Solution of Triangles
Solution of Triangles
Solution of Triangles
1.
Solution of Triangles
Introduction
Congruence of Triangles:
If ∆ABC is congruent to ∆DEF, we write ∆ABC ≅ ∆DEF. This happens
when
AB = DE, BC = EF, AC = DF and ∠A = ∠D, ∠B = ∠E, ∠C = ∠F
1.
Side-Angle-Side (SAS) Congruence
criterion:
In ∆ABC and ∆DEF if
AB = DE
∠ABC = ∠DEF
BC = EF then
∆ABC ≅ ∆DEF
2.
Angle-Side-Angle (ASA) Congruence
criterion:
In ∆ABC and ∆DEF if
∠ABC = ∠DEF
BC = EF
∠BCA = ∠EFD then
∆ABC ≅ ∆DEF
3.
Angle-Angle-Side (AAS) Congruence
criterion:
In ∆ABC and ∆DEF if
∠CAB = ∠FDE
∠ABC = ∠DEF
BC = EF then
∆ABC ≅ ∆DEF
4.
Side-Side-Side (SSS) Congruence
criterion:
In ∆ABC and ∆DEF if
AB = DE
BC = EF
Solution of Triangles
AC = DF then
∆ABC ≅ ∆DEF
2.
5. Right Angle-Hypotenuse-Side (RHS)
Congruence criterion:
In ∆ABC and ∆DEF if
π
∠ABC = ∠DEF =
2
AC = DF
BC = EF then
∆ABC ≅ ∆DEF
Q.
In the figure ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that:
(i) AC bisects each of the angle A and C.
(ii) BE = ED
(iii) ∠ABC = ∠ADC.
Can we say that AE = EC?
About Triangle:
In any triangle the three sides and the three angles are generally called
the elements of triangle. In any triangle ABC, the measures of the angles
BAC, CBA and ACB are denoted by the letter A, B and C respectively. The
side BC, CA and AB are opposite to the angles A, B and C respectively
Solution of Triangles
3.
Sine Formula:
In any triangle ABC, the ratios of the sides to sine of the opposite angles
are equal.
a b c
i.e. = = = 2R
sinA sinB sinC
where R is circumradius of ∆ABC
1
Note: ∆ = absinC
2
1 1
∆= bcsinA , ∆ = casinB
2 2
Proof: In ∆BOD
a
sinA = 2
R
a
⇒ 2R =
sinA
Similarly, others can be proved
Now area of triangle
1
∆=
2
( )( )
BC AD
1
=
2
( )(
a csinB )
1
∆= acsinB
2
Similarly, others can also be proved.
Napier’s Formula:
B − C b − c A
(i) tan = cot
2 b+c 2
C − A c −a B
(ii) tan = cot
2 c+a 2
A −B a −b C
(iii) tan = cot
Solution of Triangles
2 a +b 2
4.
b−c 2RsinB − 2RsinC
Proof: =
b+c 2RsinB + 2RsinC
B + C B-C
2cos sin
= 2 2
B + C B-C
2sin cos
2 2
A
sin
= 2 tan B − C
A
cos 2
2
B − C b − c A
⇒ tan = cot
2 b+c 2
Similarly, others can also be proved.
Cosine Formula:
b2 + c2 − a2 a2 + c2 − b2 a2 + b2 − c2
cosA = , cosB = , cosC =
2bc 2ac 2ab
cosA > 0 ⇔ ∠A is acute ⇔ b + c – a > 0 ⇔ b + c > a2
2 2 2 2 2
Proof: In ∆ADC
⇒ AD2 + DC2 = AC2
⇒ (csinB)2 + (a – c cosB)2 = b2
⇒ c2sin2B + a2 + c2cos2B – 2ac cosB = b2
⇒ a2 + c2 – b2 = 2ac cosB
a2 + c2 − b2
⇒ cosB =
2ac
Similarly, others can also be proved.
Uses:
# To solve the triangle, if two sides and the included angle (e.g. b, c, ∠A)
Solution of Triangles
are given.
# To solve the triangle, if two sides and any one angle (e.g. a, b, ∠A) are
given.
5.
Projection Formula:
a = c cos B + b cos C
b = a cos C + c cos A
c = a cos B + b cos A
Proof: In ∆ABC
BC = BD + DC
a = c cosB + b cosC
Similarly, others can also be proved.
A B C
Q. In a ∆, sides are proportional to cos
2
, cos , cos . Find angles of triangle
2 2
where A, B and C are angles of triangle.
Sol. Given
a b c
= = …(1)
A B C
cos cos cos
2 2 2
2RsinA 2RsinB 2RsinC
= = (by sine-rule)
A B C
cos cos cos
2 2 2
A B C
⇒ sin = sin = sin
2 2 2
A B C
⇒ = =
2 2 2
π
⇒ A = B = C =
3
3
Q. In a triangle ABC, a = 5, b = 7 and sinA =
4
. How many such triangles are
possible?
3 sinB sinC
4
21
⇒ sinB = (which is not possible)
20
Hence no such triangle is possible.
6.
Q. If the angles of a triangle be in the ratio 1 : 2 : 7, then the ratio of its greatest
side to the least side is ?
Sol. Let A : B : C = 1 : 2 : 7
A = θ, B = 2θ, C = 7θ
π
now A + B + C = π ⇒ 10θ = π ⇒ θ =
10
∵ C is greatest angle ⇒ c is greatest
A is least angle ⇒ a is least
a c
Now =
sinA sinC
c sinC
⇒ =
a sinA
7π 3π
sin sin
c 10 = 10 = 5+1 4
= ×
a π π 4 5−1
sin sin
10 10
c 3+ 5
⇒ =
a 2
Sol.
a2(sin2C – sin2B) + b2(sin2A – sin2C) + c2(sin2B – sin2A)
= [(asinc)2 – (csinA)2] + [(bsinA)2 – (asinB)2] + [(csinB)2 – (b sin C)2]
by sine rule asinB = bsinA
= 0 + 0 + 0 = 0 asinC = csinA
bsinC = csinB
Ans. (A)
Sol. let a2 = b2 – k, c2 = b2 + k
Solution of Triangles
B − C b − c A
Sol. tan =
2 b+c
cot
2
θ
⇒ tan =
( )
3 + 1 k − 2k
cot30° (∠B – ∠C = θ)
2 ( 3 + 1) k + 2k
θ
= 15° ⇒ θ = 30°
⇒
2
∠B – ∠C = 30°
Q. If the sides of a triangle are p, q and p2 + pq + q2 , then the biggest angle is:
Sol. Let r = p2 + pq + q2
Opposite angle to r will be biggest, let it be “θ”, then
cosθ =
2 2 2
p2 + q2 − r2 p + q − p + pq + q
=
(
2
)
2pq 2pq
−1 2π
cosθ = ⇒ θ=
2 3
Solution of Triangles
8.
cosA cosB cosC
Q. In a ∆ABC,
a
=
b
=
c
and the side a = 2 then area of the triangle is:
Q. a2 − b2 ( )
sin A − B
In any triangle if = then
2
a +b 2
sin ( A + B)
Ans. (D)
(
a2 − b2 + a2 + b2 sin A − B + sin A + B
=
) ( )
a2 − b2 − a2 − b2 (
sin A − B − sin A + B ) ( )
−a2 2sinAcosB
⇒ =
b 2 −2cosAsinB
sin2 A sinAcosB
⇒ =
sin B 2 cosAsinB
sinA cosB
⇒ =
sinB cosA
⇒ sinA cosA – sinB cosB = 0
Solution of Triangles
1
⇒ sin2A − sin2B = 0
2
⇒ cos(A + B) sin(A – B) = 0
cos(A + B) = 0 or sin (A – B) = 0
9.
π
A+B= A–B=0
2
π
C= A=B
2
Right angled ∆ Isosceles ∆
Q. b+c A b−c
For any triangle ABC. If B = 3C, show that cosC = & sin =
4c 2 2c
Sol. A + B + C = π ⇒ A + 4C = π ⇒ A = π – 4C
B − C b − c A
tan = cot
2 b+c 2
3C − C b − c π
tan = cot − 2C
2 b+c 2
b−c
tanC = tan2C
b+c
b − c 2tanC
⇒ tanC = ·
b + c 1 − tan2C
1 − tan2C =
(
2 b−c )
b+c
(
2 b−c )
(
⇒ 1 − sec2C − 1 = ) b+c
2c b+c
sec2C = 2 × ⇒ cosC =
b+c 4c
A π
Now sin = sin − 2C = cos2C = 2 cos2C – 1
2 2
b + c b−c
= 2 −1=
4c 2c
(A)
(p2
)
+ q2 sinθ
(B)
(p 2
)
+ q2 sinθ
(C)
(p2
)
− q2 sinθ
(D)
( )
2 p2 + q2 sinθ
pcosθ + qsinθ pcosθ – qsinθ pcosθ + qsinθ pcosθ − qsinθ
10.
Ans. (A)
Sol.
See the figure let BM || AD and ∠BDM = α
now in ∆BDC
tanα = p/q …(1)
In ∆BMC
p
tan(θ + α) =
MC
p
tan(θ + α) =
q − DM
tanθ + tanα p
=
1 − tanθtanα q − AB
p
tanθ +
q p
⇒ =
p q − AB
1 − tanθ
q
qtanθ + p p
⇒ =
q − ptanθ q − AB
⇒ q2tanθ + pq – AB(qtanθ + p) = pq – p2tanθ
⇒ (q2 + p2)tanθ = (qtanθ + p)AB
⇒ AB =
(p 2
)
+ q2 sinθ
qsinθ + pcosθ
4
Q. If in a ∆ABC, a = 6, b =3 and cos(A – B) =
5
then find its area ?
4
Sol. Let A – B = θ then cosθ =
5
θ
1 − tan2
⇒ 2 = 4 ⇒ tan θ = 1
θ 5 2 3
1 + tan2
2
A −B a −b C
⇒ tan = cot
2 a +b 2
1 3 C C
Solution of Triangles
11.
Q. The sides of a triangle are consecutive integers n, n + 1 and n + 2 and the larg-
est angle is twice the smallest angle. Find n?
Sol. Let a = n, b = n + 1, c = n + 2
also given C = 2A then B = π – 3A
sinA sinB sinC
now = =
a b c
⇒
sinA sin 3A
=
( ) =
sin2A
n n+ 1 n+2
⇒ sinA = nα, sin2A = (n + 2)α, sin3A = (n + 1)α
Now sin2A = 2sinA cosA
n+2
(n + 2)α = 2(nα) cosA ⇒ cosA =
2n
and 3sinA – 4sin3A = sin3A
⇒ 3sinA – 4sin3A = (n + 1)α
⇒ sinA{3 – 4sin2A} = (n + 1)α
2
n + 2
⇒ nα 3 − 4 1 −
(
= n+ 1 α
2n
)
n + 2 2
⇒ n (
− 1 = n + 1)
n
⇒ n{(n + 2)2 – n2} = n2(n + 1)
⇒ n2 – 3n – 4 = 0
⇒ (n – 4) (n + 1) = 0 ⇒ n = 4
Q. If a = 5, b = 7, c = 8 find angle B ?
52 + 82 − 72
Sol. cosB =
( )( )
2 5 8
25 + 64 − 49 1
= =
80 2
Solution of Triangles
π
⇒ B =
3
12.
Q. cosA cosB cosC a2 + b2 + c2
Prove that + + =
a b c 2abc
Sol. LHS
b2 + c2 − a2 a2 + c2 − b2 a2 + b2 − c2
= + +
2abc 2abc 2abc
a2 + b2 + c2
= = RHS Hence Proved.
2abc
B B
Q. In a ∆ABC, if the sides are a = 3, b = 5 and c = 4 then sin
2
+ cos is equal to ?
2
a2 + c2 − b2 9 + 16 − 25
Sol. cosB =
2ac
=
( )( )
2 3 4
=0
π
⇒ B =
2
B B π π
hence sin + cos = sin + cos
2 2 4 4
1 1
= + = 2
2 2
Sol.
Let PQ = PR = a
Now,
PQ PR QR
2R’ = = = , R’ = Radius of Circumcircle
sinR sinQ sinP
a a QR
2a = = =
sinR sinQ sinP
1 π
sinR = sinQ = ⇒ ∠R = ∠Q =
2 6
π π 2π
Solution of Triangles
then ∠P = π − + =
6 6 3
13.
Q. In a ∆ABC, the value of (b2 – c2) cotA + (c2 – a2) cotB + (a2 – b2) cotC is
(A) 0 (B) 1 (C) 2 (D) None of these
Ans. (A)
Sol.
LHS is
Ist term = {(2RsinB)2 – (2RsinC)2} cotA
= 4R2(sin2B – sin2C) cotA
= 4R2{sin(B + C) sin(B – C)} cotA
cosA
= 4R2[sinA. sin(B – C)]
sinA
= 4R2{sin(B – C) cos(π – (B + C))}
= – 4R2{sin(B – C) cos(B + C)}
= – 2R2 {sin2B – sin2C}
Similarly,
IInd term = – 2R2 {sin2C – sin2A}
III term = – 2R2 {sin2A – sin2B}
Sum of all terms = 0
Q. a2 + b2 + c2
Prove that cotA + cotB + cotC =
4∆
Sol. LHS
cotA + cotB + cotC
cosA cosB cosC
= + +
sinA sinB sinC
b2 + c2 − a2 a2 + c2 − b2 a2 + b2 − c2
= + +
2bcsinA 2acsinB 2absinC
b2 + c2 − a2 a2 + c2 − b2 a2 + b2 − c2
= + +
4∆ 4∆ 4∆
a2 + b2 + c2
= = RHS Hence Proved.
4∆
Solution of Triangles
14.
π
Q. Let ABC be a triangle such that ∠ACB =
6
and let a, b and c denote the length
(
(A) − 2 + 3 ) (B) 1 + 3 (C) 2 + 3 (D) 4 3
Ans. (B)
a2 + b2 − c2
Sol. cosC =
2ab
( ) ( ) (
2 2
)
2
π x2 + x + 1 + x2 − 1 − 2x + 1
⇒ cos =
6 ( )( )
2 x2 + x + 1 x2 − 1
3 ( x + x + 1) − ( 2x + 1) + ( x − 1)
2 2 2
2 2
⇒ =
2 2 ( x + x + 1)( x − 1)
2 2
3 ( x + 3x + 2 )( x − x ) + ( x − 1)
2
2 2 2
⇒ =
2 2 ( x + x + 1)( x − 1)
2 2
( )( ) ( ) ( ) ( )
2 2
3 x+1 x+2 x x−1 + x+1 x−1
⇒ =
2 (
2 x2 + x + 1 x + 1 x − 1 )( )( )
⇒
(
3 x x+2 + x −1
=
2
) ( )
2 (
2 x2 + x + 1 )
⇒ ( )
3 x2 + x + 1 = 2x2 + 2x – 1
⇒ (2 − 3 ) x + (2 − 3 ) x − ( 1 + 3 ) = 0
2
− ( 2 − 3 ) ± ( 2 − 3 ) + 4 ( 2 − 3 )( 1 + 3 )
2
x=
2 (2 − 3 )
Solution of Triangles
− ( 2 − 3 ) ± ( 2 − 3 )( 2 − 3 + 4 + 4 3 )
x=
2 (2 − 3 )
15.
x=
(
− 2− 3 ± ) (2 − 3 ) 3 (2 + 3 )
2 (2 − 3 )
− (2 − 3 ) ± 3 2 ( 3 − 1) −2
x= = ,
2 (2 − 3 ) 2 (2 − 3 ) 2 (2 − 3 )
x = ( 3 − 1)( 2 + 3 ) , − ( 2 + 3 )
x = 1 + 3,
x = – (2 + 3) : rejected (∵ 2x + 1 < 0)
Q. If a = 3,b =
1
2
( )
6 + 2 and c = 2 then find ∠A
b2 + c2 − a2
Sol. cosA =
2bc
1
= 4
(
8+4 3 +2−3 )
2 3+1 (2 )( )
=
(2 + 3 ) − 1 = 1 ⇒ A = π
2 ( 3 + 1) 2 3
Sol.
Let x2 + x + 1 = a, 2x + 1 = b, x2 – 1 = c
(i) a > 0 ⇒ x2 + x + 1 > 0
⇒ x∈R
(ii) b > 0 ⇒ 2x + 1 > 0
1
⇒ x > –
2
(iii) c > 0 ⇒ x2 – 1 > 0
⇒ x∈(–∞, –1) ∪ (1, ∞)
Solution of Triangles
16.
() − ( x + x + 1)
2 2
( 2x + 1)
2
+ x2 − 1 2
=
2 ( 2x + 1) ( x − 1) 2
( x + 3x + 2)( −x + x ) + ( x − 1)
2
2 2 2
=
2 ( 2x + 1) ( x − 1) 2
( x + 1)( x + 2)( −x )( x − 1) + ( x + 1) ( x − 1)
2 2
=
2 ( 2x + 1)( x + 1)( x − 1)
=
( −x 2
) (
− 2x + x2 − 1 )=−1 ⇒ A = 120° Hence Proved.
(
2 2x + 1 ) 2
cosC
cotC sinC
Sol. =
cotA + cotB cosA cosB
+
sinA sinB
a2 + b2 − c2
4∆ a2 + b2 − c2
= 2 =
b + c2 − a2 a2 + c2 − b2 2c2
+
4∆ 4∆
101c2 − c2
= = 50
2c2
Q. The two adjacent sides of a cyclic quadrilateral are 2 & 5 and the angle
between them is 60°. If the area of the quadrilateral is 4 3 , find remaining
two sides.
Sol.
Let AD = 5, DC = 2
Solution of Triangles
and ∠D = 60°
then ∠B = 180° – 60°
= 120°
Also let BC = x, AB = y
ar (ABCD) = ar(∆ADC) + ar(∆ABC)
17.
1 1
4 3 = 2·5 sin60° + xy sin120°
2 2
3 1 3
4 3 = 5 + xy
2 2 2
⇒ 16 3 = 10 3 + xy 3
⇒ xy + 10 = 16
xy = 6 …(1)
In ∆ADC
52 + 22 − AC2 1 29 − AC2
cos60° = ⇒ =
2×5×2 2 20
⇒ AC = 19
2
In ∆ABC
x2 + y 2 − AC2 −1 x2 + y 2 − 19
cos120° = ⇒ =
2xy 2 2 6 ( )
x2 + y2 = 13 …(2)
from (1) & (2) x = 3, y = 2 or x = 2, y = 3
hence other sides are 2, 3.
A
Q. Prove that a(cosB + cosC) = 2(b + c) sin2
2
b + c 2RsinB + 2RsinC
Sol. a
=
2RsinA
B + C B − C
2sin cos
2 2
=
sinA
A B − C B − C
2cos cos 2cos
2 2 2
= =
A A A
2sin cos 2sin
2 2 2
A B − C B + C B − C
2sin .cos 2cos. .cos
= 2 2 = 2 2
A A
2sin2 2sin2
2 2
Solution of Triangles
cosB + cosC
=
A
2sin2
2
A
⇒ a(cosB + cosC) = 2(b + c)sin2 Hence Proved.
2
18.
A
Q. Prove that a(cosC – cosB) = 2(b – c) cos2
2
b − c 2RsinB − 2RsinC
Sol. a
=
2RsinA
B + C B − C
2cos .sin
= 2 2
sinA
A B − C B − C
2sin sin 2sin
2 2 2
= =
A A A
2sin cos 2cos
2 2 2
A B − C B + C B − C
2cos sin 2sin sin
2 2 2 2
= =
A A
2cos2 2cos2
2 2
cosC − cosB
=
A
2cos2
2
A
⇒ a(cosC – cosB) = 2(b – c)cos2 Hence Proved.
2
Sol. LHS:
= (bcosA + acosB) + (bcosC + ccosB) + (ccosA + acosC)
= c + a + b Hence Proved.
Sol. RHS
= 2RsinAcosA + 2RsinB cosB + 2RsinC cosC
= R(sin2A + sin2B + sin2C)
Solution of Triangles
19.
Q. If two times the square of the diameter of the circumcircle of a triangle is
equal to the sum of the squares of its sides then prove that the triangle is
right angled.
Sol.
Given
2(2R)2 = a2 + b2 + c2
⇒ 8R2 = a2 + b2 + c2
⇒ 8R2 = (2RsinA)2 + (2RsinB)2 + (2RsinC)2
⇒ 8R2 = 4R2{sin2A + sin2B + sin2C}
1 − cos2A 1 − cos2B 1 − cos2C
⇒ 2= + +
2 2 2
⇒ cos2A + cos2B + cos2C = – 1
⇒ – 1 – 4 cosA. cosB. cosC = – 1
⇒ cosA = 0 or cosB = 0 or cosC = 0
π π π
A= or B = or C =
2 2 2
Hence triangle is right angled.
Sol.
LHS
Ist term = a2[2RsinA.cos(B–C)]
= a2R[2sin(B + C) cos(B – C)]
= a2R[sin2B + sin2C]
= a2R[2sinB cosB + 2sinC cosC]
= a2[bcosB + ccosC]
= a2b cosB + a2c cosC
Similarly
IInd term = b2c cosC + b2a cosA
IIIrd term = c2a cosA + c2b cosB
On adding all the three terms,
LHS = ab(a cosB + b cosA) + bc(b cos C + c cosB) + ca(a cosC + ccosA)
Now by projection Rule,
LHS = ab(c) + bc(a) + ca(b) = 3abc Hence Proved.
Solution of Triangles
20.
HALF angle Formulas
1. (i)
A
sin =
s−b s−c ( )( )
2 bc
Proof: As we know that,
b2 + c2 − a2
cosA =
2bc
A b2 + c2 − a2
⇒ 1 – 2sin2 =
2 2bc
A
⇒ 2bc – 4bc sin2 = b2 + c2 – a2
2
A
⇒ a2 – (b2 + c2 – 2bc) = 4bc sin2
2
A
⇒ a2 – (b – c)2 = 4bc sin2
2
A
⇒ (a + b – c) (a – b + c) = 4bc sin2
2
⇒ sin2
A
=
{(a + b + c) − 2c} {(a + b + c) − 2b}
2 4bc
⇒ sin2
A
=
(
2s − 2c 2s − 2b
)( )
2 4bc
⇒ sin2
A
=
(
s−b s−c )( )
2 bc
sin
A ( s − b)( s − c ) A
⇒ = as sin > 0
2 bc 2
(ii) sin
B
=
( s − a )( s − c )
2 ac
(iii) sin
C
=
( s − a )( s − b)
2 ab
2. (i) cos
A
=
(
s s−a ) (ii) cos
B
=
(
s s−b )
Solution of Triangles
2 bc 2 ac
(iii) cos
C
=
(
s s−c )
2 ab
21.
3. From the above formulae, we can prove:
(i) tan
A
=
( s − b)( s − c ) (ii) tan
B
=
( s − a )( s − c )
2 s (s − a ) 2 s ( s − b)
(iii) tan
C
=
( s − a )( s − b)
2 s (s − c)
= bc
( s − b)( s − c ) s ( s − a )
bc bc
∴ ∆= (
s s−a s−b s−c)( )( )
abc
Note: Relation between area and circumradius R =
4∆
abc abc c
Proof: = = = R Hence Proved.
4∆ 1 2sinC
4 absinC
2
abc A B C
Q. Prove that
s
cos ·cos ·cos = ∆
2 2 2
abc (
s s−a ) s ( s − b) s ( s − c )
Sol. LHS =
s bc ac ab
= (
s s−a s−b s−c )( )( )
= ∆ Hence Proved.
Q. The circum-radius of the triangle whose sides are 13, 12 and 5 is:
1
∆= bc sinA = 30
2
abc 13.12.5 13
now R = = = units
4∆ 4·30 2
22.
Q. If the cotangents of half the angles of a triangle are in A.P. then prove that the
sides are in A.P.
A B C
Sol. Given cot
2
, cot , cot
2 2
in A.P.
B A C
⇒ 2cot = cot + cot
2 2 2
⇒ 2
(
s s−b ) =
s s−a( ) +
(
s s−c )
( s − a )( s − c ) ( s − b)( s − c ) ( s − a )( s − b)
Multiply by (
s s−a s−b s−c )( )( )
⇒ 2s(s – b) = s(s – a) + s(s – c)
⇒ 2s2 – 2sb = s2 – sa + s2 – sc
⇒ s(2b – a – c) = 0
⇒ 2b = a + c ⇒ a, b, c in AP.
C B
Q. Prove that b cos2
2
+ c cos2 = s
2
Sol. LHS
s s − c
= b
( ) + c s ( s − b)
ab ac
1
=
a
{(
s s−c + s s−b) ( )}
1
= {2s2 – s(b + c)}
a
=
1
a
{ (
2s2 − s 2s − a )} =
as
a
= s Hence Proved.
π
Sol. c2 = a2 + b2 ⇒ c =
2
Now LHS is
Solution of Triangles
23.
Q. If the sides of a triangle are 17, 25 and 28 then find the greatest length of
altitude ?
Sol.
Greatest altitude will form on shortest side
Hence let greatest altitude is x, hence
1
Area = .17.x = ∆ …(1)
2
17 + 28 + 25
Now s = = 35
2
∆= ( )( )( )
s s − a s − b s − c = 35.10.7.18
BD AD
Proof: In ∆ABD = …(1)
sinα sinB
DC AD
In ∆ACD = …(2)
sin sinC
BD sinβ sinC
(1) ÷ (2) ⇒ = …(3)
DC sinα sinB
⇒ m sinβ sin π − +β
=
{ ( )} {∵ α + B= θ}
n sinα sin θ − α ( )
⇒
msinβ sin θ + β
=
( )
nsinα sin θ − α( )
msinβ sinθcosβ + cossinβ
Solution of Triangles
⇒ =
nsinα sinθcosα − cossinα
⇒
msinβ sinsinβ cotβ + cotθ
=
( )
nsinα (
sinsinα cotα − cotθ )
24.
⇒ m cotα – m cotθ = n cotβ + n cotθ
⇒ (m + n) cotθ = mcotα – n cotβ Hence Proved.
from (3)
⇒
( (
m sin π − θ + C ))
=
sinC
n (
sin θ − B) sinB
⇒
(
m sinθcosC + cosθsinC ) = sinC
n ( sinθcosB − cosθsinB) sinB
AD 1
Sol. = (∵ CD is median)
DB 1
⇒ m = 1, n = 1
let ∠CDB = θ
then by ‘M – N’ theorem
(m + n) cotθ = m cotα – n cotβ
2cot(90° + A) = 1.cot90° – 1.cot(C – 90°)
– 2tanA = 0 + tanC
⇒ 2tanA + tanC = 0
3π
Q. If the median AM of a ∆ABC subtends an angle
4
with the side BC, then evaluate
|cotB – cotC|
Sol.
Given AM is a median.
⇒ BM : MC = 1 : 1
m = 1, n = 1
Solution of Triangles
3π
and θ =
4
by ‘M – N’ rule
(m + n) cotθ = ncotB – mcotC
25.
3π
2 cot = 1 cotB – 1 cotC
4
2(–1) = cotB – cotC
⇒ |cotB – cotC| = |–2| = 2
Q. The base of a triangle is divided into three equal parts. If t1, t2, t3 are the
tangents of the angles subtended by these parts at the opposite vertex then
prove that
1 1 1 1 1
+ + = 4 1 + 2
t 1 t2 t2 t3 t 2
Sol.
Given that
tanθ1 = t1
tanθ2 = t2
tanθ3 = t3
and BP : PQ : QC = 1 : 1 : 1
In ∆ABC by ‘m – n’ Rule
(2 + 1) cotα = 2cot(θ1 + θ2) – 1cotθ3
3cotα = 2cot(θ1 + θ2) – cotθ3 …(1)
In ∆PAC by ‘m – n’ Rule
(1 + 1) cotα = 1 cotθ2 – 1 cotθ3
2cotα = cotθ2 – cotθ3 …(2)
(1) ÷ (2)
( )
3 2 cot θ1 + θ2 − cotθ3
=
2 cotθ2 − cotθ3
⇒ 3cotθ2 – 3cotθ3 = 4cot(θ1 + θ2) – 2cotθ3
⇒ 3cotθ2 – cotθ3 =
(
4 cotθ2cotθ1 − 1 )
cotθ2 + cotθ1
⇒ 3cot2θ2 + 3cotθ1 cotθ2 – cotθ3 cotθ2 – cotθ1cotθ3 = 4cotθ2cotθ1 – 4
⇒ 3cot2θ2 + 4 = cotθ1 cotθ2 + cotθ2 cotθ3 + cotθ1cotθ3
⇒ 4cot2θ2 + 4 = cotθ1cotθ2 + cot2θ2 + cotθ1 cotθ3 + cotθ2cotθ3
⇒ 4(1 + cot2θ2) = (cotθ1 + cotθ2) (cotθ2 + cotθ3)
1 1 1 1 1
⇒ 4 1 + 2 = + + Hence Proved.
t t 1 t 2 t 2 t 3
2
Solution of Triangles
26.
Q. A man observes that when he moves up a distance ‘c’ meters on a slope, then
angle of depression of a point on the horizontal plane from the base of the
slope is 30°, and when he moves up further a distance ‘c’ meters, the angle of
depression of that point is 45°. The angle of inclination of the slope with the
horizontal is
Sol.
Given
QD = DR = c meter
and ∠DPQ = 30°
∠RPQ = 45°
DQ c 1
now = = ⇒ m = 1, n = 1
DR c 1
and ∠RDP = 30° + θ
now by ‘M – N’ Rule
(m + n) cot(30° + θ) = mcot30° – ncot(45° – 30°)
2
tan 30° + θ (
= 1. 3 − 1. 2 + 3
) ( )
⇒ tan(30° + θ) = – 1
⇒ 30° + θ = 135° ⇒ θ = 105°
Incircle of a triangle:
∆
(i) ∆ = sr ⇒ r =
s
(ii) To express the radius of incircle in terms of sides and tangent of the
half angle.
Hence
A B C
( ) (
r = (s – a) tan = s − b tan = s − c tan
2 2
)
2
B C
asin sin
(iii) r = 2 2
A
cos
2
A B C
(iv) r = 4R sin sin sin
2 2 2
Solution of Triangles
27.
Proof:
(i) Area of triangle ABC
∆ = ar(AIB) + ar(BIC) + ar(AIC)
1 1 1
∆= c.r + a.r + b.r
2 2 2
1 ∆
∆= (a + b + c).r ⇒ ∆ = s.r ⇒ r =
2 s
(ii) (s – a)tan
A
= (s – a)
( s − b)( s − c )
2 s (s − a )
( )(
s s−a s−b s−c )( ) ∆
= = =r
s s
B C
asin sin
(iii) 2 2
A
cos
2
= a
( s − a )( s − c ) · ( s − a )( s − b) bc
=
∆
=r
ac ab (
s s−a ) s
A B C
(iv) 4Rsin sin sin
2 2 2
= 4
abc ( s − b)( s − c ) ( s − a )( s − c ) ( s − a )( s − b)
4∆ bc ac ab
=
(
s s−a s−b s−c)( )( ) = ∆2 =
∆
=r
∆ s () ∆s s
2
C
r3 = s tan
2
28.
(iii) Radii of the escribed circles in terms of one side and function of half
angles:
A B C
r1 = 4Rsin cos cos
2 2 2
A B C
r2 = 4Rcos sin cos
2 2 2
A B C
r3 = 4Rcos cos sin
2
2
2
Proof: Area of triangle ABC
∆ = ar(ABI1) + ar(ACI1) – ar(BCI1)
1 1 1
∆ = c.r1 + b.r1 – a.r1
2 2 2
1
∆ = (c + b – a)r1
2
1
⇒∆= (a + b + c – 2a)r1
2
1
∆ = (2s – 2a)r1 ⇒ ∆ = (s – a). r1
2
∆
⇒ r1 =
s−a
Hence other results can be proved by using angle formulae.
Q. 1 1 1 1
Prove that + + =
r1 r2 r3 r
1 1 1
Sol. LHS = r1r2r3 + +
r r
1 2
r3
29.
1
= r1r2r3 ·
r
∆3 s
= ·
s−a s−b s−c ∆( )( )( )
∆2
= · s2 = s2 Hence proved.
(
s s−a s−b s−c )( )( )
Q. B−C C−A A −B
Prove that (r + r1) tan
2
(
+ r + r2 tan
2
) ( )
+ r + r3 tan
2
=0
Sol. In LHS
A Ab − c A
( 2
)
Ist term = s − a tan + stan
2b + c
cot
2
(
= s−a+s ) bb +− cc
b−c
= (2s – a).
b+c
b−c
= (b + c) . =b–c
b+c
Solution of Triangles
similarly II term = c – a
III term = a – b
Hence sum of all terms = 0
30.
Q. bc − r2r3 ca − r3r1 ab − r1r2
Prove that = = =r
r1 r2 r3
∆2 ∆2
Sol. rr1 + r2r3 =
(
s s−a
+
) ( s − b)( s − c )
1 1
= ∆2 +
s s − a ( ) ( )( )
s−b s−c
( s − b )( s − c ) + s ( s − a )
= ∆2
s ( s − a )( s − b )( s − c )
= ∆ 2
(2s2 − (a + b + c) s + bc)
∆2
= = 2s2 – (2s)s + bc
= bc
⇒ rr1 + r2r3 = bc
⇒ rr1 = bc – r2r3
bc − r2r3
⇒ r =
r1
Similarly, others can be proved.
Q. A
Prove that rr1r2r3 = ∆2 = r2r12cot 2
2
∆4
Sol. rr1r2r3 =
(
s s−a s−b s−c )( )( )
∆4
= 2
= ∆2
∆
A ∆2 ∆2 1
now r2r12cot 2 = 2· ·
2 s s − a tan2 A
2
( )
2
∆2 (
s s−a )
Solution of Triangles
= ×
s2 s − a( )
2
( s − b)( s − c )
∆4
= = ∆2 Hence proved.
(
s s−a s−b s−c )( )( )
31.
Q.
The triangle shown has sides of length 13, 30
and 37. The radius of the inscribed circle is
9
(A) 7 + 2 (B)
2
7
(C) 7 − 2 (D)
2
Ans. (B)
Q. r
In any triangle, prove that cosA + cosB + cosC = 1 +
R
Sol. LHS
A B C
= 1 + 4sin sin .sin
2 2 2
r A B C
= 1 + as r = 4Rsin sin sin Hence Proved.
R 2 2 2
Q. r1 + r2
Prove that = 2R
1 + cosC
Sol. LHS
C
4Rcos
= 2 sin A cos B + cos A sin B
C 2 2 2 2
2cos2
2
2R A +B
= sin
Solution of Triangles
C 2
cos
2
2R π −C
= · sin = 2R Hence Proved.
C 2
cos
2
32.
Q. In acute angled triangle ABC, a semicircle with radius ra is constructed with
its base on BC and tangent to the other two sides rb and rc are defined similar-
ly. If r is the radius of the incircle of triangle ABC then prove that
2 1 1 1
= + +
r ra rb rc
Q. r1 r2 r3 1 1
Prove that + + = −
bc ca ab r 2R
Sol. In LHS
A B C
4Rsin cos cos
I term = 2 2 2
2RsinB.2RsinC
A B C
sin cos cos
1 2 2 2
=
R B B C C
2sin cos 2sin cos
2 2 2 2
Solution of Triangles
A
sin
1 2
=
R B C
4sin sin
2 2
33.
Similarly,
B
sin
1 2
II term =
RA C
sin4sin
2 2
C
sin
1 2
III term =
R A B
4sin sin
2 2
Sum of all terms
A B C
sin2 + sin2 + sin2
1 2 2 2
=
R A B C
4sin sin sin
2 2 2
=
(
1 1 − cosA + 1 − cosB + 1 − cosC
) ( ) ( )
R A B C
2.4sin sin sin
2 2 2
A B C
3 − 1 + 4sin sin sin
1 2 2 2
=
R A B C
8sin sin sin
2 2 2
r
2 −
1 R 1 1
= = − Hence Proved.
R r r 2R
2
R
Q. Prove that r1 + r2 + r3 – r = 4R
Sol. LHS
∆ ∆ ∆ ∆
= + + −
s − a s − b s − c s
2s − a − b
= ∆
s − s − c ( )
+ ∆
s − a s − b ( )( ) (
s s − c )
Solution of Triangles
∆c ∆c
= +
( s − a )( s − b) s s−c( )
34.
= ∆c
(
2s2 − a + b + c s + ab
)
( )(
s s − a s − b s − c ) )(
∆ ( abc ) abc
= = .4 = 4R Hence Proved.
∆2 4∆
Q. The area of a triangle is 96 sq. units and the measure of its ex radius are 8, 12,
24. Find the length of the sides.
Q. r r
If 1 − 1 1 − 1 = 2, prove that the triangle is right angled.
r2 r3
s −b s−c
Sol. 1 −
1 −
s − a
=2
s−a
⇒
( s − a ) − ( s − b) · ( s − a ) − ( s − c ) = 2
(s − a ) (s − a )
⇒ (b – a) (c – a) = 2(s – a)2
2
(
b + c − a
⇒ bc – (b + c) a + a2 = 2
)
2
Solution of Triangles
35.
Q. Prove that r2 + r12 + r22 + r32 = 16R2 – (a2 + b2 + c2)
r1 + r2 + r3 – r = 4R
on squaring we get,
r2 + r12 + r22 + r32 + 2(r1r2 + r2r3 + r3r1) – 2r(r1 + r2 + r3) = 16R2
∆2 ∆2 ∆2
now rr1 + rr2 + rr3 = + +
(
s s−a ) (
s s−b ) (
s s−c )
( )( ) ( )( ) (
s − b s − c + s − a s − c + s − a s − b
= ∆2
)( )
s s−a s−b s−c ( )( )( )
= 3s2 – 2(a + b + c)s + (ab + bc + ca)
= – s2 + (ab + bc + ca) …(2)
put value from (2) into (1) then
r2 + r12 + r22 + r32 + 2s2 – 2(–s2 + ab + bc + ca) = 16R2
2
a +b + c
r2 + r12 + r22 + r32 + 4 – 2(ab + bc + ca) = 16R
2
2
⇒ r2 + r12 + r22 + r32 = 16R2 – (a2 + b2 + c2) Hence Proved.
Q. b−c c −a a −b
Prove that + + =0
r1 r2 r3
Sol. In LHS
2R
I term = (sinB – sinC)
r1
B + C B − C
2R 2cos sin
2 2
=
A B C
4Rsin cos cos
2 2 2
Solution of Triangles
B C B C
sin cos − cos sin
= 2 2 2 2
B C
cos .cos
2 2
36.
B C
= tan − tan
2 2
C A
Similarly II term = tan − tan
2 2
A B
III term = tan − tan
2 2
⇒ Sum of all terms = 0
Sol. By using AM ≥ GM
(s−a + s−b ) ( )
2
( )(
≥ s − a s − b …(1) )
( s − b) + ( s − c ) ≥
2
( s − b)( s − c ) …(2)
(s − c) + (s − a ) ≥
2
( )(
s − c s − a …(3) )
On multiplying (1), (2) and (3) we get
c a b
. . ≥ (s – a) (s – b) (s – c)
2 2 2
Now multiply by s on both side
(
s abc )
≥ s(s – a) (s – b) (s – c)
8
abc 1
(a + b + c)
16
≥ ∆2 ⇒
4
(
a + b + c abc ≥ ∆ )
Clearly equality occurs if, s – a = s – b = s – c
⇒ a = b = c Hence Proved.
Solution of Triangles
37.
Q. Prove that (r3 + r1) (r3 + r2) sinC = 2r3 r2r3 + r3r1 + r1r2
Sol. LHS
∆ ∆ ∆ ∆
= + + sinC
s − c s − a s − c s −b
b a
= ∆2
. .sinC
( s − c )( s − a ) ( s − c )( s − b)
absinC 2∆
= ∆2 . = ∆2 .
( s − c ) ( s − a )( s − b) ( s − a )( s − b)( s − c )
2 2
2∆ ∆2
=
( s − c ) ( s − a )( s − b)( s − c )
∆2 ·∆2
= 2r3
( s − a ) ( s − b) ( s − c )
2 2 2
∆2 .s
= 2r3
( s − a )( s − b)( s − c )
(
s − a + s − b + s − c
= 2r3 ∆2
) ( ) ( )
s−a s−b s−c ( )( )( )
1 1 1
= 2r3 ∆2 + +
s − b s − c (
s−a s−c )(
s−a s−b ) ( )( ) ( )( )
1 1 1 1
⇒ − = +
s−a s s−b s−c
⇒
(
s− s−a ) = ( s − c ) + ( s − b)
s (s − a ) ( s − b)( s − c )
38.
a a
⇒ =
(
s s−a ) ( s − b)( s − c )
⇒ (s – b) (s – c) = s(s – a)
⇒ s2 – (b + c)s + bc = s2 – as
(a – b – c)s + bc = 0
a + b + c ( )
⇒ [a – (b + c)] + bc = 0
2
⇒ a – (b + c) + 2bc = 0
2 2
⇒ a2 – b2 – c2 = 0
π
⇒ a2 = b2 + c2 ⇒ ∠A =
2
⇒ Right angled ∆
Q. Given a right triangle with ∠A = 90°. Let M be the mid point of BC. If the inradii
of the triangle ACM and ABM are r1 and r2 respectively then find the range of
r1
.
r2
∆/2
and s2 = AB + BM + AM ⇒ r2 = …(2)
s2
39.
a a
r1 s2 c++
= = 2 2 = c+a …(3)
r2 s1 a a b + a
b+ +
2 2
Now let c = acosθ, b = asinθ hence
r1 1 + cos θ r 1+ 1
= ⇒ 1 < =2
r2 1 + sin θ r2 1+0
max
r1 1+0 1 r 1
and > = ⇒ 1 ∈ , 2
r2 1+ 1 2 r2 2
min
1
Q. In a triangle PQR, P is the largest angle and cosP =
3
. Further the incircle of
the triangle touches the sides PQ, QR and RP at N, L and M respectively such
that the lengths of PN, QL and RM are consecutive even integers. Then possi-
ble length(s) of side(s) of the triangle is/are
(A) 16 (B) 18 (C) 24 (D) 22
Ans. (B,D)
Sol.
Let
PN = 2n = PM
QL = 2n + 2 = QN
RM = 2n + 4 = RL
Hence
PQ = 4n + 2
QR = 4n + 6, PR = 4n + 4
Now cosP = 1/3
PQ2 + PR2 − QR2 1
=
2PQ.PR 3
( ) ( ) ( )
2 2 2
22 2n + 1 + 22 2n + 2 − 22 2n + 3 1
⇒ =
( ) (
2.2 2n + 1 .2 2n + 2 ) 3
⇒ 3(4n2 – 4) = 2(4n2 + 6n + 2)
⇒ 4n2 – 12n – 16 = 0
Solution of Triangles
⇒ n2 – 3n – 4 = 0 ⇒ n = 4
sides (PQ, QR, RP) = (18, 22, 20)
40.
Q. If the length of the perpendicular from the vertices of a triangle A, B, C on the
1 1 1 1 1 1 1
opposite sides are p1, p2, p3 then prove that + + = = + +
p1 p2 p3 r r1 r2 r3
=
(
3s − a + b + c )=s =
1
Hence Proved.
∆ ∆ r
Q. If P1, P2, P3 are the altitudes of a triangle from the vertices A, B, C and ∆ de-
1 1 1 2ab C
notes the area of the triangle then prove that + + = cos2 .
P1 P2 P3 ( )
a +b+c 2
=
( (a + b) − c ) ( (a + b) + c )
(
2∆ a + b + c )
(a + b) − c2 = 2ab + (a2 + b2 − c2 )
2
Solution of Triangles
=
2∆ ( a + b + c ) 2∆ ( a + b + c )
41.
C
2ab 2cos2
2ab + 2abcosC 2
= =
(
2∆ a + b + c )
2∆ a + b + c ( )
2ab c
= .cos2 Hence Proved.
(
∆ a +b+c ) 2
1
AD = 2b2 + 2c2 − a2
2
1
BE = 2c2 + 2a2 − b2
2
1
CF = 2a2 + 2b2 − c2
2
Proof:
In ∆ABD
AB2 + BD2 − AD2
cosB =
2AB.BD
2
a
c2 + − AD2
Solution of Triangles
2
2 2 2
a +c −b
⇒ =
2ac a
2.c.
2
a2
⇒ a2 + c2 – b2 = 2c2 + – 2AD2
2
42.
a2
⇒ 2AD2 = c2 – + b2
2
⇒ AD2 =
1
4
(2b2 + 2c2 − a2 )
1
⇒ AD = 2b2 + 2c2 − a2 Hence Proved.
2
Similarly, other sides can be proved.
Sol.
Given BE = 12, CF = 9
G is centroid of ∆ABC
now
2
BG = BE = 8
3
2
CG = CF = 6
3
1
ar(∆BGC) = BG. CG. sinθ = 24sinθ
2
ar(∆ABC) = 3. ar(∆BGC)
∆ = 72 sinθ
hence ∆max = 72
Q. In a non-right angle ∆PQR, let p, q, r denotes the lengths of the sides opposite
to angles at P, Q, R respectively. The median from R meets the side PQ at S.
The perpendicular from P meets the side QR at E, and RS and PE intersect at
O. If p = 3 , q = 1 and the radius of the circumcricle of the ∆PQR equal to 1
then which of the following options is/are correct?
(A) Area of ∆SOE = 3 / 12
7
(C) Length of RS =
2
1
(D) Length of OE =
6
43.
Ans. (B,C,D)
Sol.
In ∆PQR
Circumradius
q p
R’ = =
2sinQ 2sinP
1 3
1= =
2sinQ 2sinP
π 2π π
⇒Q= ,P = and R = (∵ ∆PQR is not right angled)
6 3 6
r 1
Now = 1 ⇒ r = 2 = 1
2sinR 2
∵ ∠θ = ∠R ⇒ ∆PQR is isosceles
Hence PE is also a median ⇒ O is centroid
Now length of median
1 1 1
PE = 2q2 + 2r2 − p2 = 2+2−3 =
2 2 2
1 1 7
RS =
2
2p2 + 2q2 − r2 =
2
( )
2 3 +2−1 =
2
1 1
OE = PE ⇒ OE =
3 6
2 2 7 7
OS =
PS = . =
3 3 2 3
ar(∆OES) = ar(∆PES) – ar(∆POS)
1 π 1 π
= PE.PS.sin − Q − .PO.PS sin − Q
2 2 2 2
1 1 1 3 1 2 1 1 3
= · · − . . .
222 2 2 3 2 2 2
3 1 1 3−2 1
= − . = =
16 3 8 16 3 16 3
radius of incircle of ∆PQR
∆
r’ =
s
Solution of Triangles
1
(
p.q.sinR )
= 2
1
2
(
p+q+r )
44.
1
3.1
= 2
( 3 + 1+ 1 )
=
3 (2 − 3 )
×
2 (2 + 3 ) (2 − 3 )
⇒ r' =
2
3
(2 − 3 )
Length Of Angle Bisector:
In ∆ABC let AP, BQ and CR are angle bisectors then
2bc A
AP = cos ; P is point where angle bisector of A meet BC
b + c 2
2ac B
BQ = cos
a + c 2
2ab C
CR = cos
a +b 2
Proof:
ar (∆ABC) = ar (∆ABP) + ar (∆ACP)
1 1 A 1 A
bc sinA = c.AP sin + b.APsin
2 2 2 2 2
A
⇒ bc sinA = (b + c) sin .AP
2
A A A
⇒ bc 2sin cos = (b + c) sin .AP
2 2 2
A
⇒ 2bc cos = (b + c) . AP
2
2bc A
⇒ AP = cos
b+c 2
Similarly, others can be proved.
Solution of Triangles
45.
The Perimeter And Area Of A Regular Polygon Of
N-sides Inscribed In A Circle
∴ AL = OA sin ∠AOL
π
= rsin
n
Perimeter of polygon = nAB
π
= 2n. AL = 2nr sin
n
Area of polygon = n (Area of triangle AOB)
nr2 2π
= sin
2 n
= n
(
OL.AB
)
2
π
= nr2tan
n
Distance of orthocenter from vertices
bcosA
⇒ AH =
sinB
= 2R cosA
where H is the orthocenter of triangle ABC,
similarly
Solution of Triangles
BH = 2R cosB
CH = 2R cosC
Distance of orthocenter from sides
DH = CH. cosB
⇒ DH = 2R cosB cosC
where D is foot of altitude from A
46.
Q. Given (AH) (BH) (CH) = 3 and AH2 + BH2 + CH2 = 7
(where H is orthocenter of ∆ABC) Find
∏ cosA
(a) the ratio
∑ cos2 A
7
cos2A + cos2B + cos2C = …(2)
4R2
1 + cos2A 1 + cos2B 1 + cos2C 7
+ + =
2 2 2 4R2
7
3 + (cos2A + cos2B + cos2C) =
2R2
7
3 + (–1 – 4cosA. cosB. cosC) =
2R2
3 7
2 – 4. 3
= {from (1)}
8R 2R2
⇒ 4R3 – 3 = 7R
⇒ 4R3 – 7R – 3 = 0
⇒ (R + 1) (4R2 – 4R – 3) = 0
⇒ (R + 1) (4R2 – 6R + 2R – 3) = 0
3
⇒ (R + 1) (2R + 1) (2R – 3) = 0 ⇒ R =
2
now dividing (1) by (2),
∏ cosA 3 4R2 3 3 1
= × = = =
2
∑ cos A 8R 3 7 14R 3 7
14
2
Sol.
(2R cosB cosC) (2R cosA cosC) (2R cosA cosB)
Solution of Triangles
47.
9 9 1
= = =
8R 3
3
3 3
8
2
3
Sol. R=
2
Pedal Triangle
In an acute-angled triangle, orthocenter of ∆ABC is the incentre of the
Pedal triangle DEF.
R
Circumradius of pedal triangle =
2
(where R is circumradius of ∆ABC)
Proof: In ∆AEF
AE2 + AF2 − EF2
cosA =
2. AE. AF
⇒ cosA =
2 ( ccosA )(bcosA )
⇒ 2bc cos3A = (b2 + c2) cos2A – EF2
⇒ EF2 = (b2 + c2) cos2A – 2bc cos3A
⇒ EF2 = cos2A {b2 + c2 – 2bc cosA}
⇒ EF2 = cos2A {b2 + c2 – (b2 + c2 – a2)}
⇒ EF2 = cos2A. a2
Solution of Triangles
⇒ EF = a cosA
Now ∠FDE = ∠FDA + ∠ADE
π π
= −A+ −A
2 2
= π – 2A
48.
Q. Prove that circumradius of ∆HBC, ∆HCA, ∆HAB, ∆ABC are same.
Sol. Circumradius of
a b c
(i) ∆ABC ⇒ R = = = …(1)
2sinA 2sinB 2sinC
BC a
(ii)
∆HBC ⇒ R1 = =
(
2sin BHC ) (
2sin B + C )
a a
= = …(2)
(
2sin π − A ) 2sinA
CA b
(iii)
∆HCA ⇒ R2 = =
(
2sin CHA ) (
2sin C + A )
b b
= = …(3)
(
2sin π − B ) 2sinB
AB c
(iv)
∆HAB ⇒ R3 = =
(
2sin AHB ) (
2sin A + B )
c c
= = …(4)
(
2sin π − C ) 2sinC
Q.
Prove that
(i) ∆P = 2∆ cosA cosB cosC; ∆P is area of pedal triangle.
R
(ii) RP =
2
(iii) rP = 2R cosA cosB cosC
1
Sol. (i) ∆p =
2
DE. EF sin(DEF)
1
= (c cosC) (a cosA) sin(π – 2B)
2
1
Solution of Triangles
49.
EF acosA
(ii) RP = =
2sin FDE( ) (
2sin π − 2A )
acosA 1 a
= = .
(
2 2sinAcosA ) 2 2sinA
R
=
2
π − 2A π − 2B π − 2C
(iii) rP = 4RP sin sin sin
2 2 2
R π π π
= 4 .sin − A sin − B sin − C
2 2 2 2
= 2R cosA cosB cosC Hence Proved.
Q Area ∆ABC = 18; Area ∆BDF = 2, length of side DF = 2 2 then find R (where D, F
are feet of altitude from A and C respectively)
Sol. Given ∆ = 18
ar(∆BDF) = 2
1
FB. DB sin(FBD) = 2
2
1
(a cosB) (c cosB) sinB = 2
2
1
⇒ acsinB . cos2B = 2 ⇒ 18 cos2B = 2
2
1
⇒ cosB =
3
also DF is one of the side of pedal triangle of ∆ABC hence
DF = b cosB = 2 2
1
⇒ b = 2 2 ⇒ b = 6 2
3
b 6 2 6 2
now R= = =
2sinB 1 2 2
Solution of Triangles
2 1− 2.
9 3
9
= units
2
50.
External Or Ex-centric Triangle
Triangle is formed by joining the 3 excentres is
called excentral triangle or excentric triangle
∆ABC is a pedal triangle for ∆I1I2I3 therefore
incentre of ∆ABC is orthocenter for ∆I1I2I3.
Note:
π A π B π C
(i) Internal angles are − , − , −
2 2 2 2 2 2
(ii) Sides of excentric ∆ are
C
I1 I2 = 4R cos
2
A
I2 I3 = 4R cos
2
B
I1 I3 = 4R cos
2
Distance between Incentre and excentre:
A
II1 = 4R sin
2
B
II2 = 4R sin
2
C
II3 = 4R sin
2
Proof: In ∆IBD
B ID r
sin = ⇒ BI = …(1)
2 BI B
sin
2
In ∆IBI1
BI
cos(BII1) =
II1
r
B
π c sin 2
⇒ cos − =
2 2 II1
{∠BII1 = ∠BCI1 ∵ B, I1, C, I are concyclic}
Solution of Triangles
C r
⇒ sin =
2 B
II1sin
2
51.
r A
⇒ II1 = = 4R sin Hence Proved.
B C 2
sin .sin
2 2
Now in ∆II1I2 by sine rule
I1I2 II1
=
( )
sin I1II2
sin
A
2
A
I1I2 4Rsin
⇒ = 2
A +B A
sin π − sin
2 2
{∵ I, C, I2, A are concyclic}
A +B
⇒ I1I2 = 4R sin
2
C
⇒ I1I2 = 4R cos Hence Proved.
2
Proof: O → Circumcenter
H → Orthocenter
In ∆AOH
OA2 + AH2 − OH2
cosθ =
2OA.AH
( )
2
R2 + 2RcosA − OH2
cos{A – (π – 2C)} =
2.R.2RcosA
R2 + 4R2cos2 A − OH2
cos(C – B) =
4R2cosA
4R2cosA. cos(C – B) = R2 + 4R2cos2A – OH2
⇒ OH2 = R2 + 4R2 cosA{cosA – cos(C – B)}
Solution of Triangles
52.
(B)
Distance between incentre & circumcenter
is given by R2 − 2Rr
Proof: O → Circumcenter
I → Incenter
In ∆AOI
OA2 + IA2 − OI2
cosθ =
2 . OA .IA
⇒ OI2 = OA2 + IA2 – 2OA. IA cosθ
2
r r A π
= R2 + − 2R. cos − + C
sin A sin A 2 2
2 2
2 B C B C C −B
= R + 4Rsin 2 sin 2 − 2R.4Rsin 2 sin 2 cos 2
B C B C C − B
= R2 + 8R2 sin sin 2sin sin − cos
2 2 2 2 2
B C B + C
= R2 + 8R2sin sin −cos
2 2 2
A B C
= R2 − 8R2sin sin sin
2 2 2
= R2 – 2Rr
⇒ OI = R2 − 2Rr Hence Proved.
(C) Distance between incenter and orthocenter is
2r2 − 4R2cosAcosBcosC
Proof: I → incentre
H → orthocenter
B C
AI = 4Rsin sin
2 2
AH = 2R cosA
Now in ∆AIH,
cosθ =
2AI. AH
53.
B C B C C −B
= 16R2sin2 sin2 + 4R2cos2 A − 2.4Rsin sin .2RcosA.cos
2 2 2 2 2
B C B C C B C B
= 4R2cos2A + 16R2sin2 sin2 − 16R2cosAsin sin cos cos + sin sin
2 2 2 2 2 2 2 2
B C B C
= 4R2cos2A + 16R2sin2 sin2 – 4R2cosA sinB sinC – 16R2cosA sin2 sin2
2 2 2 2
B C
= 4R2cosA{cosA – sinB sinC} + 16R2sin2
2
sin2
2
(1 − cosA )
B C A
= 4R2cosA(– cosB cosC) + 16R2sin2 sin2 2sin2
2 2 2
2
A B C
= – 4R cosA cosB cosC + 2 4Rsin sin sin
2
2 2 2
= – 4R2cosA cosB cosC + 2r2
⇒ IH = 2r2 − 4R2cosAcosBcosC Hence Proved.
Q. Prove that in any triangle H, G, O are collinear and G divides the line HO in 2 : 1
where H is orthocenter, O is circumcenter and G is centroid.
Sol.
Let O → Circumcentre
H → Orthocentre
and OD || HM
also Q is point of intersection of AD and OH
Now In ∆AQH & ∆DQO
∠DQO = ∠AQH (Vertically opposite)
∠ODQ = ∠HAQ (alternate)
⇒ ∆DOQ ∼ ∆AHQ
Now OD = RcosA
AH = 2RcosA
AQ HQ AH 2RcosA
By similarity = = =
QD QO OD RcosA
AQ HQ 2
⇒ = =
QD QO 1
Solution of Triangles
54.
Q. 1 2
Prove that: (OG)2 = R2 – (a + b2 + c2) where O is circumcenter & G is centroid
9
1 1
Sol. OG =
3
OH = R 1 − 8cosAcosBcosC
3
1 2
(OG)2 =
9
( {
R 1 − 4 2cosAcosB cosC} )
1 2
= R (1 – 4{cos(A + B) + cos(A-B)}cosC)
9
1 2
= R (1 – 4{– cosC + cos(A – B)}.cosC)
9
1 2
= R {1 + 4cos2C + 4cos(A – B) cos(A + B)}
9
1 2
= R {1 + 4(1 – sin2C) + 4cos2A – 4sin2B}
9
1 2
= R {5 – 4sin2C + 4(1 – sin2A) – 4sin2B}
9
1 2
= R {9 – (2sinA)2 – (2sinB)2 – (2sinC)2}
9
1 2 a2 b2 c2
=
9
R 9 − 2 − 2 − 2 = R2 −
R R R
1 2
9
(
a + b2 + c2 ) Hence Proved.
Ambiguous cases:
Q. Two sides of a triangle are of lengths 6 and 4 and the angle opposite to
smaller side is 30°. How many such triangles are possible? Find the length of
their third side and area.
⇒ = ⇒ sinB =
1 sinB 3
2
Now two values of B are possible hence two triangles are possible
55.
b c
Now =
sinB sinC
c=
(
sin A + B ) .b
sinB
sinA.cosB cosAsinB
c= + b
sinB sinB
= (sinA cotB + cosA)b
1 1 3
= ±
2
+
2 2
4 = 6 ± 1 ( ) 2 unit
1
Area = bc sinA
2
=
1
2
4 ( 6±1 ) 2×
1
2
= ( 6±1 ) 2 sq. unit
ð
Q. If b = 3, c = 4 and B = 3 , then find number of triangles that can be constructed.
b c
Sol. =
sinB sinC
3 4 2
⇒ = ⇒ sinC = >1
3 sinC 3
2
⇒ No such triangle is possible.
b c
Sol. =
sinB sinC
5 5 2
⇒ = ⇒ sin C = 1
1 sinC
2
Solution of Triangles
π
C=
2
Hence only one triangle is possible
56.
5
Q. If b = 3, c =
2
and C = 30° then find the number of triangles that can be
constructed.
b c
Sol. =
sinB sinC
5
3 3
⇒ = 2 ⇒ sinB =
sinB 1 5
2
Two values of B are possible, hence two such triangles are possible
a b
Sol. =
sinA sinB
7 8 4
⇒ = ⇒ sinB =
1 sinB 7
2
Two values of B are possible, hence two such triangles are possible.
57.