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Properties of Triangles
Key points:
1.
Sine rule : In ΔABC,
a
b
c
=
=
= 2R Where R is the circum-radius.
sin A sin B sin C
⇒ a = 2R sin A, b = 2R sin B, c = 2R sin C
a : b : c = sin A : sin B : sin C.
2.
Cosine rule : In ΔABC,
a2 = b2 + c2 – 2bc cos A
b2 = c2 + a2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
or
cos A =
b2 + c2 − a 2
c2 + a 2 − b2
; cos B =
;
2bc
2ca
a 2 + b2 − c2
⇒ cos A : cos B : cos C.
2ab
cos C =
= a(b2 + c2 – a2) : b(c2 + a2 – b2) : c(a2 + b2 – c2)
3.
Projection rule : In ΔABC
a = b cos C + c cos B,
4.
b = c cos A + a cos C,
c = a Cos B + bcos A
Mollwiede’s rule : In ΔABC
a−b
=
c
A−B
2 ’
C
cos
2
sin
a+b
=
c
A−B
2
C
sin
2
cos
Similarly the other two can be written by symmetry.
5.
Tangent rule (or) Napier’s analogy : In ΔABC
tan
a−b
A−B
C
=
cot ;
a+b
2
2
tan
b−c
B−C
A
=
cot ;
b+c
2
2
tan
c−a
C−A
B
=
cot
c+a
2
2
6.
Half angle formulae :
i.
sin
A
=
2
(s − b)(s − c)
;
bc
sin
B
=
2
(s − c)(s − a)
;
ca
ii.
cos
A
=
2
s(s − a )
;
bc
cos
B
=
2
s(s − b)
C
; cos =
ac
2
sin
C
=
2
s(s − c)
ab
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(s − a)(s − b)
.
ab
iii. tan
A
=
2
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(s − b )(s − c ) = (s − b )(s − c ) ;
(s − a )(s − c ) =
B
tan =
2
s(s − a )
s(s − b )
Δ
tan
C
=
2
(s − a )(s − b )
s(s − c )
cot
B
=
2
s(s − b )
=
(s − c )(s − a )
b2 + c2 − a 2
;
4Δ
7.
cot A =
8.
Area of ΔABC is given by
i.
Δ=
Δ
s(s − b )
;
Δ
cot B =
cot
cot
C
=
2
c2 + a 2 − b2
;
4Δ
A
=
2
Δ
s(s − a )
s(s − a )
=
;
(s − b )(s − c )
Δ
s(s − c )
=
(s − a )(s − b )
cot C =
s(s − c )
Δ
a 2 + b2 − c2
.
4Δ
1
1
1
ab sin C = bc sin A = ca sin B
2
2
2
s(s − a )(s − b )(s − c ) .
ii. Δ =
iii. Δ =
(s − a )(s − b ) iv.
=
(s − a )(s − c ) ;
abc
4R
iv. Δ =2R2 sin A sin B sin C
v.
Δ = rs
vi. Δ =
9.
r r1 r2 r3
If ‘r’ is radius of in circle and r1, r2, r3 are the radii of ex-circles opposite to the vertices A, B,
C of ΔABC respectively then
i. r =
Δ
Δ
Δ
Δ
, r1 =
, r2 =
, r3 =
s
s−a
s−b
s−c
ii. r = 4R sin
A
B
C
sin sin
2
2
2
r1 = 4R sin
A
C
cos cos
2
2
r2 = 4R cos
A
B
C
sin cos
2
2
2
r3 = 4R cos
A
B
C
cos sin
2
2
2
A
B
C
= (s − b ) tan = tan (s − c )
2
2
2
B
A
C
r1 = s tan = (s − b ) cot = (s − c ) cot
2
2
2
B
A
C
r2 =s tan
= (s − c ) cot = (s − a ) cot
2
2
2
B
A
C
r3 = s tan = (s − a ) cot = (s − b ) cot
2
2
2
iii. r = (s − a ) tan
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1
1
1
1
=
+
+
r
r1
r2
r3
2
ii) rr1r2r3 = Δ
10. i)
11. i) r1r2 + r2r3 +r3r1 = s2
ii) r(r1 + r2 + r3) = ab + bc + ca – s2.
i) (r1 – r)(r2 + r3) = a2
(r2 – r)(r3 + r1) = b2
(r3 – r)(r1 + r2) = c2
ii) a = (r2 + r3)
12. i) r1 – r = 4R sin2
rr1
,
r2 r3
b = (r3 + r1)
rr2
,
r3 r1
A
,
2
r2 – r = 4R sin2
B
,
2
ii) r1 + r2 = 4R cos2
C
,
2
13. r1 + r2 + r3 = 4R,
r2 + r3 = 4R cos2
c = )r1 + r2)
r3 – r = 4R sin2
A
,
2
r + r2 + r3 – r1 = 4R cosA,
rr3
r1r2
C
2
r3 + r1 = 4R cos2
B
2
r + r3 + r2 – r2 = 4R cosB,
r + r1 + r2 – r3 = 4R cosC
14. In an equilateral triangle of side ‘a’
3a 2
i) area =
4
iii) r = R / 2
ii) R = a /
3
iv) r1 = r2 + r3 = 3R / 2
v) r : R : r1 = 1 : 2 : 3
PROBLEMS
VSAQ’S
1. If the lengths of the sides of a triangle are 3,4,5 find the circum radius of the triangle.
Sol. sides of the triangle are 3,4,5
Therefore, the triangle is rt. Angled triangle and its hypotenuse is 5
Circum radius =
2. Show that
1
5
. ( hypotenuse ) =
2
2
∑ a ( sin B − sin C ) = 0
Solutoin: -
∑
2 R sin A ( sin B − sin C ) ∵ a = 2 R sin A
2 R {sin A ( sin B − sin C ) + sin B ( sin C − sin A ) + sin C ( sin A − sin B )} = 0
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3. If a = 3 + 1 cms ∠B = 30 ∠C = 450 then find C
Solution : 0
∠B = 300
∠C = 450
but A + B + C = 1800 ⇒ A + 750 = 1800
A = 1050
a
c
3 +1
C
3 +1
C
=
⇒
=
⇒
=
0
0
sin A sin C
sin105
sin 45
⎛ 3 + 1⎞ ⎛ 1 ⎞
⎟
⎜
⎟ ⎜
⎝ 2 2 ⎠ ⎝ 2⎠
4. If a = 2 , b = 3, c = 4 then find cos A
Solution : -
Cosine Rule we known that cos A =
b2 + c2 − a 2
2bc
9 + 16 − 4
21 7
⇒ cos A =
=
2× 3× 4
24 8
63
5. If a = 26, b = 30 cos C =
then find C
65
Solution : cos A =
From cosine rule c 2 = a 2 + b 2 −2ab cos c
63
( ) ( 30 ) 65
c 2 = ( 26 ) + ( 30 ) − 2 26
2
2
⇒ c 2 = 676 + 900 − 1512
5
c 2 = 64 ⇒ c = 8
6. If the angles are in the ratio 1 : 5 : 6 then find the ratio of sides
Solution : -
Let A = x
B = 5x
C = 6x
We know that A + B + C = 1800 ⇒ 12 x = 1800 ⇒ x = 150
∴ A = 150 B = 5 × 150
C = 6 × 150
A = 150 B = 750 C = 900
Ratio of sides = a : b : c = 2 R sin A : 2 R sin B : 2 R sin C
= sin A : sin B : sin C
= sin150 : sin 750 : sin 900 =
=
7.
(
)(
3 −1 :
)
3 −1 3 +1
:
:1
2 2
2 2
3 +1 :2 2
Prove tht 2 {bc cos A + c a cos B + ab cos C} = a 2 + b 2 + c 2
L.H.S 2bc cos A + 2ca cos B + 2ab cos c
From cosine rule cos A =
b2 + c2 − a 2
a 2 + c2 − b2
cos B =
2bc
2ac
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a +b −c
2ab
2
b + c2 − a2
2
cos c =
{
∴ 2bc
2
2
2bc
} + 2ac ⎧ a + c
2
⎨
⎩
{
a 2 + b2 − c 2
− b2 ⎫
⎬ + 2ab
2ac
2ab
⎭
2
}
b2 + c 2 − a 2 + a 2 + c 2 − b2 + a 2 + b2 − c 2 = a 2 + b2 + c 2
8. .
Prove that
a 2 + b 2 − c 2 tan B
=
c 2 + a 2 − b 2 tan C
(
)
⎛ b ⎞
2
2
4
4
R
⎜
⎟
sin B
2ac
b
2R
= 2⎝ 2 ⎠ 2 =
× 2 2
=
tan B =
cos B a + c − b
2 R a + c − b2
2R a 2 + c 2 − b 2
2ac
4Δ
4Δ
tan B = 2 2
|||ly tan C = 2
2
a +c −b
a + b2− c2
4Δ
2
tan B a + c 2 − b 2 a 2 + b 2 − c 2
=
= 2 2 2
R.H.S
4Δ
tan C
a +c −b
2
2
2
a +b −c
(
9.
)
{∵ abc = 4 RΔ}
Prove that ( b + c ) cos A + ( c + a ) cos B + ( a + b ) cos C = a + b + c
Solution : -
( 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 ) + ( b cos c + c cos B ) + ( c cos A + a cos c )
c +a +b {from projection rule}
=a+b+c
9.
Prove that ( b − a cos c ) sin A = a cos A sin c
Solution : -
From projection rule b = a cos c + c cos A
L.H.S = ( a cos C + c cos A − a cos C ) sin A = ccos A sin A
= 2 Rsin C cos Asin A
10.
= ( 2 R sin A) cos A sin C
= a cos A sin C
If 4, 5 are two sides of a triangle and the include angle is 600 find the area
Solution: -
A = 600
1
1
Area of triangle Δ = bc sin A = × 4 × 5 sin 600 = 5 3
2
2
Given b = 4
C=5
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Show that b cos 2
11.
C
B
+ cos 2 = S
2
2
Solution : -
We known that cos
C
=
2
S ( S − C ) cos B
=
ab
2
S ( S − b)
ac
S (S − C)
S ( S − b)
C
B
+ c cos 2 = b
+ c
2
2
ab
ac
S
s
s
= {s − c + s} = [ 2 s − b − c ] = {a + b + c − b − c }
a
a
a
=S
a
b
c
If
then show that triangle ABC is equilateral
=
=
cos A cos B cos C
b cos 2
12.
Solution: a
b
c
2 R sin A 2 R sin B 2 R sin C
=
=
⇒
=
=
cos A cos B cos A
cos A
cos B
cos C
tan A = tan B = tan C ⇒ A = B = C
∴ Triangle ABC is equilateral
13. Show that Σ a cot A = 2(R + r).
Sol.
Σa cot A = Σ 2R sin A
cos A
sin A
= 2RΣ cos A
A
B
C⎞
⎛
= 2R ⎜1 + 4sin sin sin ⎟ (∵ from transformations)
2
2
2⎠
⎝
A
B
C⎞
⎛
= 2 ⎜ R + 4R sin sin sin ⎟ = 2(R + r)
2
2
2⎠
⎝
14. In ΔABC, prove that r1 + r2 + r3 – r = 4R.
Sol. r1 + r2 + r3 – r = 4R sin
A
B
C
A
B
C
A
B
C
cos cos + 4R cos sin cos + 4R cos cos sin
2
2
2
2
2
2
2
2
2
A
B
C
sin sin
2
2
2
A⎡
B
C
B
C⎤
A⎡ B
C
B
C⎤
= 4R sin ⎢cos cos − sin sin ⎥ + 4R cos ⎢sin cos + cos sin ⎥
2⎣
2
2
2
2⎦
2⎣
2
2
2
2⎦
−4R sin
= 4R sin
A
A
⎛ B+C⎞
⎛ B+C⎞
cos ⎜
⎟ + 4R cos sin ⎜
⎟
2
2
⎝ 2 ⎠
⎝ 2 ⎠
A
A
A
A
sin + 4R cos cos
2
2
2
2
A
A⎤
⎡
= 4R ⎢sin 2 + cos 2 ⎥
2
2⎦
⎣
= 4R sin
A
A
⎛
⎞
= 4R ⎜∵ sin 2 + cos 2 = 1⎟
2
2
⎝
⎠
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15. Show that
c − b cos A cos B
=
b − cos A cos C
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Hint: Apply projection formula i.e., C = b coA + a cosB
b = a cos C + C cos A
16. Prove that a {b cos C − c cos B} = b 2 − c 2
Solution : a {b cos c − c cos B} = ab cos c − ac cos B
Write cos C =
a 2 + b2 − c2
a 2 + c2 − b2
and simplify to get R.H.S
cos B =
ab
ac
17. In ΔABC, show that Σ(b + c) cos A = 2s.
Sol. L.H.S.
= (b + c)cos A + (c + a)cos B + (a + b)cos C
= (b cos A + a cos B) + (c cos B + b cos C) + (a cos C + c cos A)
= c + a + b = 2s = R.H.S.
18. In ΔABC, if (a + b + c), (b + c – a) = 3bc, find A.
Sol. 2s(2s – 2a) = 3bc
⇒
s(s − a) 3
A 3
= ⇒ cos 2 =
bc
4
2 4
⇒ cos
∴
A
3
=
= cos 30°
2
2
A
= 30° ⇒ A = 60°
2
19. In ΔABC, prove that
Sol. L.H.S. =
=
1 1 1 1
+ + = .
r1 r2 r3 r
1 1 1 s−a s−b s−c
+ + =
+
+
r1 r2 r3
Δ
Δ
Δ
3s − (a + b + c) 3s − 2s s 1
=
= = = R.H.S.
Δ
Δ
Δ r
20. Show that rr1r2 r3 = Δ 2 .
Sol. L.H.S. = rr1r2 r3 =
Δ Δ
Δ
Δ
⋅
⋅
⋅
s s−a s−b s−c
Δ4
= 2 = Δ 2 = R.H.S.
Δ
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r
21. In an equilateral triangle, find the value of .
R
r
Sol. =
R
4R sin
A
B
C
sin sin
2
2
2 = 4sin 2 30°
R
3
⎛1⎞ 1
= 4 ⎜ ⎟ = (∵ A = B = C = 60°)
2
⎝2⎠
22. If rr2 = r1r3, then find B.
Sol. rr2 = r1r3 ⇒
Δ Δ
Δ
Δ
⋅
=
⋅
s s−b s−a s−c
⇒ (s − a)(s − c) = s(s − b)
⇒
(s − c)(s − a)
B
= 1 ⇒ tan 2 = 1
s(s − b)
2
B
B
= 1 ⇒ = 45° ⇒ B = 90°
2
2
⇒ tan
23. If A = 90°, show that 2(r + R) = b + c.
Sol. L.H.S. = 2r + 2R = 2(s – a) tan
A
+ 2R ⋅1
2
= 2(s – a) tan 45° + 2R sin A
= (2s – 2a)1 + a (∵ A = 90°)
=b+c
= R.H.S.
24. In a triangle ABC express
∑
r1 cot
A
in terms of S
2
Solution : -
∑ r cot A / 2 = r cot A / 2
1
1
+ r2 cot B / 2 + r3 cot
C
2
r1 = s tan A / 2 r2 = s tan B / 2 r3 = s tan C / 2
= s tan
A
C
C
cot A / 2 + s tan B / 2 cot B / 2 + s tan cot
2
2
2
= s + s + s = 3s
25. Show that
r1
3
∑ ( s − b )( s − c ) = r
Solution : -
r1
r1
r2
r3
∑ ( s − b )( s −c ) = ( s − b )( s − c ) + ( s − c )( s − a ) + ( s − a )( s − b )
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Δ
Δ
Δ
But r1 =
r2 =
r3 =
s−a
s−b
s−c
=
=
Δ
Δ
Δ
+
+
( s − a )( s − b )( s − c ) ( s − a )( s − b )( s − c ) ( s − a )( s − b )( s − c )
3Δs
3 Δs
3
3
= 2 =
=
s ( s − a )( s − b )( s −c ) Δ
⎛Δ⎞ r
⎜ ⎟
⎝s⎠
SAQ’S
26. Prove that in a triangle ABC a cos A + b cos B + C cos c = 4 R sin A sin B sin C
Solution : -
L.H.S = a cos A + b cos B + C cos C= 2 R sin A cos A + 2 sin B cos B + 2 R sin C cos C
{
= R sin 2 A + sin 2 B + sin 2 C
}
Given that A + B + C = 1800
∴ L.H .S = R {sin ( A + B ) cos ( A − B ) + sin 2C}
= R {2 sin C cos ( A − B ) + 2 sin C cos C} = 2 R sin C {cos ( A− B ) + cos C}
{
(
= 2 R sin C cos ( A − B ) + cos 1800 − A + B
)}
= 2 R sin C {cos ( A − B ) − cos ( A+B )} = 4 R sin A sin B sin C
27. Prove that
∑
a3 sin ( B − C ) = 0
Solution : -
∑
a 3 sin ( B − C ) = ∑
8R3
∑
( 2 R sin A )
3
sin ( B − C ) = 8 R 3
sin 2 A sin A sin ( B − C ) = 8 R 3
{∵ A = 180
∑
0
(
)
(
sin 3 A sin ( B − C )
(
)
sin 2 A sin 1800 − B + C sin ( B − C )
−B+C
= 8 R 3 ∑ sin 2 A sin ( B + C ) . sin ( B − C ) = 8 R 3
{
∑
}
∑
(
sin 2 A sin 2 B − sin 2 C
)
(
)
= 8 R 3 sin 2 A sin 2 B − sin 2 C + sin 2 B sin 2 C − sin 2 A + sin 2 C sin 2 A − sin 2 B
=0
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)}
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a sin ( B − C ) b sin ( C − A ) C sin ( A − B )
28. In a triangle ABC prove that
=
=
b2 − c2
C 2 − a2
a 2 − b2
Solution : a sin ( B − C )
b −c
2
2
=
2 R sin A sin ( B − C ) ⎧∵ a = 2 R sin A : b = 2 R sin B ⎫
⎨
⎬
4 R 2 sin 2 B − sin 2 C ⎩C = 2 R sin C from sin e Rule ⎭
{
}
1 sin ( B + C ) . sin ( B − C )
{∵ sin A = sin ( B + C ) In a triangle ABC}
2R
sin 2 B − sin 2 C
1 sin 2 B − sin 2 C
1
=
2
2
2 R sin B − sin C 2 R
|||ly we have prove that
29. Prove that
b sin ( C − A )
c −a
2
2
=
C sin ( A − B ) 1
1
and
=
2R
a 2 − b2
2R
a cos A b
cos B c cos C
+
=
+
=
+
bc
a
ca
b
ab
c
Solution : -
a cos A a 2 + bc cos A
+
=
bc
a
abc
b2 + c2 − a 2
We knonw that cos A =
2bc
a 2 + bc cos A
=
abc
a2 +
bc ⎡⎣b 2 + c 2 − a 2 ⎤⎦
2a 2 + b 2 + c 2 − a 2
2bc
=
2abc
abc
a 2 + b2 + c2
2abc
b +
2
b cos B b + ac cos B
+
=
=
ca
b
abc
c cos c a 2 + b 2 + c 2
+
=
ab
c
2abc
2
ly
|||
|||ly
(
ac a 2 + c 2 − b 2
2ac
abc
)
a 2 + b2 + c 2
=
2abc
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30
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1 + cos ( A − B ) cos C a 2 + b 2
Prove that
=
1 + cos ( A − C ) cos B a 2 + c 2
Solution : -
In a triangle ABC A + B + C = 1800 ⇒ C = 1800 − A + B
(
B = 1800 − A + C
)
1+ cos ( A − B ) cos C
1 + cos ( A − C ) cos B
=
=
(
1 + cos ( A− B ) cos 1800 − A + D
(
1 + cos ( A − C ) . cos 1800 − A + C
{
1 − cos ( A − C ) cos ( A + C ) 1 − {cos
1 − cos ( A − B ) cos ( A + B )
=
)
)
}
C}
1 − cos 2 A − sin 2 B
2
A − sin
2
a2
b2
+
1 − cos A + sin B sin A + sin B 4 R 2 4 R 2 a 2 + b 2
=
=
= 2
= 2
a
c2
a + c2
1 − cos 2 A + sin 2 C sin 2 A + sin 2 C
+
4R2 4R2
2
2
2
31. If C = 600 then show that (i)
2
a
b
+
=1
b+c c+a
(ii)
b
a
+ 2
=0
2
c −a
c − b2
2
Solution : -
Given C = 600
C 2 = a 2 + b 2 − 2ab cos C
⎛1⎞
C 2 = a 2 + b 2 − 2 ab ⎜ ⎟
⎝2⎠
2
2
2
C + ab = a + b
a
b
+
=
b+c c+a
ac + a 2 + bc + b 2 a 2 + b 2 + ac + bc
=
( b + c )( c + a ) bc + ab + c 2 + ac
But a 2 + b 2 = c 2 + ab
c 2 + ab + ac + bc
=1
bc + ab + c 2 + ac
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{
(
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} {
)(
b c − b2 + a c2 − a 2
b
a
+ 2
=0⇒
c2− a2
c − b2
c2 − a 2 c 2 − b2
(ii)
2
)
}
{
}
c 2 ( a + b ) − ( a + b ) a 2 + b 2 − ab
bc 2 − b3 + ac 2 − a 3
= 2
=
c − a 2 c2 − b2
c2 − a 2 c2 − b2
(
=
)(
)
(
)(
)
( a + b ) ⎡⎣c 2 − {a 2 + b2 −ab} ⎤⎦ ( a + b ) ⎡⎣c 2 − ( c 2 +
(c
2
− a2
) (c − b )
2
2
=
(c
2
− a2
)(
)
ab − ab ⎤
⎦
2
2
c −b
)
=0
(iii)
In a triangle ABC
1
1
3
show that c = 600
+
=
a+c b+c a+b+c
Solution : -
Given
a+b+c a+b+c
+
=3
a+c
b+c
1
1
3
+
=
a+c b+c a+b+c
( a + c ) + b + a + (b + c ) = 3 ⇒
a+c
b+c
b+c
b
a
a +c
=3
+
+
+
a − c a+ c b + c b + c
b
a
b 2 + bc + a 2 + ac
+
=1⇒
=1
a+c b+c
ab + ac + bc + c 2
{
a 2 + b 2 − c 2 = ab ⇒ 2ab cos C = ab ∵ a 2 + b 2 − c 2 = 2abc cos C
cos C =
}
ab
1
ab 1
= ⇒ C = 600
= ⇒ C = 600 cos C =
2ab 2
2ab 2
32. If a : b : c = 7 : 8 : 9 then find cos A = cos B = cos C
12
b 2 + c 2 − a 2 64k 2 + 81k 2 − 49k 2
96k 2
=
=
Solution : - cos A =
2bc
2 ( 8k )( 9k )
2 × 8k × 9 k
a 2 + c 2 − b 2 49k 2 + 81k 2 − 64k 2
6633
11
=
cos B =
=
=
2ac
2 × 7 k × 9k
2 × 63 21
cos C =
a 2 + b 2 − c 2 49k 2 + 64k 2 − 81k 2
32k 2
2
=
=
=
2ab
2 × 7 k × 8k
20k × 8k 7
∴ cos A = cos B = co C =
2 11 2 2
11
2
=
= = × 21 = × 21 = × 21
3 21 7 3
21
7
= 14 :11: 6
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cos A cos B cos C a 2 + b 2 + c 2
+
+
=
33. Show that
a
b
c
2abc
LHS
cos A cos B cos C bc cos A + ca cos B + ab cos C
+
+
=
a
b
c
abc
=
2bc cos A + 2ca cos B + 2ab cos C
2bac
=
b2 + c2 − a 2 + a 2 + c2 − b2 + a 2 + b2 − c2
2abc
⎧⎪∵ 2bc cos A = b 2 + c 2 − a 2 : 2ac cos B = a 2 + c 2 − b 2 ⎫⎪
⎨
⎬
2
2
2
⎪⎩ 2ab cos c = a + b + c
⎪⎭
=
a 2 + b2 + c2
2abc
⎛ A− B ⎞
⎛ A+ B⎞
34. Prove that ( b − a ) cos c + c ( cos B − cos A ) = c sin ⎜
⎟ cos ec ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
Solution : - ( b − a ) cos c + c {cos B − cos A}
b cos c − a cos c + cos B − cos A = ( b cos c + cos B ) − ca cos C + cos A
a − b { from projection Rule}
= 2 R {sin A − sin B}
⎧
⎛ A+ B ⎞
⎛ A − B ⎞⎫
= 2 R ⎨2 cos ⎜
⎟ sin ⎜
⎟⎬
⎝ 2 ⎠
⎝ 2 ⎠⎭
⎩
⎧
C
C
⎛ A − R ⎞⎫
C
C ⎫ ⎛ A− B ⎞
⎧
2 R ⎨2sin sin ⎜
2 R ⎨2sin cos ⎬ sin ⎜
⎟ ⎬ cos
⎟
2
2
2
2⎭ ⎝ 2 ⎠
⎝ 2 ⎠⎭
⎩
⎩
=
C
C
cos
cos
2
2
⎛ A− B⎞
2 R sin C sin ⎜
⎟
⎝ 2 ⎠ = c sin Csin ⎛ A − B ⎞ cos ec ⎛ A + B ⎞
=
⎜
⎟
⎜
⎟
⎛ A+ B⎞
⎝ 2 ⎠
⎝ 2 ⎠
sin ⎜
⎟
⎝ 2 ⎠
⎧∵ In a triangle ABC
⎫
⎪
⎪
C⎬
⎨C ⎛ 0 A + B ⎞ A + B
0
⎪ 2 = ⎜ 90 − 2 ⎟ ; 2 = 90 − 2 ⎪
⎝
⎠
⎩
⎭
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( s − a )( s − b ) + c ( s − b )( s − c )
C
A
35. Express a sin 2 + c sin 2 = a
2
2
bc
ab
( s − b ){s − a + s − c}
b
( s − b ){2s − ac} = ( s − b ) { a
+b+ c − a − c
b
b
36. If b + c = 3a then find the value of cot
} = ( s − b)
B
c
cot
2
2
Solution : -
cot
s ( s − c)
( s − a )( s − c ) ( s − a )( s − b )
s ( s − b)
=
=
s ( s − b)
B
c
cot =
2
2
.
s ( s− c )
( s − a )( s − c ) ( s − a )( s − b )
=
s
2s
=
s − a 2(s − a)
b + c + a 3a + a 4a
=
=
=2
b + c − a 3a −a 2a
37. Show that ( b − c ) cos 2
2
A
A
2
+ ( b + c ) sin 2 = a 2
2
2
Solution : -
(b − c )
{b
2
2
cos 2
A
A
2
+ ( b + c ) sin 2
2
2
+ c 2 − 2bc} cos 2
A
A
+ {b 2 + c 2 + 2bc} sin 2
2
2
A
A⎫
A
A⎫
A
A⎫
⎧
⎧
⎧
b 2 ⎨cos 2 + sin 2 ⎬ + c 2 ⎨cos 2 + sin 2 ⎬ − 2bc ⎨cos 2 − sin 2 ⎬
2
2⎭
2
2⎭
2
2⎭
⎩
⎩
⎩
b 2 + c 2 − 2bc cos A = a 2 (from cosine rule)
|||ly prove that (i) ( c − a ) cos 2
2
(ii) ( a − b ) cos 2
2
C
B
2
+ ( c + a ) sin 2 = b 2
2
2
c
C
2
+ ( a + b ) sin 2 = c 2
2
2
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38. In ΔABC, prove that r + r1 + r2 – r3 = 4R cos C.
Sol.
A
B
C
A
B
C
A
B
C
A
B
C
r + r1 + r2 − r3 = 4R sin sin sin + 4R sin cos cos + 4R cos sin cos − 4R cos cos sin
2
2
2
2
2
2
2
2
2
2
2
2
= 4R sin
A⎡
B
C
B
C⎤
A⎡ B
C
B
C⎤
cos cos + sin sin ⎥ + 4R cos ⎢sin cos − cos cos ⎥
⎢
2⎣
2
2
2
2⎦
2⎣
2
2
2
2⎦
= 4R sin
A
A
⎛ B−C⎞
⎛ B−C⎞
cos ⎜
⎟ + 4R cos sin ⎜
⎟
2
2
⎝ 2 ⎠
⎝ 2 ⎠
⎛ B+C⎞
⎛ B−C⎞
⎛ B+C⎞ ⎛ B−C⎞
= 4R cos ⎜
⎟ cos ⎜
⎟ +4R sin ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎝ 2 ⎠ ⎝ 2 ⎠
⎡ ⎛ B+C⎞
⎛ B−C⎞
⎛ B + C ⎞ ⎛ B − C ⎞⎤
= 4R ⎢cos ⎜
⎟ cos ⎜
⎟ + sin ⎜
⎟ sin ⎜
⎟⎥
⎝ 2 ⎠
⎝ 2 ⎠ ⎝ 2 ⎠⎦
⎣ ⎝ 2 ⎠
⎛ B+C B−C⎞
= 4R cos ⎜
−
⎟
2 ⎠
⎝ 2
⎛ B+C−B+C⎞
= 4R cos ⎜
⎟
2
⎝
⎠
2C
= 4R cos
= 4R cos C
2
39. If r1 + r2 + r3 = r, then show that ∠C = 90°.
Sol. r1 + r2 = r − r3 ⇒
r1 + r2
=1
r − r3
…(1)
A
B
C
A
B
C
cos cos +4R cos sin cos
2
2
2
2
2
2
C⎡ A
B
A
B⎤
= 4R cos ⎢sin cos + cos sin ⎥
2⎣
2
2
2
2⎦
r1 + r2 = 4R sin
= 4R cos
C ⎡ A + B⎤
sin
2 ⎢⎣
2 ⎥⎦
C
C
⋅ cos
2
2
C
= 4R cos 2
2
= 4R cos
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A
B
C
A
B
C
r − r3 = 4R sin sin sin − 4R cos cos sin
2
2
2
2
2
2
C⎡ A
B
A
B⎤
= 4R sin ⎢sin sin − cos cos ⎥
2⎣
2
2
2
2⎦
= 4R sin
C⎡
⎛ A + B ⎞⎤
− cos ⎜
⎟⎥
⎢
2⎣
⎝ 2 ⎠⎦
= 4R sin
C⎡
C⎤
− sin ⎥
⎢
2⎣
2⎦
= −4R sin 2
C
2
C
4R sin 2
r − r3
2 = tan 2 C
=
r1 + r2 4R cos 2 C
2
2
C
∴ tan 2 = tan 45°
From(1)
2
C
= 45° ∴∠C = 90°
2
40. Prove that 4(r1r2 + r2r3 + r3r1) = (a + b + c)2.
Δ
Δ
Δ
Δ
Δ ⎤
⎡ Δ
⋅
+
⋅
+
⋅
Sol. 4(r1r2 + r2r3 + r3r1) = 4 ⎢
⎣ S − a S − b S − b S − c S − c S − a ⎥⎦
⎡ S−c+S−a +S− b ⎤
= 4Δ 2 ⎢
⎥
⎣ (S − a)(S − b)(S − c) ⎦
⎡ 3S − (a + b + c) ⎤
= 4Δ 2 ⎢
⎥
⎣ (S − a)(S − b)(S − c) ⎦
⎡
⎤
S2
= 4Δ 2 ⎢
⎥
⎣ S(S − a)(S − b)(S − c) ⎦
S2
= 4Δ 2 = 4S2
Δ
2
= (2S) 2 = (a + b + c) 2
⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ abc 4R
41. Prove that ⎜ − ⎟ ⎜ − ⎟ ⎜ − ⎟ = 3 = 2 2 .
r S
⎝ r r1 ⎠ ⎝ r r2 ⎠ ⎝ r r3 ⎠ Δ
Sol.
1 1 S S−a S−S+ a a
− = −
=
=
r r1 Δ
Δ
Δ
Δ
Similarly we get
1 1 b
1 1 c
− = and − =
r r2 Δ
r r3 Δ
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⎛ 1 1 ⎞⎛ 1 1 ⎞⎛ 1 1 ⎞
L.H.S. = ⎜ − ⎟ ⎜ − ⎟ ⎜ − ⎟
⎝ r r1 ⎠ ⎝ r r2 ⎠ ⎝ r r3 ⎠
a b c abc
=
Δ Δ Δ Δ3
4R ⋅ Δ 4R
4R
=
= 2 =
= R.H.S.
3
Δ
Δ
(rS) 2
=
42. Prove that
Solution : -
r1 ( r2 + r3 )
r1r2 + r2 r3 + r3r3
=a
LHS
Δ ⎧ Δ
Δ ⎫
+
⎨
⎬
s −a ⎩ s + b s − c ⎭
Δ2
Δ2
Δ2
+
+
( s − a )( s − b ) ( s − b )( s − c ) ( s − c )( s − a )
Δ2
×
( s − a )( s − b )( s − c )
aΔ
s ( s − a )( s − b )( s − c )
=
( s − a ) ( s −b ) ( s − c )
Δ
=
Δ2
{s − b + s − c}
( s − a )( s − b )( s − c )
Δ2
( s − a )( s − b ) + ( s − c )
( s − a )( s − b )( s − c )
×
a
3s − 2 s
aΔ
=a
Δ
43. Prove that r(r1 + r2 + r3) = ab + bc + ca – S2.
Sol. L.H.S. = r(r1 + r2 + r3)
=
Δ⎛ Δ
Δ
Δ ⎞
+
+
⎜
⎟
S ⎝S−a S− b S−c ⎠
⎡ (S − b)(S − c) + (S − a)(S − c) ⎤
⎥
Δ 2 ⎢⎢
+(S − a)(S − b)
⎥
=
S ⎢
(S − a)(S − b)(S − c)
⎥
⎢
⎥
⎣
⎦
Δ2
= 2
Δ
⎡S2 − Sc − Sb + bc + S2 − Sc − Sa ⎤
⎢
⎥
2
⎣⎢ +ac + S − Sb − Sa + ab
⎦⎥
= 3S2 − 2S(a + b + c) + bc + ca + ab
= 3S2 − 4S2 + bc + ca + ab
= ab + bc + ca − S2
= R.H.S.
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C
C
44. Show that (r1 + r2 ) tan = (r3 − r) cot = c .
2
2
Sol. (r1 + r2 ) tan
C
2
C
C
tan
2
2
C
sin
C
2
= 4R cos 2
2 cos C
2
C
C
= 4R sin cos
2
2
= 2R sin C = c
= 4R cos 2
...(1)
C
C
C
= 4R sin cot
2
2
2
C
cos
C
2
= 4R sin 2
2 sin C
2
C
C
= 4R sin cos
2
2
= 2R sin C = c
...(2)
(r3 − r) cot
From (1) and (2)
(r1 + r2 ) tan
C
C
= (r3 − r) cot = c
2
2
45. In a ΔABC, show that the sides a, b, c are in A.P. if and only if r1, r2, r3 are in H.P.
Sol. r1, r2, r3 are in H.P. ⇔
1 1 1
, , are in A.P.
r1 r2 r3
s−a s−b s−c
,
,
are in A.P.
Δ
Δ
Δ
⇔ s − a,s − b,s − c are in A.P.
⇔
⇔ –a, –a, –c are in A.P.
⇔ a, b, c are in A.P.
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46. If A, A1 , A2 , A3 are the areas of incircle and excircle of a triangle respectively then prove
1
1
1
1
+
+
=
A1
A2
A3
A
that
Solution : -
Here A = π r 2 A1 = π r12 ; A2 = π r22 ; A3 = π r32
LHS
1
1
1
1
1
1
+
+
=
+
+
2
2
A1
A2
A3
π r2
π r32
π r1
1 ⎧ 1 1 1 ⎫ 1 ⎧ s − a s −b s − c ⎫
+
+
⎨ + + ⎬=
⎨
⎬
Δ
Δ ⎭
π ⎩ r1 r2 r3 ⎭
π ⎩ Δ
1 ⎧ 3s − ( a + b + c ) ⎫
1 s
1
1
=
×
⎨
⎬=
Δ
π Δ
π ⎛Δ⎞
Δ ⎩
⎭
⎜ ⎟
⎝s⎠
1
1
1
1
=
=
2
π r
A
πr
.
LAQ’S
47. In a triangle ABC prove that cot A + cot B + cot C =
a 2+ b2 + c2
4Δ
Solution : cot A =
cos A
sin A
cos A =
b2 + c2− a 2
2bc
b2 + c2 − a 2
b2+ c2− a 2
b2+ c2− a 2
2bc
∴ cot =
=
=
sin A
2bc sin A
⎧1
⎫
4 ⎨ bc sin A⎬
⎩2
⎭
cot A =
b2 + c2 − a 2
4Δ
|||ly cot B =
1
⎧
⎫
⎨∵ Δ = bc sin A⎬
2
⎩
⎭
a2+ c2 − b2
a 2 + b2 − c2
cot c =
4Δ
4Δ
cot A + cot B+ cot C =
=
=
b2 + c2 − a 2 a 2 + c2 − b2
a 2 + b2 − c2
+
+
4Δ
4Δ
4Δ
b2 + c2 − a 2 + a 2 + c2 − b2 + a 2 + b2 − c2
4Δ
a 2+ b2 + c2
4Δ
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48. If cot
A
B
C
: cot
cot = 3 : 5 : 7 show that a : b : c = 6 : 5 : 4
2
2
2
Solution : -
cot
A
B
C s ( s − a) s ( s − c)
: cot : cot =
:
= 6:5:4
2
2
2
Δ
Δ
∴s − a : s − b : s − c = 6:5 = 4
Let s − a = 3k s − b = 5k s − c = 7 k
( s − a ) + ( s − b ) + ( s − c ) = 3k + 5k + 7k
3s − 2 s = 15k ⇒ s = 15k
s − a = 3k ⇒ 15k − a = 3k ⇒ a = 12k
s − b = 5k ⇒ 15k − 5k = b ⇒ b = 10k
s − c = 7 k ⇒ 15k − 7 k = c ⇒ c = 8k
a : b : c = 12k :10k : 8k ⇒ a : b : c = 6 : 5 : 4
A
B⎫
C
⎧
49. In a triangle ABC show that ( a + b + c ) ⎨ tan + tan ⎬ = 2c cot
2
2⎭
2
⎩
⎧ ( s − b )( s − c ) ( s − c )( s − a ) ⎫
A
B⎤
⎡
+
Solution : - ( a + b + c ) ⎢ tan + tan ⎥ = 2s ⎨
⎬
Δ
Δ
2
2⎦
⎣
⎩
⎭
2s ( s − c )
{s − b + s − a}
Δ
C
C
= 2 cot {2 s − b −a} ⇒ 2 cot
a + b +c− b − a
2
2
=
{
= 2c cot
}
C
2
50. Show that cos A + cos B + cos C = 1 +
r
.
R
Sol. L.H.S. = cos A + cos B + cos C
= 1 + 4sin
= 1+
= 1+
A
B
C
sin sin
2
2
2
4R sin
A
B
C
sin sin
2
2
2
R
(∵ From transformations Prove This In Exam )
r
= R.H.S.
R
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A
B
C
r
51. Show that cos 2 + cos 2 + cos 2 = 2 +
.
2
2
2
2R
Sol. L.H.S. = cos 2
A
B
C
+ cos 2 + cos 2
2
2
2
A
B
C
sin sin
2
2
2
( from transformationsPROVE THIS IN EXAM )
= 2 + 2sin
= 2+
= 2+
4R sin
A
B
C
sin sin
2
2
2
2R
r
= R.H.S.
2R
A
B
C s2
52. Prove that (i) cot + cot + cot =
2
2
2 Δ
Solution : -
A
B
C s ( s − a) s ( s − b) s ( s − c)
+ cot + cot =
+
+
2
2
2
Δ
Δ
Δ
S
s
s
= {s− a + s − b + s − c} = {3s − ( a+ b + c )} = ( 3s − 2 s )
Δ
Δ
Δ
2
s
s
= (s) =
Δ
Δ
A
B
C bc + ca + ab − s 2
Prove that tan + tan + tan =
2
2
2
Δ
cot
(i)
Solution : tan
A
B
C
+ tan + tan
2
2
2
( s − b )( s − c ) + ( s − c )( s − a ) + ( s − a )( s − b )
Δ
Δ
Δ
s 2 − cs − bs + bc + s 2 − as − cs + ac + s 2 − bs − as + ab
Δ
3s 2 − 2as − 2bs − 2cs + bc + ca + ab
Δ
bc + ca + ab + 3s 2 − 2 s ( a + b + c )
Δ
=
bc + ca + ab + 3s 2 − 2 s ( 2 s )
Δ
bc + ca + ab + 3s 2 − 4 s 2 bc + ca + ab − s 2
=
Δ
Δ
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A
B
C
+ cot + cot
2
2
2
cot A + cot B + cot C
cot
(iii)
Solution : cot
A
B
C s ( s − a) s ( s − b) s ( s − c)
+ cot + cot =
+
+
Δ
Δ
Δ
2
2
2
s
s
s
s2
s
a
s
b
s
c
s
a
b
c
s
s
−
+
−
+
−
=
3
−
−
−
=
×
3
−
2
=
{
} {
}
(
)
Δ
Δ
Δ
Δ
b2 + c2 − a 2
cos A
b2 + c2 − a 2
b2+ c2 − a 2
2bc
cot A =
=
=
=
bc sin A
sin A
sin A
⎧1
⎫
Δ ⎨ bc sin A⎬
⎩2
⎭
=
b2 + c 2 − a 2 ⎧ 1
⎫
⎨∵ bc sin A = Δ ⎬
4Δ
⎩ 2
⎭
cot A + cot B + cot C =
=
b2 + c2 − a 2 c2 + a 2 − b2
a 2 + b2 − c2
+
+
4Δ
4Δ
4Δ
b2 + c 2 − a 2 + c2 + a 2 − b2 + a 2 + b2 − c2 a 2 + b2 + c2
=
4Δ
4Δ
A
B
C
s2
+ cot
+ cot
4Δ
s2
Δ
2
2
2 =
× 2 2
=
2
2
2
a +b +c
cot A + cot B + cot C
Δ a + b + c2
4Δ
cot
( 2s )
(a + b + c)
4s 2
= 2
= 2
= 2
2
2
2
2
a +b +c
a +b +c
a + b2 + c2
2
53. Show that (i)
Solution : -
2
⎛ A− B⎞
∑ ( a + b ) tan ⎜⎝ 2 ⎟⎠ = 0
⎛ A− B⎞
∑ ( a + b ) tan ⎜⎝ 2 ⎟⎠ from Napiries Rules
C
⎛ A− B⎞ a−b
tan ⎜
cot
⎟=
2
⎝ 2 ⎠ a+b
∴∑
∑
s
Δ
A− B⎞
⎟=∑
⎝ 2 ⎠
( a + b ) tan ⎛⎜
(a − b) s ( s − c) =
Δ
s
Δ
⎛
⎞
⎝
⎠
( a + b ) ⎜⎜ aa +− bb ⎟⎟ cot 2c
∑ ( a − b )( s − c )
∑ s (a − b) − c (a − b) = 0
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s
⎡ s ( a − b ) + s ( b − c ) + x ( c − a ) ⎤⎦ − {c ( a − b ) + b ( c − a ) + a ( b − c )} = 0
Δ⎣
(ii)
b−c
A b+c
A
cot +
tan = 2 cos ec ( B − C )
2 b−c
2
b+c
Solution : -
b−c
b+c
cot A / 2 +
tan A / 2
b+c
b−c
b−c
A
⎛B−C⎞
cot = tan ⎜
⎟
b+ c
2
⎝ 2 ⎠
From Napiers Rule
b+c
A
⎛B−C⎞
tan = cot ⎜
⎟
b−c
2
⎝ 2 ⎠
⎛B−C⎞
⎛B−C⎞
∴ LHS = tan ⎜
⎟ + cot ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎛B−C⎞
⎛B−C⎞
2⎛ B −C ⎞
2⎛ B −C ⎞
sin ⎜
⎟ cos ⎜
⎟ sin ⎜
⎟ + cos ⎜
⎟
⎝ 2 ⎠ +
⎝ 2 ⎠=
⎝ 2 ⎠
⎝ 2 ⎠
=
⎛ B− C ⎞
⎛B−C⎞
⎛ B− C ⎞
⎛B−C⎞
cos ⎜
sin ⎜
cos ⎜
⎟
⎟
⎟ sin ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎝ 2 ⎠
⎝ 2 ⎠
2
2
=
= 2 cos ec ( B − C )
⎛B−C⎞
⎛ B − C ⎞ sin ( B − C )
2sin ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
It sin θ =
54. . (i)
Solution : - sin θ =
2 bc
A
a
then show that cos θ =
cos
b+c
2
b+c
a
⇒ cos 2 θ = 1 − sin 2 θ
b+c
∴ cos θ = 1 −
cos θ =
2
=
(b + c )
(b + c )
2bc cos A + 2bc
(b + c )
cos 2 θ =
cos θ =
2
b 2 + c 2 + 2bc − a 2
2
=
2bc × 2 cos 2
(b + c )
2
(b + c ) − a2
θ=
2
(b + c )
2
a2
2
⇒ cos
(b
=
2
2
)
+ c 2 − a 2 + 2bc
( b+ c )
2
2bc {1 + cos A}
(b + c )
2
A
2
2
2 bc
A
cos
b+c
2
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(ii)
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2 bc
A
If a = ( b + c ) cos θ then prove that sin θ =
cos
b+c
2
Solution: - a = ( b + c ) cos θ ⇒
sin θ = 1 − cos θ = 1 −
2
a
= cos θ
b+c
sin θ =
2
(b + c )
b 2 + c 2 + 2bc − a 2
(b + c )
(b + c ) − a2
θ=
2
(b + c )
2
a2
2
2
⇒ sin
⇒ sin
2
2
(b
θ=
2
)
+ c 2 − a 2 + 2bc
(b + c )
2bc {1 + cos A}
2bc cos A + 2bc
2
⇒
θ
=
sin 2 θ =
sin
2
2
(b + c )
(b + c )
2
2
A⎞
⎛
2bc ⎜ 2 cos 2 ⎟
2 bc
A
2⎠
⎝
sin 2 θ =
⇒ sin θ =
cos
2
b− c
2
(b + c )
(iii)
If a = ( b − c ) sec θ prove that tan θ =
2 bc
A
sin
b−c
2
Solution : -
sec θ =
a
a2
⇒ tan 2 θ = sec 2 θ − 1 =
=1
2
b−c
(b − c )
tan θ =
2
tan θ =
2
tan 2 θ =
a2− (b − c )
(b − c )
2
2
⇒ tan θ =
2
{
2bc − b 2 + c 2 − a 2
(b − c )
(b − c )
}
(b − c )
2
2
2bc − 2bc cos A
2
a 2 − b 2 − c 2 + 2bc
=
2bc [1 − cos A]
(b − c )
2
A⎞
⎛
2bc ⎜ 2sin 2 ⎟
2 bc
A
2⎠
⎝
tan 2 θ =
⇒ tan θ =
sin
2
b−c
2
(b − c )
(iv)
For any angle θ show that a cos θ = b cos ( C + θ ) + c cos ( B − θ )
Soluton : -
RHS = b cos ( c + θ ) + c cos ( B − θ ) = b {cos c cos θ − sin c sin θ }
+ c {cos B cos θ + sin B sin θ }
= b cos c cos θ − b sin sin θ + c cos B cos θ + csin B sin θ
= cos θ {b cos c + c cos B} −
bc
cb
/ sin θ +
sin θ = a cos θ
2R
2R
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55. If the angles of a triangle ABC are in AP and b : c = 3 : 2 then show that A = 750
Solution : -
Angles of a triangle ABC are in AP
∴ 2 B = A + C ⇒ 3B = A + B + C but A + B + C = 1800
∴ 3B = 1800 ⇒ B = 600
Given that b : c = 3 : 2
2 R sin B = 2 R sin C = 3 : 2 ⇒ sin B = sin C = 3 = 2
sin 600 = sin C = 3 : 2
3
3
× 2=
: sin C = 3 : 2 =
2
2
⇒ sin c =
3 sin C
1
⇒ C = 450
2
A + B + C = 1800 ⇒ A + 600 + 450 = 1800 ⇒ A = 750 c (proved)
56. If
a 2 + b2
sin C
then prove that triangle ABC is either isosceles or right angled
=
2
2
a −b
sin ( A − B )
Solution : -
Given
(
(
a 2+ b2
sin C
=
2
2
sin ( A − B )
a −b
)
)
(
)
⇒ a 2 + b 2 sin ( A− B ) = a 2 − b 2 sin C
Using sine rule we have
{
}
{
}
4R 2 sin 2 A + sin 2 B sin ( A − B ) = 4 R 2 sin 2 A − sin 2 B sin C
{sin
2
}
A + sin 2 B sin ( A − B ) − sin ( A − B ) sin ( A + B ) sin C = 0
But in triangle ABC sin ( A + B ) = sin C
(
)
∴ sin 2 A + sin 2 B sin ( A− B ) − sin ( A − B )( sin C ) ( sin C )= 0
{
}
sin ( A − B ) sin 2 A + sin 2 B − sin 2 C = 0
sin ( A − B ) = 0 or sin 2 A + sin 2 B = sin 2 C
A = B or a 2 + b 2 = c 2
∴ triangle either isosceles or right angled
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57. If cos A + cos B + cos C =
3
then show that the triangle is equilateral
2
Solution: - cos A + cos B + cos C =
3
3
⎛ A+ B⎞
⎛ A− B⎞
⇒ 2 cos ⎜
⎟ cos ⎜
⎟ + cos C =
2
2
⎝ 2 ⎠
⎝ 2 ⎠
C⎞
3
⎛
⎛ A− B⎞
2 cos ⎜ 900 − ⎟ cos ⎜
⎟ + cos C =
2⎠
2
⎝
⎝ 2 ⎠
2sin
C
⎛ A− B⎞
2 C
cos ⎜
= 3/ 2
⎟ + 1 −2sin
2
2
⎝ 2 ⎠
2sin
C
1
⎛ A− B⎞
2 C
cos ⎜
=
⎟ − 2sin
2
2 2
⎝ 2 ⎠
4sin
C
C
C
⎛ A− B⎞
⎛ A− B⎞
2 C
cos ⎜
= 1 ⇒ 1 + 4sin 2 − 4sin cos ⎜
⎟ − 4sin
⎟=0
2
2
2
2
⎝ 2 ⎠
⎝ 2 ⎠
2
C⎞
C
A −B⎞
⎛
⎛
2 ⎛ A− B ⎞
2⎛ A− B⎞
⎜ 2sin ⎟ − 2 ⎜ 2sin cos
⎟ + cos ⎜
⎟ − cos ⎜
⎟ +1= 0
2⎠
2
2 ⎠
⎝
⎝
⎝ 2 ⎠
⎝ 2 ⎠
2
⎧
C
⎛ A − B ⎞⎫
2⎛ A− B⎞
⎨2sin − cos ⎜
⎟ ⎬ + sin ⎜
⎟=0
2
⎝ 2 ⎠⎭
⎝ 2 ⎠
⎩
∴ 2sin
C
A− B
⎛ A− B ⎞
− cos ⎜
=0
⎟ = 0 and sin
2
2
⎝ 2 ⎠
∴ 2sin
C
⎛ A −B ⎞
= cos ⎜
⎟ and A − B= 0
2
⎝ 2 ⎠
∴ 2sin
C
C
= 1 ⇒ = 300 ⇒ c = 600
2
2
A = B ∴ A = B = 600
Hence trangile is equilateral
58. It cos 2 A + cos 2 B + cos 2 C = 1 then show that triangle ABC is right angled
Solution: -
cos 2 A + cos 2 B + cos 2 C = 1 ⇒ cos 2 A + cos 2 B − 1 + cos 2 C = 0
cos 2 A − sin 2 B + cos 2 C = 0 ⇒ cos ( A − B ) cos ( A − B ) + cos 2 C = 0
(
)
cos 1800 − c . cos ( A − B ) + cos 2 C = 0
− cos C cos ( A − B ) + cos C = 0
− cos C {cos ( A − B ) − cos C} = 0
− cos C {cos ( A − B ) − cos C} = 0
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{
(
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− cos c cos ( A − B ) − cos 1800 − A + B
)} = 0
− cos c {cos ( A − B ) + cos ( A + B )}= 0
− cos c {cos ( A − B ) + cos ( A + B )}= 0
2 cos A cos B cos C = 0
⇒ cos A = 0 or cos B = 0 (or ) cos C= 0
A = 900 (or ) B = 900 or C = 900
∴ Triangle is right angled triangle
59. If a 2 + b 2 + c 2 = 8 R 2 then prove that the triangle is right angled
Solutin :
- Given a 2 + b 2 + c 2 = 8 R 2⇒ 4 R 2 {sin 2 A + sin 2 B + sin 2 C}= 8 R 2
sin 2 A + sin 2 B + sin 2 C = 2 ⇒ 1 − cos 2 A + sin 2 B + sin 2 C = 2
{
}
1 − cos 2 A − sin 2 B + 1 − cos 2 C = 2
− cos ( A − B ) cos ( A − B ) − cos 2 C= 0
cos C cos ( A − B ) − cos 2 = 0 ⇒ cos C {cos ( A − B ) − cos C} = 0
cos C {cos ( A− B ) + cos ( A + B )} = 0 ⇒ 2 cos A cos B cos C = 0
cos A = 0 or cos B = 0 (or ) cos C = 0 ⇒ A = 900 ( or ) B = 900 ( or ) C = 900
60. If cot
A
B
C
, cot , cot are in AP then prove that a, b, c are in AP
2
2
2
Solution ; cot
A
B
C
, cot cot are in AP
2
2
2
2 cot
2s ( s − b ) s ( s − a ) s ( s − c )
B
A
C
=
+
= cot + cot ⇒
Δ
Δ
Δ
2
2
2
⇒ 2 ( s − b ) = ( s − a ) + ( s − c ) ⇒ a − b + c = 2s − a −c
a + c − b = a + b + c − a − c ⇒ a + c = 2b
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61. If sin 2
A
B
C
are in HP then show that a, b, c are in HP
, sin 2 , sin 2
2
2
2
Solution : sin 2
A
B
C
, sin 2 , sin 2
2
2
2
are in HP
( s − b )( s − c ) , ( s − c )( s − a ) , ( s − a )( s − b )
bc
ac
ab
are in HP
bc
ac
ab
,
,
are in HP
( s − b )( s − c ) ( s − a )( s − c ) ( s − a )( s− b )
Multiplying with
( s − a )( s − b )( s − c )
abc
we have
bc ( s −a )( s − b )( s − c ) ac ( s − a ) ( s − b ) ( s − c ) ab ( s − a ) ( s− b ) ( s − c )
and P
,
,
abc ( s − b ) ( s − c )
abc ( s − a ) s − c
abc s − a ( s − b )
(
)
(
)
s − a s − b s− c
are in AP
,
,
a
b
c
Adding ‘1’ to every term we here
s−a
s−b
s− c
+ 1,
+ 1,
+ 1 are in AP
a
b
c
s s s
1 1 1
, , are in AP ⇒ , , are in AP
a b c
a b c
a, b, c are in HP
62. Prove that a 2 cot A + b 2 cot B + c 2 cot C =
abc
R
Solution: -
a 2 cot A + b 2 cot B + c 2 cot C
4 R 2 sin 2 A ×
cos A
cos B
cos C
+ 4 R 2 sin 2 B
+ 4 R 2 sin 2 C
sin A
sin B
sin C
2 R 2 {sin 2 A + sin 2 B + sin 2C}
2 R 2 {2sin ( A + B ) cos ( A − B ) + sin 2C}
2 R 2 {2sin C cos ( A − B ) + 2sin C cos C}
2 R 2 ⎡⎣ 2sin C {cos ( A − B ) + cos C}⎤⎦ = 4 R sin C {cos ( A − B ) − cos ( A + B )}
4 R sin C sin A sin B =
2
{
2 2 R 2 sin A sin B sin C
R
}
=
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( 2 R sin A) ( 2 R sin B ) ( 2 R sin C ) = abc
R
R
A
B
C
Δ
+ b cos 2 + c cos 2 = s +
2
2
2
R
63. Show that a cos 2
Solution: a cos 2
A
B
C
+ b cos 2 + c cos 2
2
2
2
a (1 + cos A )
2
+
b (1 + cos B )
2
+
c (1 + cos C )
2
a + a cos A + b + b cos B + C + c cos C ( a + b + c ) +
=
2
2
( a + b + c ) + {a cos A + b cos B + c cos C}
2
2 s + 2 R sin A cos A + 2 R sin B cos B + 2 R sin C cos C
2
2S + R ( sin 2 A + sin 2 B + sin 2C )
2
2S + R {2sin ( A + B ) cos ( A − B ) + sin 2C}
2
2S + R {2sin C cos ( A − B ) + 2sin C cos C}
2
2S + 2 R sin C {cos ( A − B ) + cos C}
2
2S + 2 R sin C {cos ( A − B ) − cos ( A + B )}
2
2 S + 4 R sin A sin B sin C
= S + 2 R sin A sin B sin C
2
S + 2 R 2 sin A sin B sin C
Δ
=S+
R
R
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A
B
C
r
64. Show that sin 2 + sin 2 + sin 2 = 1 −
.
2
2
2
2R
Sol. L.H.S. = sin 2
A
B
C
+ sin 2 + sin 2
2
2
2
1 − cos A 1 − cos B 1 − cos C
+
+
2
2
2
3 1
= − (cos A + cos B + cos C)
2 2
3 1⎡
A
B
C⎤
= − ⎢1 + 4sin sin sin ⎥
2 2⎣
2
2
2⎦
=
(∵ from transformations)
A
B
C⎤
⎡
4R sin sin sin ⎥
3 1⎢
2
2
2
= − ⎢1 +
⎥
2 2⎣
R
⎦
=
3 1⎡
r⎤
− ⎢1 + ⎥
2 2⎣ R⎦
=
3 1 r
r
− −
= 1−
= R.H.S.
2 2 2R
2R
65. Show that i. Δ = r1r2
Sol. i)R.H.S. r1r2
= r1r2
rr1
4R − r1 − r2
. ii. a = ( r2 + r3 )
r2 r3
r1 + r2
4R − r1 − r2
r1 + r2
4R − (r1 + r2 )
r1 + r2
⎛
2 C⎞
⎜∵ r1 + r2 = 4R cos
⎟
2⎠
⎝
= r1r2
C⎞
⎛
4R ⎜1 − cos 2 ⎟
2⎠
⎝
C
4R cos 2
2
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C
2 = r r tan C
= r1r2
C 12
2
cos 2
2
Δ
Δ
C
=
⋅
tan
S−a S−b
2
sin 2
Δ2
(S − b)(S − a)
=
(S − a)(S − b)
S(S − c)
Δ2
Δ2
=
=
= Δ = R.H.S.
S(S − a)(S − b)(S − c) Δ
ii) RHS = ( r2 + r3 )
=
=
{
Δ2
s (s − a)
Δ2
( s − b )( s − c )
Δ ⎞
rr1 ⎛ Δ
=⎜
+
⎟
r2 r3 ⎝ s − b s − c ⎠
Δ {s −c + s − b}
( s − b )( s − c )
Δ . ( 2 s − b −c )
( s − b )( s − c )}
2
( s − b )( s − c )
Δ2
.
s (s − a)
Δ2
( s − b )( s − c )
s ( s −a )
=
Δ
{a + b + c − b − c} = a
s ( s − a )( s − b )( s − c )
66. Prove that r12 + r22 + r32 + r 2 = 16R 2 − (a 2 + b 2 + c2 ) .
Sol.
(r1 + r2 + r3 − r) 2 = r12 + r22 + r32 + r 2 − 2r(r1 + r2 + r3 ) + 2(r1r2 + r2 r3 + r3r1 )...(1)
But [r1 + r2 + r3 − r] = 4R and r1r2 + r2 r3 + r3r1 = S2
16R 2 = [r1 + r2 + r3 − r]2
16R 2 = r12 + r22 + r32 + r 2 − 2r(r1 + r2 + r3 ) + 2(r1r2 + r2 r3 + r3r1 )
= r12 + r22 + r32 + r 2 − 2r(r1 + r2 + r3 ) + 2S2
16R 2 = r12 + r22 + r32 + r 2 − 2(rr1 + rr2 + rr3 ) + 2S2 Consider 2(rr1 + rr2 + rr3) =
⎡ Δ2
Δ2
Δ2 ⎤
= 2⎢
+
+
⎥
⎣ S(S − a) S(S − b) S(S − c) ⎦
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= 2Δ
=
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2 [ (S − b)(S − c) + (S − a)(S − c) + (S − a)(S − b) ]
S(S − a)(S − b)(S − c)
2Δ 2 2
⎡S − Sc − Sb + bc + S2 − Sc − Sa + ac + S2 − Sb − Sa + ab ⎤
2 ⎣
⎦
Δ
= 2 ⎡⎣3S2 − 2S(a + b + c) + ab + bc + ca ⎤⎦
= 2 ⎡⎣3S2 − 4S2 + ab + bc + ca ⎤⎦
= 2 ⎡⎣ab + bc + ca − S2 ⎤⎦
= 2 [ ab + bc + ca ] − 2S2
From (2)
⇒ r12 + r22 + r32 + r 2
= 16R 2 + 2r(r1 + r2 + r3 ) − 2S2
= 16R 2 − 2S2 + 2(ab + bc + ca) − 2S2
= 16R 2 − 4S2 + 2(ab + bc + ca)
2
⎛a+b+c⎞
= 16R 2 − 4 ⎜
⎟ + 2(ab + bc + ca)
2
⎝
⎠
= 16R 2 − ⎣⎡(a + b + c) 2 − 2(ab + bc + ca) ⎦⎤
= 16R 2 − (a 2 + b 2 + c 2 )
67. If P1, P2, P3 are the altitudes from the vertices A, B, C to the opposite side of a triangle
respectively, then show that
(i)
1 1 1 1
1 1 1 1
(abc) 2 8Δ3
+ + = (ii)
+ − =
(iii) P1P2 P3 =
=
P1 P2 P3 r
P1 P2 P3 r3
abc
8R 3
Sol. We know that
1
1
1
Δ = aP1 , Δ = bP2 , Δ = cP3
2
2
2
2Δ
2Δ
2Δ
⇒ P1 =
, P2 =
and P3 =
a
b
c
1 1 1
a
b
c
i)
+ + =
+
+
P1 P2 P3 2Δ 2Δ 2Δ
=
ii)
a + b + c 2S 1
=
=
2Δ
2Δ r
1 1 1
a
b
c
+ − =
+
−
P1 P2 P3 2Δ 2Δ 2Δ
=
a + b − c 2S − 2c S − c 1
=
=
=
2Δ
2Δ
r3
Δ
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iii) P1P2 P3 =
2 Δ 2 Δ 2 Δ 8Δ
⋅
⋅
=
a b c
abc
3
68. If a = 13, b = 14, c = 15, show that R =
65
21
, r = 4, r1 =
, r2 = 12 and r3 = 14.
8
2
Sol. Given that a = 13, b = 14, c = 15
a + b + c 13 + 14 + 15 42
=
=
= 21
2
2
2
S − a = 21 − 13 = 8,S − b = 21 − 14 = 7
S=
S − c = 21 − 15 = 6
Δ = S(S − a)(S − b)(S − c)
= 21(8)(7)(6) = 21× 16 × 21
= 21× 21× 4 × 4 = 21× 4 = 84
abc 13 × 14 ×15 65
=
=
4Δ
4 × 84
8
Δ 84
=4
r= =
S 21
Δ
84 21
=
=
r1 =
S−a 8
2
Δ
84
=
= 12
r2 =
S−b 7
Δ
84
=
= 14
r3 =
S−c 6
R=
69. If r1 = 2, r2 = 3, r3 = 6 and r = 1, prove that
a = 3, b = 4 and c = 5.
Sol. Δ 2 = rr1r2 r3 = 1 ⋅ 2 ⋅ 3 ⋅ 6 = 36
Δ 2 = 36 ⇒ Δ = 6
Δ 6
= ⇒S=6
(∵ r = 1)
S S
Δ
Δ
r1 =
⇒ S−a =
S−a
r1
r=
∴a = S −
r2 =
Δ
6
= 6− = 6−3 = 3
r1
2
Δ
Δ
⇒ S−b =
S−b
r2
∴b = S −
Δ
= 6−3 = 3
r2
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r3 =
Δ
Δ
⇒ S−c =
S−c
r3
∴c = S −
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Δ
= 6 −1 = 5
r3
70. Show that a 2 cot A + b 2 cot B + c2 cos C =
abc
.
R
Sol. L.H.S. a 2 cot A + b 2 cot B + c 2 cos C
= 4R 2 sin 2 A
cos A
cos B
+ 4R 2 sin 2 B
sin A
sin B
+4R 2 sin 2 C
cos C
(by sine rule)
sin C
= 2R 2 (2sin A cos A + 2sin Bcos B + 2sin C cos C)
= 2R 2 (sin 2A + sin 2B + sin 2C)
= 2R 2 (4sin A sin Bsin C)
1
(2R sin A)(2R sin B)(2R sin C)
R
abc
=
= R.H.S.
R
=
71. In ΔABC, if
Sol.
1
1
3
+
=
, show that C = 60°.
a+c b+c a+b+c
1
1
3
+
=
a+c b+c a+b+c
⇒
b+c+a+c
3
=
(a + c)(b + c) a + b + c
⇒ 3(a + c)(b + c) = (a + b + 2c)(a + b + c)
⇒ 3(ab + ac + bc + c 2 )
= (a 2 + b 2 + 2ab) + 3c(a + b) + 2c2
⇒ ab = a 2 + b 2 − c2 = 2ab cos C
(from cosine rule)
⇒ cos C =
1
⇒ C = 60°
2
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a 2 + b2 + c2
72. Prove that cot A + cot B + cot C =
.
4Δ
Sol. L.H.S. = Σ cot A = Σ
cos A
sin A
⎛ b2 + c2 − a 2 ⎞
= Σ⎜
⎟ (by cosine rule)
⎝ 2bc sin A ⎠
=Σ
b2 + c2 − a 2
4Δ
1
⎡
⎤
⎢⎣∵ Δ = 2 bc sin A ⎥⎦
=
1 2 2 2 2
[b + c − a + c + a 2 − b 2 + a 2 + b 2 − c 2 ]
4Δ
=
a 2 + b2 + c2
= R.H.S.
4Δ
73. In ΔABC, if a cos A = b cos B, prove that the triangle is either isosceles or right angled.
Sol. a cos A = b cos B
⇒ 2R sin A cos A = 2R sin B cos B
⇒ sin 2A = sin 2B = sin(180° – 2B)
Hence 2A = 2B or 2A = 180° – 2B
⇒ A = B or A = (90° – B)
⇒ a = b or (A + B) = 90°
⇒ a = b or C = 90°
∴ The triangle is isosceles or right angled.
A
B
C
: cot : cot = 3 : 5 : 7 , show that a : b : c = 6 : 5 : 4 .
2
2
2
A
B
C
Sol. cot : cot : cot = 3 : 5 : 7
2
2
2
s(s − a) s(s − b) s(s − c)
⇒
⋅
⋅
= 3:5: 7
Δ
Δ
Δ
⇒ (s − a) : (s − b) : (s − c) = 3 : 5 : 7
74. If cot
s−a s−b s−c
=
=
= k (say)
3
5
7
Then s – a = 3k, s – b = 5k, s – c = 4k
⇒
Adding these equations,
3s – (a + b + c) = 3k + 5k + 7k = 15k
⇒ 3s – 2s = 15k ⇒ s = 15k
Hence a = 12k, b = 10k, c = 8k
∴ a : b : c = 12k : 10k : 8k = 6 : 5 : 4
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75. Prove that a cos(B – C) + b cos (C – A) + c3 cos(A – B) = 3 abc.
3
3
Sol. L.H.S. = Σ a3 cos(B – C)
= Σa 2 (2R sin A) cos(B − C)
= RΣa 2 ⋅ [2sin(B + C) cos(B − C)]
= RΣa 2 (sin 2B + sin 2C)
= RΣa 2 (2sin Bcos B + 2sin C cos C)
= Σ[a 2 (2R sin B) cos B + a 2 (2R sin C) cos C]
= Σ(a 2 b cos B + a 2 c cos C)
= (a 2 b cos B + a 2c cos C) + (b 2c cos C + b 2a cos A) + (c 2a cos A + c 2 b cos B)
= ab(a cos B + b cos A) + bc(b cos C + c cos B)
+ca(c cos A + a cos C)
= ab(c) + bc(a) + ca(b)
= 3abc = R.H.S.
76. If p1, p2, p3 are the altitudes of the vertices A, B, C of a triangle respectively, show that
1
1
1 cot A + cot B + cot C
+ 2+ 2 =
.
2
Δ
p1 p 2 p3
Sol. Since p1, p2, p3 are the altitudes of ΔABC, we have
1
1
1
Δ = ap1 = bp 2 = cp3
2
2
2
2Δ
2Δ
2Δ
⇒ p1 =
, p2 =
, p3 =
a
b
c
Now
=
1
1
1 a 2 + b2 + c2
+
+
=
p12 p 22 p32
4Δ 2
1
(cot A + cot B + cot C) = R.H.S.
Δ
[∵ cot A + cot B + cot C =
a 2 + b2 + c2
]
4Δ
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77. If (r2–r1)(r3–r1) = 2r2r3. Show that A = 90°.
Sol. (r2 – r1)(r3 – r1) = 2r2r3
⎡ Δ
Δ ⎤⎡ Δ
Δ ⎤
⇒⎢
−
−
⎥
⎢
⎥
⎣ (s − b) (s − a) ⎦ ⎣ (s − c) (s − a) ⎦
Δ
Δ
=2
(s − b) (s − c)
⎡ s−a −s+ b ⎤ ⎡ s−a −s+c ⎤
⇒ Δ⎢
⎥⋅Δ ⎢
⎥
⎣ (s − b)(s − a) ⎦ ⎣ (s − c)(s − a) ⎦
=
2Δ 2
(s − b)(s − c)
⇒ (b − a)(c − a) = 2(s − a) 2
⎛ b+c−a ⎞
⇒ (b − a)(c − a) = 2 ⎜
⎟
2
⎝
⎠
2
⇒ 2(bc − ca − ab + a 2 )
= b 2 + c 2 + a 2 + 2bc − 2ca − 2ab
⇒ 2a 2 = b 2 + c2 + a 2
⇒ b2 + c2 = a 2
Hence ΔABC is right angled and A = 90°.
78. In a triangle ABC prove that
⎛B−C⎞
∑ ( r + r ) tan ⎜⎝ 2 ⎟⎠ = 0
1
Solution: A
B
C
A
B
C
cos cos + 4 R sin sin sin
2
2
2
2
2
2
A
⎛B−C⎞
= 4 R sin cos ⎜
⎟
2
⎝ 2 ⎠
r1r2 = 4 R sin
∑
∑
⎛ B −C ⎞
sin ⎜
⎟
B−C⎞
A
⎛B−C⎞
⎝ 2 ⎠
=
4
R
sin
cos
( r1 + r ) tan ⎛⎜
⎟ ∑
⎜
⎟
2
⎝ 2 ⎠
⎝ 2 ⎠
⎛B−C⎞
cos ⎜
⎟
⎝ 2 ⎠
B+C⎞
⎛
⎛B−C⎞
3R 2sin ⎜ 900 −
⎟ . sin ⎜
⎟
2 ⎠
⎝
⎝ 2 ⎠
∑
⎧
⎛ B +C ⎞
⎛ B − C ⎞⎫
2 R ⎨2 cos ⎜
⎟ sin ⎜
⎟⎬
⎝ 2 ⎠
⎝ 2 ⎠⎭
⎩
∑
∑
2 R {sin B − sin C} = ∑ 2 R sin B − 2 R sin C
b−c= b−c+c−a+a−b=0
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1 1 1 1 a 2 + b2 + c2
79. Show that 2 + 2 + 2 + 2 =
.
r
r1 r2 r3
Δ2
Sol. L.H.S. =
s2
Δ2
1
= 2
Δ
1
= 2
Δ
=
+
1 1 1 1
+ + +
r 2 r12 r22 r32
(s − a) 2 (s − b) 2 (s − c) 2
+
+
Δ2
Δ2
Δ2
⎡s 2 + (s − a) 2 + (s − b) 2 + (s − c) 2 ⎤
⎣
⎦
⎡s 2 + s 2 − 2as + a 2 + s 2 − 2bs + b 2 + s 2 − 2cs + c 2 ⎤
⎣
⎦
=
1
⎡ 4s 2 − 2s(a + b + c) + a 2 + b 2 + c2 ⎤
2 ⎣
⎦
Δ
=
1
a 2 + b2 + c2
⎡ 4s 2 − 2s(2s) ⎤ +
⎦
Δ2 ⎣
Δ2
=
a 2 + b2 + c2
= R.H.S.
Δ2
80. Show that
Sol. L.H.S. =
=
r1 r2 r3 1 1
+ + = −
.
bc ca ab r 2R
r1 r2 r3
1
[ar1 + br2 + cr3 ]
+ + =
bc ca ab abc
1 ⎡
A⎤
s
A
Σa ⋅ s tan ⎥ =
Σ 2R sin A tan
⎢
abc ⎣
2 ⎦ abc
2
abc ⎞
⎛
⎜∵ Δ =
⎟
4R ⎠
⎝
⎡
A ⎞⎤
⎛
sin ⎟ ⎥
⎢
⎜
s
A
A
2 (∵ r = Δ / s)
=
Σ ⎢ 2R ⋅ 2sin cos ⋅ ⎜
⎟⎥
A
abc ⎢
2
2 ⎜ cos ⎟ ⎥
⎢⎣
2 ⎠ ⎥⎦
⎝
Rs ⎛ 2 A ⎞ s ⎛ 1 − cos A ⎞
Σ ⎜ sin
⎟ = Σ⎜
⎟
abc ⎝
2⎠ Δ ⎝
2
⎠
1
= (1 − cos A + 1 − cos B + 1 − cos C)
2r
1
= [3 − (cos A + cos B + cos C) ]
2r
=4
=
1 ⎡ ⎛
A
B
C ⎞⎤
3 − ⎜1 + 4sin sin sin ⎟ ⎥
⎢
2r ⎣ ⎝
2
2
2 ⎠⎦
⎡ ⎛
A
B
C ⎞⎤
4R sin sin sin ⎟ ⎥
⎢
⎜
1
2
2
2
= ⎢2 − ⎜
⎟⎥
⎠⎦
2r ⎣ ⎝
R
=
1 ⎡
r⎤ 1 1
2− ⎥ = −
= R.H.S.
⎢
2r ⎣ R ⎦ r 2R
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81. If r : R : r1 = 2 : 5 : 12, then prove that the triangle is right angled at A.
Sol. If r : R : r1 = 2 : 5 : 12, then r = 2k, R = 5k and r1 = 12k for some k.
r1 − r = 12k − 2k = 10k = 2(5k) = 2R
⇒ 4R sin
⇒ 2sin
⇒ sin
⇒
A⎡
B
C
B
C⎤
cos cos − sin sin ⎥ = 2R
⎢
2⎣
2
2
2
2⎦
A
1⎡
A⎤
⎛ B+C⎞
⎛ B+C⎞
2 A
cos ⎜
= ⎢∵ cos ⎜
⎟ = 1 ⇒ sin
⎟ = sin ⎥
2
2 2⎣
2⎦
⎝ 2 ⎠
⎝ 2 ⎠
A
1
=
= sin 45°
2
2
A
= 45° ⇒ A = 90°
2
Hence the triangle is right angled at A.
82. Show that r + r3 + r1 – r2 = 4R cos B.
Sol. r + r3
= 4R sin
C⎡ A
B
A
B⎤
sin sin + cos cos ⎥
⎢
2⎣
2
2
2
2⎦
= 4R sin
C
⎛A−B⎞
cos ⎜
⎟
2
⎝ 2 ⎠
r1 – r2
= 4R cos
C⎡ A
B
A
B⎤
sin cos − cos sin ⎥
⎢
2⎣
2
2
2
2⎦
= 4R cos
C ⎛A−B⎞
sin ⎜
⎟
2
⎝ 2 ⎠
∴ r + r3 + r1 – r2
⎡ C
C ⎛ A − B ⎞⎤
⎛A−B⎞
= 4R ⎢sin cos ⎜
⎟ + cos sin ⎜
⎟⎥
2
2
⎝ 2 ⎠
⎝ 2 ⎠⎦
⎣
B B⎞
⎛C A B⎞
⎛
= 4R sin ⎜ + − ⎟ = 4R sin ⎜ 90° − − ⎟
2 2⎠
⎝2 2 2⎠
⎝
= 4R cos B
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83. A lamp post is situated at the middle point M of the side AC of a triangular plot ABC
with BC = 7 m, CA = 8 m and AB = 9 m. Lamp post subtends an angle 15° at the point
B. Find the height of the lamp post.
Sol.
A
N
M
15°
C
B
MN is the height of the lamp post.
Let MN = h (?)
Given that ∠NBM = 15°
b2 + c2 − a 2
2abc
64 + 49 − 81 16 × 2 32 2
=
=
=
=
2×8× 7
16 × 7 112 7
2
∴ cos C =
7
Let BM = x
In ΔABC, cos C =
In ΔBCM, cos C =
7 2 + 42 − x 2
2×7× 4
2 49 + 16 − x 2
=
7
7×8
16 = 65 − x 2
x 2 = 65 − 16 ⇒ x = 7
h
In ΔBMN : tan15° =
x
h = x tan15° = 7(2 − 3)
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84. Two ships leave a port at the same time. One goes 24 km per hour in the direction N
45°E and other travels 32 km per hour in the direction S 75° E. Find the distance
between the ships at the end of 3 hours.
Sol.
E
B
S
96
72
60°
45°
N
75°
P
S
P is the position of the port.
A is the position of the North-East traveled ship after 3 hours is = 72 km
Position of the South-East traveled ship after 3 hours is 3 × 32 = 96 km
From the data ∠APB = 60°
In ΔAPB,
cos P =
AP 2 + BP 2 − AB2
2AP BP
cos 60° =
(72) 2 + (96) 2 − AB2
2 × 72 × 96
1 722 + 962 − AB2
=
2
2 × 72 × 96
1=
5184 + 9216 − AB2
72 × 96
1=
14400 − AB2
6912
6912 = 14400 − AB2
AB2 = 14400 − 6912
AB2 = 7488
AB = 7488 = 86.53 = 86.4 km
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85. A tree stands vertically on the slant of the hill. From a point A on the ground 35 meters
down the hill from the base of the tree, the angle of elevation of the top of the tree is
60°. If the angle of elevation of the foot of the tree from A is 15°, then find the height of
the tree.
Sol.
B
h D
y
C
35m
60°
15°
x
A
BD is the height of the tree and A is the point of observation.
Let CD = y
AC = x
Given that, ∠CAD = 15°, ∠CAB = 60° and AD = 35 m.
In ΔCAD, sin15° =
y = 35sin15° =
cos15° =
x=
y
35
35( 3 − 1)
2 2
…(1)
x
35
3 +1
× 35
2 2
In ΔCAB, tan 60° =
h=x 3=
…(2)
h
x
3 +1
× 35 × 3
2 2
Height of the tree = h – y
3 +1
3 −1
× 35 3 −
× 35 =
2 2
2 2
3 +1 ⎡
3 + 3 − 3 + 1⎤⎦
2 2 ⎣
35 × 4
=
= 35 2 m
2 2
=
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86. The upper 3/4th portion of a vertical pole subtends an angle Tan–13/5 at a point in the
horizontal plane through its foot and at a distance 40 m from the foot. Given that the
vertical pole is at a height less than 100 m from the ground, find its height.
Sol.
A
3(100 − x)
4
C
θ=tan–1(3/5)
α
100–x
100 − x
4
B
AB is the height of the tree.
AD is the 3/4th part of upper part of the tree.
DB is the 1/4th lower part of the tree.
Let AB = 100 – x
C is the point of observation.
In ΔBCD,
3
3
Let ∠DCA : θ = tan −1 ⇒ tan θ =
5
5
tan α =
100 − x 1 100 − x
×
=
4
40
160
tan θ + tan α
1 − tan θ tan α
3 100 − x
+
100 − x
160
= 5
−x
3
100
40
1− ×
5
160
tan(θ + α) =
100 − x 480 + 5(100 − x)
=
40
800 − 3(100 − x)
100 − x 480 + 500 − 5x
=
40
800 − 300 + 3x
100 − x 980 − 5x
=
40
500 + 3x
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[100 − x][500 + 3x] = 40[980 − 5x]
50000 + 300x − 500x − 3x 2 = 39200 − 200x
⇒ 3x 2 + 500x − 400x = 50000 − 39200
3x 2 = 10800
x2 =
10800
= 3600
3
x = 3600 = 60
Height of the tree = 100 – x = 40 m.
87. Let an object be placed at some height h cm and let P and Q be two points of
observation which are at a distance 10 cm apart on a line inclined at angle 15° to the
horizontal. If the angles of elevation of the object from P and Q are 30° and 60°
respectively then find h.
Sol.
A
30°
E h
120°
P
60°
30°
18°
B
A is the position of the object.
Given that AB = h cm
P and Q are points of observation.
Given that, PQ = 10 cm
We have,
∠BPE = 15°, ∠EPA = 30°, ∠EQA = 60°
In ΔPQA,
P = 30°, Q = 120° and A = 30°
∴ By sine rule,
AP
PQ
=
sin120° sin 30°
AP
10
=
sin(180° − 60°) 1/ 2
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AP
AP
= 20° ⇒
= 20°
sin 60°
3/2
3
= 10 3
2
AB
In ΔPBA, sin 45° =
AP
1
h
=
2 10 3
AP = 20 ×
h=
10 3 5 ⋅ 2 ⋅ 3
=
= 5 2 3 = 5 6 cm
2
2
88. The angle of elevation of the top point P of the vertical tower PQ of height h from a
point A is 45° and from a point B is 60°, where B is a point at a distance 30 meters
from the point A measured along the line AB which makes an angle 30° with AQ. Find
the height of the tower.
Sol.
P
15°
h
B
A
60°
C
30° 45°
Q
D
In the figure
PQ = h, ∠PAQ = 45°
∠BAQ = 30° and ∠PBC = 60°
Also, AB = 30 m
∴ ∠BAP = ∠APB = 15°
This gives, BP = AB = 30 and
h = PC + CD = BP sin 60° + AB sin 30°
= 15 3 + 15 = 15( 3 + 1) meters.
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89. Theorem : - In a triangle ABC prove that
(i) sin
A
=
2
( S − b )( S − c )
(ii) cos
bc
A
=
2
S (S − a)
( S − b )( S − C ) = Δ = ( S − b )( S − C )
S (S − a)
S (S − a)
Δ
S (S − a)
S ( S− a )
Δ
=
=
(iv) cot A / 2 =
Δ
( S − b )( S − c ) ( S − b )( S − c )
(iii) tan
A
=
2
Proof (i)
From cosine rule we know that
a 2 + b 2 + c 2 = 2bc cos A ⇒ 2bc cos = b 2 + c 2 − a 2
cos A =
b2 + c2 − a 2
2bc
We know that 2sin 2
A
= 1 − cos A
2
b 2 + c 2 − a 2 2bc − b 2 − c 2 + a 2
=1−
=
2bc
2bc
2
2
2
2
a − {b + c −2bc}
a2 − (b − c )
2 A
2 A
⇒ sin
=
∴ 2sin
=
2
2bc
2
4bc
sin 2
A ( a + b − c )( a − b + c )
=
2
4bc
∵ a + b + c = 2 S we have 2 S − 2C= a + b − c
∴ sin 2
A 2 ( S − C ) 2 ( S − b)
A
=
⇒ sin =
2
4 bc
2
( S − b )( S − C )
bc
Proof (ii)
2 cos 2
A
b2 + c2− a 2
= 1+ cos A = 1 +
2
2bc
2
A 2bc + b 2 + c 2 − a 2
2 A (b + c ) − a
=
2 cos
⇒ 2 cos
2
2bc
2
2bc
2
2
cos 2
A ( b + c − a )( b + c + a )
=
2
4bc
Since a + b + c = 2 S ; 2 S − 2a = b + c − a
∴ cos 2
A 2 (S − a) 2 S
A
⇒ cos =
=
2
4 bc
2
S (S − a)
bc
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A
A
2 =
Proof (iii) tan =
2 cos A
2
sin
tan
A
=
2
tan
A
2
( S − b )( S − c )
b−c
S (S − a)
b−c
( S − b )( S − c )
S (S − a)
=
( S − b )( S − c ) ( S − b )( S− c ) =
S ( S − a ) ( S − b )( S − c )
( S − b )( S − c ) × S ( S− a ) =
S (S − a)
S (S − a)
( S − b )( S − c )
( S − b) ( S − c)
=
Δ
S ( S − a )( S − b )( S − c )
S ( S − a )( S − b )( S − c )
S (S − a)
=
Δ
S (S − a)
Proof of (iv)
By taking reciprocal of tan A / 2 we get cot A / 2
List of formulae related to half angles
sin
A
2
cos
A
=
2
tan
A
=
2
tan
B
=
2
tan
C
=
2
cot
A
=
2
cot
B
=
2
cot
C
=
2
( S − b )( S − c )
sin B / 2 =
bc
S (S − a)
cos
bc
B
=
2
( S − c )( S − a )
ac
S ( S − b)
ac
( S − b )( S− C ) = Δ = ( S − b )( S − c )
S (S − a)
S (S − a)
Δ
( S − c )( S − a ) = Δ = ( S − c )( S − a )
S ( S − b)
S ( S − b)
Δ
( S − a )( S − b ) = Δ = ( S − c )( S − a )
S ( S − c)
S ( S − c)
Δ
S (S − a)
=
Δ
=
S (S − a)
Δ
( S − b )( S − c ) ( S − b )( S − c )
S ( S − b)
S ( S − b)
Δ
=
=
Δ
( S −a )( S − c ) ( S − a )( S − c )
S ( S− C )
=
Δ
( S − a )( S − b ) ( S − a )( S − b )
=
S (S − C)
Δ
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sin
C
=
2
cos
C
2
( S − a )( S − b )
ab
S (S − c)
ab