SOT : RAV Sir
Solution of Triangle Sine Rule : In any triangle ABC, the sines of the angles are proportional to the opposite sides i.e.
a b c . sin A sin B sin C
A B cos ab 2 Example # 1 : In any ABC, prove that = . C c sin 2 a b c Solution. Since = = = k (let) sin A sin B sin C
a = k sinA, b = k sinB and c = k sinC
L.H.S. =
ab c
=
k(sin A sin B ) k sin C
C A B A B A B A B cos cos sin cos cos 2 2 2 2 2 = = = = R.H.S. C C C C C sin cos sin cos sin 2 2 2 2 2 Hence L.H.S. = R.H.S.
Proved
Example # 2 : In any ABC, prove that (b2 – c 2) cot A + (c 2 – a2) cot B + (a2 – b2) cot C = 0 Solution.
Since
a = k sinA, b = k sinB and c = k sinC (b2 – c 2) cot A = k 2 (sin2B – sin2C) cot A = k 2 sin (B + C) sin (B – C) cot A
= k 2 sin A sin (B – C)
cos A sin A
= – k 2 sin (B – C) cos (B + C) =– = –
( cos A = – cos (B + C))
k2 [2sin (B – C) cos (B + C)] 2
k2 [sin 2B – sin 2C] 2
k2 [sin 2C – sin 2A] 2 k2 and (a2 – b2) cot C = – [sin 2A – sin 2B] 2 adding equations (i), (ii) and (iii), we get (b2 – c 2) cot A + (c 2 – a2) cot B + (a2 – b2) cot C = 0 Similarly
RESONANCE
(c 2 – a2) cot B = –
..........(i) ..........(ii) ..........(iii) Hence Proved
1
SOT : RAV Sir Self Practice Problems In any ABC, prove that
(1)
A A a sin B = (b + c) sin . 2 2
(2)
a 2 sin(B C) b 2 sin(C A ) c 2 sin( A B) + + =0 sin B sin C sin C sin A sin A sin B
(3)
A B tan tan c 2 2 = . A B ab tan tan 2 2
Cosine Formula : In any ABC (i)
cos A =
(ii)
cos B =
b 2 c2 a 2 2bc
or a² = b² + c² 2bc cos A = b2 + c 2 + 2bc cos (B + C)
c2 a 2 b 2 2 ca
(iii)
cos C =
a 2 b 2 c2 2a b
Example # 3 : In a triangle ABC if a = 13, b = 8 and c = 7, then find sin A.
Solution :
64 49 169 b2 c 2 a2 = 2 .8 .7 2bc
cosA =
cosA = –
sinA = sin
1 2 2 = 3
A=
3 2
2 3
Ans.
Example # 4 : In a ABC, prove that a (b cos C – c cos B) = b2 – c 2
Solution :
=
a2 b2 c 2 2ab
cosC
2 2 a 2 b 2 c 2 2 c a c b b L.H.S. = a 2ab 2ac
=
cos B =
a2 b2 c 2 (a 2 c 2 b 2 ) – = (b2 – c 2) 2 2
= R.H.S. Hence L.H.S. = R.H.S.
RESONANCE
&
a2 c 2 b2 2ac
Since
Proved
2
SOT : RAV Sir a b c a Example # 5 : If in aABC, A = 60°, then find the value of 1 1 . c c b b
Solution.
a b c a c ab bc a (b c )2 a 2 (b 2 c 2 a 2 ) 2bc 1 1 = = = c c b b c b bc bc
b2 c 2 a2 b2 c 2 a2 = +2 =2 +2 2 bc bc = 2cosA + 2 = 3
{ A = 60°}
a b c a 1 1 = 3 c c b b
Self Practice Problems : (4)
The sides of a triangle ABC are a, b,
(5)
In a triangle ABC, prove that
a 2 ab b 2 , then prove that the greatest angle is 120°.
a(cosB + cosC) = 2(b + c) sin2
A . 2
Projection Formula : In any ABC (i) a = b cosC + c cosB (ii)
b = c cosA + a cosC
(iii)
c = a cosB + b cosA
Example # 6 : In a triangle ABC, prove that Solution.
Note:
a(b cosC – c cosB) = b2 – c 2
L.H.S. = a (b cosC – c cosB) = b (a cosC) – c (a cosB) ............(i) From projection rule, we know that b = a cosC + c cosA a cosC = b – c cosA & c = a cosB + b cosA a cosB = c – b cosA Put values of a cosC and a cosB in equation (i), we get L.H.S. = b (b – c cos A) – c(c – b cos A) = b2 – bc cos A – c 2 + bc cos A = b2 – c 2 = R.H.S. Hence L.H.S. = R.H.S. Proved
We have also proved a (b cosC – c cosB) = b2 – c 2 by using cosine – rule in solved *Example.
Example # 7 : In a ABC, prove that (b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c. Solution.
L.H.S. = = = = = Hence L.H.S. =
RESONANCE
(b + c) cos A + (c + a) cos B + (a + b) cos C b cos A + c cos A + c cos B + a cos B + a cos C + b cos C (b cos A + a cos B) + (c cos A + a cos C) + (c cos B + b cos C) a+b+c R.H.S. R.H.S. Proved
3
SOT : RAV Sir Self Practice Problems In a ABC, prove that (6)
B 2 C c cos 2 = a + b + c. 2 b cos 2 2
(7)
cos B c b cos A = . cos C b c cos A
(8)
cos A cos B cos C a2 b2 c 2 + + = . c cos B b cos C a cos C c cos A a cos B b cos A 2abc
Napier’s Analogy - tangent rule : In any ABC (i)
tan
BC A bc = cot 2 2 bc
(iii)
tan
ab AB C = cot ab 2 2
(ii)
tan
c a B CA = cot c a 2 2
Example # 8 : Find the unknown elements of the ABC in which a = Solution.
3 + 1, b =
3 – 1, C = 60°.
a=
3 + 1, b = 3 – 1, C = 60° A + B + C = 180°
A + B = 120°
ab C A B = From law of tangent, we know that tan cot a b 2 2
=
.......(i)
( 3 1) ( 3 1) ( 3 1) ( 3 1)
cot 30° =
2 3
cot 30°
A B =1 tan 2
A B = = 45° 4 2
A – B = 90° From equation (i) and (ii), we get A = 105° and B = 15° Now,
From sine-rule, we know that
a sin C ( 3 1) sin 60 c= = = sin A sin105
c=
6
c=
6 , A = 105°, B = 15°
RESONANCE
2
.......(ii)
a b c = = sin A sin B sin C
3 2 3 1
( 3 1)
sin105° =
2 2 Ans.
4
3 1 2 2
SOT : RAV Sir Self Practice Problem 7 A , then find the value of tan . 25 2
(9)
In a ABC if b = 3, c = 5 and cos (B – C) =
(10)
A B B C CA tan tan If in a ABC, we define x = tan , y = tan and 2 2 2 2 C A B tan z = tan , then show that 2 2
Answer :
x + y + z = – xyz.
1 3
(9)
Trigonometric Functions of Half Angles :
(i)
sin
(s b) (s c) (s c) (s a ) A B C = , sin = , sin = ca b c 2 2 2
(ii)
cos
s (s a ) s (s b ) A B C = , cos = , cos = ca b c 2 2 2
(iii)
tan
A = 2
(s a ) (s b) ab
s (s c) ab
(s b) (s c) abc (s b)(s c ) = = , where s = is semi perimeter and s (s a ) s (s a ) 2
is the area of triangle.
(iv)
sin A =
2 bc
s(s a )(s b)(s c) =
2 bc
Area of Triangle () =
1 1 1 ab sin C = bc sin A = ca sin B = s (s a ) (s b) (s c) 2 2 2
Example # 9 : In a ABC if a, b, c are in A.P., then find the value of tan Solution :
A C . tan 2 2
Since
tan
A = s(s a) 2
tan
2 A C . tan = 2 s (s a)(s c ) 2 2
tan
sb b A C . tan = =1– s s 2 2
........(i)
it is given that a, b, c are in A.P. abc 3b s= = 2 2 b 2 = put in equation (i), we get s 3
RESONANCE
tan
and tan
2 A C . tan =1– 3 2 2
C = s(s c ) 2
tan
2 = s (s – a) (s – b) (s – c)
2b = a + c
1 A C . tan = 3 2 2
Ans.
5
SOT : RAV Sir Example # 10 : In a ABC if b sinC(b cosC + c cosB) = 42, then find the area of the ABC. Solution :
b sinC (b cosC + c cosB) = 42 From projection rule, we know that a = b cosC + c cosB put in (i), we get ab sinC = 42
1 ab sinC 2
=
= 21 sq. unit
........(i) given
........(ii)
from equation (ii), we get
C A B tan = 2c cot . Example # 11 : In any ABC, prove that (a + b + c) tan 2 2 2 Solution :
A B tan L.H.S. = (a + b + c) tan 2 2
tan
(s b)(s c ) (s a)(s c ) L.H.S. = (a + b + c) s(s a) s(s b)
A 2
=
(s b)(s c ) s(s a)
= 2s
=2
=2
= 2c
sc s
and tan
B = 2
(s a)(s c ) s(s b)
sb sa s b s a
sbsa s( s c ) (s a)(s b)
2s= a + b + c
2s – b – a = c
cot
c s( s c ) (s a)(s b)
s(s c ) (s a)(s b)
= 2c cot
s(s c ) (s a)(s b)
C = 2
C = R.H.S. 2
Hence L.H.S. = R.H.S.
Proved .
m - n Rule : In any triangle ABC if D be any point on the base BC, such that BD : DC :: m : n and if BAD= , DAC = , CDA = , then (m + n) cot m cot n cot n cot B m cot C
RESONANCE
6
SOT : RAV Sir Example # 12 : If the median AD of a triangle ABC is perpendicular to AB, prove that tan A + 2tan B = 0. Solution :
From the figure, we see that = 90° + B (as is external angle of ABD)
Now if we apply m-n rule in ABC, we get (1 + 1) cot (90º + B) = 1. cot 90° – 1.cot (A – 90°) – 2 tan B = cot (90° – A) – 2 tan B = tan A tan A + 2 tan B = 0 Hence proved. Example # 13 : The base of a triangle is divided into three equal parts. If t1, t2, t3 be the tangents of the angles subtended by these parts at the opposite vertex, prove that
1 1 1 4 1 2 = t1 t 2 t2 Solution :
1 1 . t 2 t3
Let point D and E divides the base BC into three equal parts i.e. BD = DE = EC = d (Let) and let , and be the angles subtended by BD, DE and EC respectively at their opposite vertex. t1 = tan, t2 = tan and t3 = tan Now in ABC again
BE : EC = 2d : d = 2 : 1 from m-n rule, we get (2 + 1) cot = 2 cot ( + ) – cot 3cot = 2 cot ( + ) – cot
in ADC DE : EC = d : d = 1 : 1 if we apply m-n rule in ADC, we get (1 + 1) cot = 1. cot – 1 cot 2cot = cot – cot .........(ii)
from (i) and (ii), we get
2 cot( ) cot 3 cot = cot cot 2 cot
3cot – 3cot = 4cot ( + ) – 2 cot 3cot – cot = 4 cot ( + ) cot . cot 1 3cot – cot = 4 cot cot 3cot2+ 3cot cot – cot cot – cot cot = 4 cot cot – 4 4 + 3cot2 = cot cot + cot cot + cot cot 4 + 4cot2 = cot cot + cot cot + cot cot + cot2 4(1 + cot2) = (cot + cot) (cot + cot)
1 1 1 4 1 2 = tan tan tan
1 1 1 1 1 4 1 2 = t1 t 2 t 2 t 3 t2
RESONANCE
.........(i)
1 1 tan tan
Hence proved.
7
SOT : RAV Sir Self Practice Problems : (11)
In a ABC, the median to the side BC is of length
1 11 6 3
unit and it divides angle A into the
angles of 30° and 45°. Prove that the side BC is of length 2 unit.
Radius of Circumcirlce : If R be the circumradius of ABC, then R =
a b c abc = = = 2 sin A 2 sin B 2 sin C 4
Example # 14 : In a ABC, prove that sinA + sinB + sinC = Solution :
In a ABC, we know that
s R
a b c = = = 2R sin A sin B sin C
a b c , sinB = and sinC = . 2R 2R 2R
sin A =
sinA + sinB + sinC =
abc 2s = 2R 2R
sinA + sinB + sinC =
s . R
a + b + c = 2s
Example # 15 : In a ABC if a = 13 cm, b = 14 cm and c = 15 cm, then find its circumradius. Solution :
R=
=
s=
=
R=
abc 4
.......(i)
s( s a)( s b)( s c )
abc = 21 cm 2
21 8 7 6 = 72 42 32
131415 65 = cm 8 484
Example # 16 : In a ABC, prove that s = 4R cos Solution.
= 84 cm 2 R=
65 cm. Ans. 8
A B C . cos . cos . 2 2 2
In a ABC,
A = 2
s( s a) B , cos = bc 2
cos
R.H.S. = 4R cos
=
Hence
RESONANCE
abc .s
s( s b) C and cos = ca 2
s( s c ) abc and R = ab 4
A B C . cos . cos . 2 2 2
s(s a)(s b)(s c ) (abc )2
=s
=
s( s a)( s b)( s c )
= L.H.S. R.H.S = L.H.S. proved.
8
SOT : RAV Sir
Example # 17 : In a ABC, prove that
Solution :
1 1 1 1 4R + + – = . sa sb sc s
4R 1 1 1 1 + + – = sa sb sc s 1 1 1 1 + L.H.S. = s a s b s c s
=
2s a b (s s c ) + (s a)(s b) s(s c )
=
c c + (s a)(s b) s(s c )
2s = a + b + c
2s 2 s(a b c ) ab s( s c ) ( s a)(s b) =c =c 2 s(s a)( s b)(s c ) 2s 2 s(2s) ab abc 4R 4R L.H.S. = c = = 2 = 2 2
L.H.S. =
abc 4
R=
abc = 4R
4R = R.H.S.
Self Practice Problems : In a ABC, prove the following : (11) a cot A + b cotB + c cot C = 2(R + r). (12)
s s s r 4 1 1 1 = . a b c R
(13)
If , , are the distances of the vertices of a triangle from the corresponding points of contact with the incircle, then prove that
y = r2 y
Radius of The Incircle : If ‘r’ be the inradius of ABC, then
(i)
r=
s B C sin 2 2 and so on A cos 2
(ii)
r = (s a) tan
(iv)
r = 4R sin
a sin (iii)
r=
RESONANCE
A B C = (s b) tan = (s c) tan 2 2 2
A B C sin sin 2 2 2
9
SOT : RAV Sir
Radius of The Ex- Circles : If r1, r2 , r3 are the radii of the ex-circles of ABC opposite to the vertex A, B, C respectively, then
(i)
r1 =
; ; r2 = r3 = sa sb sc
(ii)
r1 = s tan
(iii)
r1 =
a cos B2 cos C2 cos A2
(iv)
r1 = 4 R sin
and so on
Example # 18 : In a ABC, prove that Solution :
L.H.S
A 2
;
r2 = s tan
B 2
;
r3 = s tan
C 2
A B C . cos . cos 2 2 2
r1 + r2 + r3 – r = 4R = 2a cosecA
= r1 + r2 + r3 – r =
+ + – sa sb sc s
s b s a s s c 1 1 1 1 = + = sa sb sc s (s a)(s b) s(s c ) s( s c ) ( s a)(s b) c c = = c ( s a)( s b) s( s c ) s(s a)( s b)(s c )
2s 2 s(a b c ) ab abc = = c 2
= 4R = 2a cosec A Hence L.H.S.
= R.H.S. = R.H.S.
a + b + c = 2s
R=
a = 2R = a cosec A sin A
abc 4
proved
Example # 19 : If the area of a ABC is 96 sq. unit and the radius of the escribed circles are respectively 8, 12 and 24. Find the perimeter of ABC. Solution :
= 96 sq. unit r1 = 8, r2 = 12 and r3 = 24
r1 =
sa
s – a = 12
.........(i)
r2 =
sb
s–b=8
.........(ii)
r3 =
sc
s–c=4
.........(iii)
adding equations (i), (ii) & (iii), we get 3s – (a + b + c) = 24 s = 24 perimeter of ABC = 2s = 48 unit. Ans.
RESONANCE
10
SOT : RAV Sir Self Practice Problems In a ABC, prove that (14)
r 1r 2 + r 2r 3 + r 3r 1 = s 2
(15)
rr1 + rr2 + rr3 = ab + bc + ca – s 2
(16)
c r1 r r2 r + = r . a b 3
(17)
If A, A1, A2 and A3 are the areas of the inscribed and escribed circles respectively of a ABC,
1
then prove that
A
=
1 A1
1
+
A2
+
1 A3
.
Length of Angle Bisectors, Medians & Altitudes :
(i)
Length of an angle bisector from the angle A = a =
(ii)
Length of median from the angle A = m a =
1 2
&(iii)
Length of altitude from the angle A = Aa =
2 a
2 2 2 NOTE : ma m b m c =
2 bc cos A 2 bc
;
2 b2 2 c2 a 2
3 2 (a + b2 + c 2) 4
Example # 20 : AD is a median of the ABC. If AE and AF are medians of the triangles ABD and ADC respectively, and AD = m 1, AE = m 2 , AF = m 3 , then prove that m 22 + m 32 – 2m 12 = Solution :
In ABC AD2 =
1 (2b2 + 2c 2 – a2) = m 12 4
In ABD, AE2 = m 22 =
.........(i)
1 a2 (2c 2 + 2AD2 – ) 4 4
.........(ii)
2 2AD 2 2b 2 a 4 by adding equations (ii) and (iii), we get
Similarly in ADC, AF2 = m 32 =
RESONANCE
a2 . 8
m 22 + m 32 =
1 4
1 4
........(iii)
2 4AD 2 2b 2 2c 2 a 2
2 2 1 2b 2c 2 a 2 = AD + 2 4
2 2 2b 2c 2 a 2 a 2
= AD2 +
1 4
= AD2 +
a2 a2 1 (2b2 + 2c 2 – a2) + = AD2 + AD2 + 4 8 8 11
SOT : RAV Sir 2
= 2AD2 +
2
a a = 2m 12 + 8 8
m 22 + m 32 – 2m 12 =
a2 8
AD2 = m 12
Hence Proved
Self Practice Problem : (18)
In a ABC if a = 5 cm, b = 4 cm, c = 3 cm. ‘G’ is the centroid of triangle, then find circumradius of GAB. Answer :
(18)
5 12
13
The Distances of The Special Points from Vertices and Sides of Triangle : (i)
Circumcentre (O)
:
OA = R and O a = R cos A
(ii)
Incentre ()
:
A = r cosec
(iii)
Excentre (1)
:
1 A = r1 cosec
(iv)
Orthocentre (H)
:
HA = 2R cos A and Ha = 2R cos B cos C
(v)
Centroid (G)
:
GA =
1 3
A and a = r 2 A and 1a = r1 2
2b 2 2c 2 a 2 and Ga =
2 3a
Example # 21 : If x, y and z are respectively the distances of the vertices of the ABC from its orthocentre, then prove that Solution :
(i)
abc a c b + + = xyz x z y
(ii) x + y + z = 2(R + r)
and
x = 2R cosA, y = 2R cosB, z = 2R cosC a = 2R sinA, b = 2R sinB, c = 2R sinC
a c b + + = tanA + tan B + tan C x z y
.........(i)
&
abc xyz = tanA. tanB. tanC
........(ii)
We know that in a ABC
From equations (i) and (ii), we get
x + y + z = 2R (cosA + cosB + cosC)
in a ABC
A B C x + y + z = 2R 1 4 sin . sin . sin 2 2 2
tanA = tanA
abc a c b + + = xyz x z y
cosA + cosB + cosC = 1 + 4sin
A B C = 2 R 4R sin . sin . sin 2 2 2
RESONANCE
and
A B C sin sin 2 2 2
r = 4R sin
B C A sin sin 2 2 2
x + y + z = 2(R + r)
12
SOT : RAV Sir Self Practice Problems
A B C tan tan . 2 2 2
(19)
If be the incentre of ABC, then prove that A .B .C = abc tan
(20)
If x, y, z are respectively be the perpendiculars from the circumcentre to the sides of ABC,
abc a c b + + = . 4 xyz x z y
then prove that
Orthocentre and Pedal Triangle : The triangle KLM which is formed by joining the feet of the altitudes is called the Pedal Triangle. (i)
Its angles are 2A, 2B and 2C.
(ii)
Its sides are a cosA = R sin 2A, b cosB = R sin 2B and c cosC = R sin 2C
(iii)
Circumradii of the triangles PBC, PCA, PAB and ABC are equal.
Excentral Triangle : The triangle formed by joining the three excentres 1, 2 and 3 of ABC is called the excentral or excentric triangle. (i)
ABC is the pedal triangle of the 1 2 3.
(ii)
Its angles are
(iii)
Its sides are 4R cos
(iv)
1 = 4 R sin
(v)
Incentre of ABC is the orthocentre of the excentral 1 2 3.
A B C , . and 2 2 2 2 2 2
A ; 2
A B C , 4R cos and 4R cos . 2 2 2 2 = 4 R sin
B C ; 3 = 4 R sin . 2 2
Distance Between Special Points : (i)
Distance between circumcentre and orthocentre OH2 = R2 (1 – 8 cosA cos B cos C)
(ii)
Distance between circumcentre and incentre O2 = R2 (1 – 8 sin
(iii)
Distance between circumcentre and centroid OG2 = R2 –
RESONANCE
A B C sin sin ) = R2 – 2Rr 2 2 2
1 2 (a + b2 + c2) 9
13
SOT : RAV Sir Example # 22 : If is the incentre and 1, 2, 3 are the centres of escribed circles of the ABC, prove that (i) 1. 2 . 3 = 16R2r (ii) 12 + 232 = 22 + 312 = 32 + 122 Solution : (i) We know that 1 = a sec
A B , 2 = b sec 2 2
12 = c. cosec 2 3 = a cosec
and 3 = c sec
C 2
C , 2 A B and 31 = b cosec 2 2 A B C . sec . sec 2 2 2
1 . 2 . 3 = abc sec
a = 2R sin A, b = 2R sinB and c = 2R sinC
equation (i) becomes
1. 2 . 3 = (2R sin A) (2R sin B) (2R sinC) sec
......(i)
A B C sec sec 2 2 2
A A B B C C 2 sin cos 2 sin cos 2 sin cos 2 2 2 2 2 2 = 8R3 . A B C cos . cos . cos 2 2 2 = 64R3 sin
A B C sin sin 2 2 2
1 . 2 . 3 = 16R r 2
r = 4R sin
A B C sin sin 2 2 2
Hence Proved
1 + 23 = 2 + 31 = 3 + 12 2
(ii)
2
2
2
2
a2 A A + a2 cosec 2 = A A 2 2 sin 2 cos 2 2 2
1 + 23 = a2 sec 2
a = 2 R sinA = 4R sin
2
2
2
A A cos 2 2
A A . cos 2 2 2 2 = 16R 2 A 2 A sin . cos 2 2
16 R 2 sin 2
1 + 23 = 2
2
Similarly we can prove 22 + 312 = 32 + 122 = 16R Hence 12 + 232 = 22 + 312 = 32 + 122
2
Self Practice Problem : (21)
In a ABC, if b = 2 cm, c =
3 cm and A =
, then find distance between its circumcentre 6
and incentre. Answer :
RESONANCE
(21)
2 3 cm
14
