Download (b) When three angles are given

Mathematics
Session
Properties of Triangle - 2
Session Objectives
Session Objective
Solution of Right-angled Triangle
Solution of a Oblique Triangle
(a)
When three sides are given
(b)
When three angles are given
(c)
When two sides and the included angle
between them are given
(d)
Two angles and one of the corresponding
sides are given
(e)
When two sides and an angle opposite to
one of them is given
Introduction
A triangle has three sides and three
angles. If three parts of a triangle, at
least one of which is side, are given then
other three parts can be uniquely
determined.
Finding other unknown parts, when three
parts are known is called ‘solution of
triangle’.
Solution of Right-angled Triangle
(a)
Given a side and an acute angle
Let the angle A (acute) and side c of a
right angle at C be given
B  90  A, b  c cos A, a  c sin A
B
or b = c sinB, a = c cosB
a
C
c
b
A
Solution of Right-angled Triangle
(b)
Given two sides
Let a and b are the sides of ABC .C is the
right angle. Then we can find the remaining
angles and sides by the following way
A
c  a2  b2 , etc.
b
tanA  ; B  90  A ;
a
b
C
c
a
B
Solution of a Oblique Triangle
(a) When three sides are given
(i) If the given data is in sine, use
the following formulae
A
sin 
2
 s – b  s – c 
bc
B
, sin 
2
 s – c  s – a 
ca
(ii) If the given data is in cosine, use
the following formulae.
A
cos 
2
s  s  a
bc
B
, cos 
2
s s  b
ac
, etc
, etc.
Solution of a Oblique Triangle
(iii)If the given data is in tangent,
then we use
A
tan 
2
 s  b  s  c 
,
s s  a
B
tan 
2
 s  c  s  a 
' etc
s s  b
(iv) If the lengths of the sides a, b and c are small, the
angle of triangle can also be obtained by cosine rule.
(v) For logarithmic computation, we define the following.
Lcos   10  logcos , Lsin   10  logsin ,etc.
Solution of a Oblique Triangle
(b)
When three angles are given:
In this case, the sides cannot be
determined uniquely. Only ratio of the
sides can be determined by sine rule
and hence there will be infinite number
of such triangles.
Solution of a Oblique Triangle
(c)
When two sides and the
included angle between
them are given:
If two sides b and c and the included
angle A are given, then (B – C) can be
found by using the following formula:
BC bc
A
tan

cot
2
bc
2
If b < c, then we use
C B c b
A
tan

cot
2
c b
2
if
b  c
Solution of a Oblique Triangle
(d)Two angles and one of the
corresponding sides are given:
The value of the other side and remaining
angle can be found by
a
b
c


and A  B  C  180o
sin A sinB sinC
Solution of a Oblique Triangle
(e)
When two sides and an
angle opposite to one of them is
given:
In this case, either no triangle or one
triangle or two triangles are possible
depending on the given parts.
Therefore, this case is known as ambiguous case.
Let a, b and the angle A are given.
b2  c 2  a 2
 cos A 
2bc


 c2   2bcos A  c  b2  a2  0
Solution of a Oblique Triangle
This is a quadratic equation in c.
Let c1 and c2 be two values of c
 c

2bcos A  4b2 cos2 A  4 b2  a2
2
 bcos A  b2 cos2 A  b2  a2

 bcos A  a2  b2 1  cos2 A
c  bcos A  a2  b2 sin2 A


Solution of a Oblique Triangle
 c1  bcos A  a2  b2 sin2 A
.....(i)
and c2  bcos A  a2  b2 sin2 A .....(ii)
Case I: When a < b sinA
if a  bsin A  a2  b2 sin2 A
or a2  b2 sin2 A  0
Hence, from (i) and (ii), c1 and c 2 become imaginary.
No triangle is possible.
Solution of a Oblique Triangle
Case II: When a = b sinA
If a  bsin A  a2  b2 sin2 A
 a2  b2 sin2 A  0
 From (i) and (ii),
c1  c 2  bcos A
But c1 and c 2 will be positive when A is acute
angle. In this case, only one triangle is possible
provided A is acute.
Solution of a Oblique Triangle
Also a = b sinA
a

b
sin A
 sinB  1
k sin A

 k sinB
sin A
 B  90
The triangle is right-angled in this case.
Case III:
When a > b sinA
If a  bsin A  a2  b2 sin2 A
 From (i) and (ii), c1 and c 2
 a2  b2 sin2 A  0
are real.
Solution of a Oblique Triangle
But triangle is possible only when c1 and c 2
are positive. For this we have to consider
the following cases:
(a)
When a > b
 a  b  a2  b2

 a2  b2 cos2 A  sin2 A

 a2  b2 sin2 A  b2 cos2 A

a2  b2 sin2 A  bcos A
 bcos A  a2  b2 sin2 A  0
 c1  0 and c 2  0,
Hence only one triangle
is possible.
Solution of a Oblique Triangle
(b)
When a = b
a  b  a2  b2

 a2  b2 cos2 A  sin2 A

 a2  b2 sin2 A  b2 cos2 A

a2  b2 sin2 A  bcos A
 c1  0 and c 2  0
Hence, only one triangle is possible.
Solution of a Oblique Triangle
(c) When a < b
a  b  a2  b2

 a2  b2 cos2 A  sin2 A

 a2  b2 sin2 A  b2 cos2 A
 a2  b2 sin2 A  bcos A
 c1  0
and c 2  0
 Two triangles are possible. (Ambiguous case)
Thus, when a, b and A are given, two triangles are possible
when a > b sinA and a < b
[For ambiguous case]
Class Test
Class Exercise - 1
If the sides of a triangle are
x2  x  1, 2x  1 and x2  1.
Prove that its largest angle is 120°.
Solution
Let a  x2  x  1, b  2x  1 and c  x 2  1
 a, b, c are the sides of the triangle,
 a > 0, b > 0, c > 0
1 1 1
 a  x  x  1  x  2.x.    1
2 4 4
2
2
2
2
 3
1

  x    
  0 for all x.
2

 2 
b  2x  1  0
1
 x
2
Solution Cont.
c  x2  1  0   x  1 x  1  0
 x  1 or x  1
Hence, a, b, c are positive, when x > 1Now
x2  x  1  2x  1  x2  x  x  x  1  0
as x > 1
a – c = x2 + x + 1 – x2 + 1 = x + 2 > 0
as x > 1
 a is the largest side  A is the largest angle.
b2  c 2  a2
 cos A 
2bc
Solution Cont.
 2x  1
2


   x  x  1
2  2x  1  x2  1

2
 x 1
2
2
2
1  2x  x 2  2x3


2  2x  1 x 2  1
1  2x  1  x2 
1  2x   x 1  2x 



2
2  2x  1  x  1
2 1  2x   x2  1
2

1
 A  120
2
(Pr oved)
Class Exercise - 2
The angles of a triangle are in the
ratio 1 : 2 : 7. Show that the ratio
of the greatest side to the least
sides is 5  1 : 5  1 .



Solution
Let the angles of the triangle be
x°, 2x°, 7x°.
 x + 2x + 7x = 180°  x = 18°
 A = 18°, B = 36°, C = 126°
 Least side is a and the greatest side is c.
a
c
c sinC


 
sin A sinC
a sin A
c sin126 sin 180  54  sin54
 


a
sin18
sin18
sin18
Solution Cont.

c cos36


a sin18

5 1


4

5 1

5 1
5 1
4
Proved.
Class Exercise - 3
Let L sin   10  logsin . The number
of triangle such that log b + 10 = log
c + L sinB is
(a) one
(b) two
(c) infinite
(d) None of these
Solution
 log b + 10 = log c + L sin B
 log b + 10 = log c + 10 + log sinB
 log b = log (c sinB)
 b = c sinB
 Only one triangle is possible
Class Exercise - 4
In the ambiguous case, if the
remaining angles of the triangles
formed with a, b and A be
sinC1 sinC2
B1, C1 and B2 , C2 , then

is
sinB1 sinB2
(a) 2 cosA
(b) cosA
(c) 2 sinA
(d) sinA
Solution
The two triangles formed are
AB1C and AB2C.
Let AB1  c1
AB2  c2
In AB1C
C
c1
b

sinC1 sinB1
sinC1 c1


............(1)
sinB1 b
In AB2C,
C
1
C
2
b
a
A
B1
L
B2
Solution Cont.
c2
sinC2 c 2
b



......(ii)
sinC2 sinB2
sinB2
b
sinC1 sinC2 c1  c 2
From (i) and (ii) ,


....(iii)
sinB1 sinB2
b
Triangles are formed with a, b and A,
b2  c 2  a2
 cos A 
2bc


 c2   2bcos A  c  b2  a2  0
Solution Cont.
Here two values of c are c1 and c2 .
 c1  c2  2b cosA (Sum of roots)
From (iii),
sinC1 sinC2 2bcos A


 2cos A
sinB1 sinB2
b
Hence answer is (a).
Class Exercise - 5
In ambiguous case, where b, c, B are
given and a2  2a1,prove that
3b  c 1  8sin2 B.
Solution
b, c and B are given,
c 2  a2  b2
 cosB 
2ca


 a2   2c cosB a  c2  b2  0
This is quadratic equation in a.
 a

2c cosB  4c 2 cos2 B  4 c 2  b2
2
 c cosB  c2 cos2 B  c2  b2

Solution Cont.
a  c cosB  b2  c 2 sin2 B
 a1  c cosB  b2  c 2 sin2 B
and a2  c cosB  b2  c 2 sin2 B
It is given that a2  2a1
2
2
2

2c
cosB

2
b

c
sin
B
 c cosB  b  c sin B
2
2
2
Solution Cont.

 c2 cos2 B  9 b2  c2 sin2 B



 c2 1  sin2 B  9b2  9c2 sin2 B


 9b2  c 2 1  8sin2 B  3b  c 1  8sin2 B (Pr oved)
(Negative sign is neglected as ‘b’ is the
length of side of triangle).
Thank you