled triangle
ABD
AB2 = AD2 + BD2
= (AC2 – CD2) + (BC +CD)2
= AC2 – CD2 + BC2 +2BC.CD + CD2
= AC2 +BC2 +2 BC.CD cosC
[ ∵from Δ ACD, cos (𝜋 - C) =
𝐷𝐶
𝐴𝐶
⇒–cosC =
𝐷𝐶
𝐴𝐶
⇒ DC = - AC cosC ]
= AC2 +BC2 - 2 BC.AC cosC
∴ AB2 = AC2 +BC2 - 2 BC.AC cosC
∴ c2 = b2 +a2 -2ab cosC
Hence
cosC =
a 2 +b 2 −c 2
2ab
Case III: when ∡C is a right angled triangle in fig (iii), then
A
AB2 = AC2 +BC2
= AC2 + BC2 -2AC. BC cosC
(∵ ∡C = 90 and cos90 = 0)
= b2 +a2 -2ab cosC
∴ c2 = b2 +a2 -2ab cosC
B
Hence in all cases, we have
C =D
c2 = b2 +a2 -2ab cosC
∴ CosC =
a 2 +b 2 −c 2
2ab
Similarly, drawing perpendiculars from B and C on
the opposite sides, we get
CosA =
b 2 +c 2 −a 2
2bc
and CosB =
Hence cosine law verified.
c 2 +a 2 −b 2
2ca
Sine law
In any triangle ABC,
012 3
4
=
012 5
6
=
012 7
8
where symbols have their usual meanings.
Proof: Consider any triangle ABC, whose sides BC, CA ad AB denoted by a , b and c respectively.
Then the area of the triangle ABC is given by the following formulae:
∆ ABC =
∆ ABC =
∆ ABC =
:
,
:
,
:
,
ab sin 𝐶
bc sin 𝐴
A
ac sin 𝐵
Since all these three formulae represent the area of same triangle ABC. So,
:
,
:
,
:
,
ab sin 𝐶 = bc sin 𝐴 = ac sin 𝐵
b
c
or, ab sin 𝐶 = bc sin 𝐴 = ac sin 𝐵
ab 012 7 bc 012 3 ac 012 5
or,
=
=
468
468
468
012 7 012 3 012 5
or,
=
=
8
4
6
012 3 012 5 012 7
Hence,
=
=
4
6
8
B
C
a
4
6
8
Sine Law :- In any triangle ABC,
=
=
= 2R where R is the
012 3
012 5
012 7
radius of a circumscribing triangle.
Proof :- We construct three different figures in which angle A of the triangle
ABC is considered to be acute , obtuse and right.
Let O be the centre of the circum-circle of DABC and R be the circum-radius. Join BO and
produce it to meet the circumference at D and also join DC, as shown in figure (i) and (ii).
If the angle A is right-angled then C & D represents the same point. Clearly BO = R, so that
BD = 2R.
In Fig. (i)
Since BD is a diameter so ÐBCD = 90° (angle at the semi-circle) and ÐBDC = ÐBAC
[lying in the same segment of the circle]
4
012 3
=
57
012 ÐBAC
=
57
012 ÐBDC
=
57
;<
;=
= BD = 2R
In Fig. (ii)
Since BD is a diameter so ÐBCD = 90° (angle at the semi-circle)
and ÐBDC + ÐBAC = 180° (sum of opposite angle of a cyclic quadrilateral)
ÐBDC = 180° − ÐBAC
Taking sin on both sides
sin ÐBDC = sin (180° − ÐBAC)
= sin ÐBAC= sin A
Now,
4
012 3
57
= 012 ÐBDC
=
57
;<
;=
= BD = 2R
In figure (iii)
Here, the vertex C and point D are coincide with each other. Then, BC is a diameter of
the circle so ÐBAC = 90° 𝑎 = 2R
4
012 3
=
,@
012 90°
Thus, in all cases we obtained
= 2R
4
012 3
= 2R .
Similarly, considering the angles B and C instead of the angle A we are able to prove
6
8
= 2R and 012 8 = 2R respectively.
012 5
Hence,
4
012 3
=
6
012 5
=
8
012 7
= 2R . Proved
The Projection Law
In any triangle ABC,
Alternative method
a = b cos C + c cos B
We have by sine law,
b = c cos A + a cos C
c = a cos B + b cos A
Proof : (i)
b cos C + c cos B = a
We have by cosine law,
4A B6A C8 A
,46
cos C =
and cos B = =
L.H.S. = b cos C + c cos B
=
=
=
8 A B4A C6A
,48
4A B6A C8 A
8 A B4A C6A
b . ,46
+ c × ,48
4A B6A C8 A 8 A B4A C6A
+
,4
,4
:
(𝑎, + 𝑏 , − 𝑐 ,+ 𝑐 , + 𝑎,
,4
:
,
2𝑎
,4
=
= a
= R.H.S.
or,
4
=
012 3
2R
a = 2R sin A
= 2R sin(180 – (B + C)) [∵ A + B + C = 180° ]
= 2R sin(B + C)
= 2R (sin B cos C + cos B sin C)
= 2R sin B cos C + 2R sin C cos B
= b cos C + c cos B
[∵ b = 2R sin B & c = 2R sin C]
Remaining results can be proved in similar manner.
− 𝑏,)
The Tangent Law
In any triangle ABC,
i) tan
ii) tan
iii) tan
5C7
,
7C3
,
3C5
,
=
=
=
6C8
6B8
8C4
8B4
4C6
4B6
7C3
,
3
cot ,
5
cot
,
7
cot
,
8C4
8B4
5
cot ,
Proof of (ii) : tan
=
: We have, by sine law,
a = 2R sin A and c = 2R sin c
Now, c – a = 2R sin C - 2R sin A
= 2R (sin C - sin A )
7B3
7C3
= 4R cos , sin ,
And
c + a = 2R sin C + 2R sin A
= 2R (sin C + sin A )
7B3
7C3
= 4R sin
cos
,
,
<IJ
<KJ
4R
GH0
012
8C4
A
A
Now ,
=
8B4 4R 012<IJ GH0<KJ
A
A
7B3
7C3
= cot , tan ,
7C3 8C4
7B3
or, tan
=
tan
,
8B4
,
8C4
5
=
tan(90° − ) [ ∵
8B4
,
7C3 8C4
5
∴ tan , = 8B4 cot , .
7B3
=
,
90° −
5
,
]
The Half Angle formula :3
1. (i) sin , =
(RC6)(RC8)
R(RC4)
,
68
3
3. (i) tan , =
(ii) sin , =
68
3
2. (i) cos =
5
(ii) cos
(RC6)(RC8)
R(RC4)
5
,
(R C4)(R C8)
=
5
(ii) tan , =
Where s is semi perimeter i.e s =
4B6B8
,
48
R(R C6)
48
(RC4)(RC8)
R(RC6)
7
(iii) sin , =
(RC4)(RC6)
46
7
R(RC8)
,
46
(iii) cos =
7
(iii) tan , =
(RC4)(RC6)
R(R C8)
5
s (s – b)
,
48
Proof :- cos =
We know, cos B =
c2 + a2 − b2
,48
or, 2ac cos B = c2 + a2 - b2
or, 2ac + 2ac cos B = c2 + a2 +2ac - b2
or, 2ac (1 + cos B) = (a + c)2 – b2
or,
or,
or,
or,
,5
4ac cos ,
,5
4ac cos ,
,5
4ac cos
,
,5
4ac cos
,
or,
= 2s ( 2s – 2b )
or,
= 4s ( s – b )
or,
48
.
48
c2 + a2 − b2
,48
or, − 2ac cos B = b2 − (c2 + a2 )
or, 2ac − 2ac cos B = b2 − (a2 − 2ac + c2 )
or, 2ac (1- cos B) = b2 − (a − c)2
= 2s (a + b + c – 2b)
,
∴ cos =
,
We know, cos B =
or,
s( s–b)
(s − a) (s – c)
Proof : sin =
= ( a + b + c ) (a +c –b )
5
5
,5
4ac sin ,
,5
4ac sin
,
,5
4ac sin
,
,5
ac sin
,
= ( b − a + c ) (b +a -c)
= (2s − 2a ) (2s − 2c)
= 4 (s − a) ( s − c )
= (s − a) (s − c )
5
(s − a ) ( s−c )
,
48
∴ sin =
Area of Triangle
In any triangle ABC, the area (D) is given by,
i) D =
ii) D =
:
:
:
,
,
𝑎𝑏 sin 𝐶 = 𝑏𝑐 sin 𝐴 =
,
468
S@
𝑎𝑐 sin 𝐵
, where R is circum-radius.
iii) s (s − a)(s − b) (s – c) where s = semi-perimeter of triangle.
iv) D = 2R2 sin A sin B sin C
:
v) D = S 2𝑎, 𝑏 , + 2𝑏 , 𝑐 , + 2𝑐 , 𝑎, − 𝑎S − 𝑏 S − 𝑐 S
i)
Proof: Consider a triangle ABC and draw AD ^ BC
We know,
D=
D=
:
base × height
,
:
BC × AD
. . . (i)
,
In right angled DACD
sinC =
3U
37
AC sinC = AD
Now, from (i)
D=
\
D=
:
BC × AC sinC
,
:
a × b sinC
,
Other results can be proved in similar manner.
ii)
We have,
:
,
D = = a × b sin C
By sine law,
4
6
8
=
=
012 3 012 5 012 7
= 2R
From last two ratios,
8
012 7
= 2R
c = 2R sin C
sin C =
8
,@
Now,
D=
\D=
(iii)
468
S@
:
a
,
×b
8
,@
We have,
:
,
:
8
8
= 2absin cos
,
,
,
8
8
= ab sin cos
,
,
D = ab sin C
= ab
= ab
(s − a ) ( s – b)
s ( s–c)
46
46
s (s − a )(s − b) ( s – c )
(46)A
= s (s − a )(s − b) ( s – c )
[ using half angle formula ]
iv)
We have from (ii)
468
D=
S@
,@ 012 3 ,@ 012 5 ,V 012 7
=
S@
W@A ×V012 3 012 5 012 7
=
S@
(v)
= 2R2sin 𝐴 sin 𝐵 sin 𝐶
We have,
D
= s (s − a )(s − b) ( s – c )
(4B6B8)
,
=
(4B6B8)
,
=
=
:
S
(
(
4B6B8
−
,
6B8C4
,
a
4B6B8
)(
,
4B8C6
)( ,
)(
− b) (
4B6B8
,
4B6C8
,
)
2𝑎,𝑏 , +2𝑏 ,𝑐 ,+2𝑐 ,𝑎, − 𝑎S − 𝑏 S − 𝑐 S
–c)