Transcript Sine Law

There is no doubt that
the 3 PTRs are
extremely useful when
solving problems
modeled on a right
triangle.
Unfortunately, the world
does not consist only of
right triangles…
As a matter of fact, right
triangles end up being
more of a rarity than
commonplace.
Does that mean when we
come across a situation
that can only be modeled
with a non-right triangle
that we abandon our
pursuit?….
No Way!!!!
There exists 2 Laws of Trigonometry
that allow one to solve problems
that involve non-right Triangles:
A triangle is uniquely determined
by two angles and a particular
side
C
b
A
a
O2
O1
c
B
If a corresponding angle and side
are known, they form an
“opposing pair”
C
b
A
a
O2
O1
c
B
The Sine Law can be used to
determine an unknown side or
angle given an “opposing pair”
C
b
A
a
O2
O1
c
B
Find the length of b
C
b
A
5
65o
30o
c
B
Construct CN with height h
C
b
5
h
A
30o
65o
c
N
B
By the right triangle SIN ratio
Sin 30o = h
Sin 65o = h
b
5
C
b
5
h
A
30oo
65o
c
N
B
Solve both equations for h
Sin
o
30
=h Xb
Sin
o
65
b
o
bSin30
=h
h=
=h X5
5
o
5Sin65
Because the equations
are equal
bSin30o = 5Sin65o
bSin30o = 5Sin65o
b = 5Sin65o
o
Sin30
b = 9.1
Consider the general case:
C
b
h
A
c
N
Sin A = h
Sin B = h
b
a
bSinA = aSinB
a
B
bSinA = aSinB
a
a
bSinA = SinB
a
bSinA = SinB
ab
b
SinA = SinB
a
b
Extend this to all 3 sides of a
triangle, and the Sine Law is
generated!
SinA = SinB = SinC
a
b
c
Find the length of a
C
a =
Sin73o
57o
a = 27.4
a
A
73o
c
24
24
Sin57o
N
Find h
h
5.9O
10.3O
2.9 km
Find h
1. Find O
O = 180O – 5.9O – 10.3O
= 163.8O
O
5.9O
10.3O
2.9 km
X
Find X
SIN 10.3O
=
2.9
SIN163.8O
X = 1.86km
X
163.8O
5.9O
10.3O
2.9 km
Find h
SIN 5.9O = h
1.86 km
h = 191.2 m
1.86 km
h
5.9O
10.3O
2.9 km
The Ambiguous
Case
Find A
11
48o
SinA =
11
9
A
Sin48o
9
A = 65.3o
Does that make
sense?
No Way!!!
Side 9 can also be drawn as:
11
48o
Could A be
o
65 in this
case?
9
A
This type of discrepancy is called
the “Ambiguous Case”
Be sure to check the diagram to
see which answer fits:
o
O, or 180 - O
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