**9
- 3 The Law of Sines**
** **
*Sometimes you are
working
with triangles that are not right triangles. The normal trig
functions
are defined for a right triangle and are not directly useful in solving
non-right triangles. We can however, use trig to determine
the law of sines. Study the following figure:*
* *
*How do we determine
a
means to solve this non-right triangle? Let's drop down a
perpendicular
from *__/__ B. Call it h.
*Notice we have
formed
two right triangles. The one on the left has the following
relationship
from basic trig:*
*Sin A = h/c*
*c Sin A = h*
*While the right
triangle
on the right has the relationship:*
*Sin C = h/a*
*a Sin C = h*
*Using the
transitive
property (you do remember, of course!):*
*c Sin A = a Sin C*
*Dividing by ac
yields:*
*(Sin A)/a = (Sin
C)/c*
*We can drop
perpendiculars
from the other two vertices and get the other relationships. This
then is the law of sines:*
__Law
of Sines__
Demo:
Law of Sines (Manipula Math)
*Notice that the
above
represents simple proportions and in order to solve we must know 3
parts.
Any two angles and a side will work. Two sides and a non-included
angle can work also. This is an ambiquous case and may result in
no solutions, one or two solutions. We will talk about this case
a little later.*

__Sample
problems__
*Give angles to the nearest
tenth
of a degree and lengths to 3 significant digits.*
*1) Using the triangle
above,
*__/__ A = 50^{o}, __/__ B = 65^{o} and a =
12.
Solve the triangle.
**Solution:**
**Find **__/__ C
**180 - (65 + 50) = 65**
**Thus
**__/__ C = 65^{o} which makes
the
triangle isosceles
**To find b:**
**(Sin 65)/b = (Sin
50)/12**
**b Sin 50 = 12 Sin 65**
**b = (12 Sin 65)/(sin
50)**
**b = 14.2**
**Since **__/__ C = __/__
B, c = b, so c = 14.2
** **
**2) Solve the triangle if
**__/__
B = 30^{o}, __/__ C = 70^{o} and b = 10
**Solution:**
**Find **__/__ A
**180 - (30 + 70) = 80**
__/__ A = 80^{o}
**To find a:**
**(Sin 80)/a = (Sin
30)/10**
**a Sin 30 = 10 Sin 80**
**a = (10 Sin 80)/Sin 30**
**a = 19.7**
**To find c:**
**(Sin 70)/c = (Sin
30)/10**
**Try to use exact
values
if possible. Notice, I didn't use side a.**
**c Sin 30 = 10 Sin 70**
**c = (10 Sin 70)/Sin 30**
**c = 18.8**
**Note that the largest
side
is opposite the biggest angle, smallest side opposite the smallest
angle,
etc.**

__Ambiquous Case (SSA)__
**Two sides and a
non-included
angle!!**
**In the pictures below, **__/__ A, b, and a
are
given.
** **
*Notice, the
shortest
distance possible to form a triangle is the distance bSin A from basic
trig. The shortest distance is a perpendicular.*
*Case 1:
If the length of a is shorter than b Sin A, it is impossible to form a
triangle. Result: no solution.*
* *
*Case 2: If
a is longer than b Sin A but shorter than b, two triangles can be
formed.
Result: 2 triangles*
*Case 3: If
a is the length of b Sin A, a right triangle can be formed.*
*Result: one
triangle*
* *
*Case 4: If
a is equal to or longer to b, then one triangle can be formed.*
*Result: one
triangle.*
*The blue
lines
above represent the solutions!!*

__Sample Problems__
*1) Tell how many
solutions
each has.*

* *

* a)
a = 8, b = 4, *__/__ B = 30^{o}

*
Solution: Calculate the shortest distance: 8 Sin 30 =
4.
Since b = 4 we can form 1 triangle and it is a right triangle.*
*
b)
a = 6, b = 4, *__/__ A = 45^{o}

*
Solution: Again, calculate the shortest distance: 6 Sin 45
= 4.2. Since b is less than 4.2, no triangle can be formed.*

* *

* c)
b = 7, c = 6.5, *__/__ C = 60^{o}

*
Solution: Guess what to calculate! 7 Sin 60 = 6.06.
Since
c is in between the shortest distance and side b, it will form two
triangles.*

*
d)
a = 5, c = 10, *__/__ C = 70^{o}

*
Solution: Calculate the shortest distance: 5 Sin 70 =
4.7.
Since c is larger than a, only one triangle can be formed. (We
really
didn't have to calculate the shortest distance, but it's good
practice!!)*

__Story
problems with Law of Sines__
** **

*1) When the angle of
elevation of the sun is 62*^{o}, a telephone pole tilted
at
an angle of 7^{o} away from the sun casts a shadow 30 feet long
on the ground. To tenths place, approximate the length of the
phone
pole.
*Solution:*
*Draw a diagram!!*
*The other base
angle
is (90 - 7) = 81*^{o} Why?
*The vertex angle is
180
- ( 62 + 81) = 180 - 143 = 37*^{o}
*Use the law of
sines
to calculate the height of the pole!!*
*(Sin 62)/x = (Sin
37)/30*
*x Sin 37 = 30 Sin 62*
*x = (30 Sin 62)/
Sin
37*
*x = 44.0*
*The pole is about
44.0
feet!!*
* *
*2) To find the
distance
between two points A and B that are on opposite sides of a river, a
surveyor
measures a distance on the same side of the river as point A. The
distance to this point is 240 feet and call it point C. He then
measures
the angles from A to B as 62*^{o} and measures the angle from C
to B as 55^{o}. To tenths place, approximate the distance
from A to B.

* *
*Solution:*
*Again, draw a
picture!!*
*First, we need *__/__
B.
*180 - (62 + 55) =
180
- 117 = 63*^{o}
*We can now use the
law
of sines:*
*The side we need is
side
c*
*(Sin 55)/c = (Sin
63)/240*
*c Sin 63 = 240 Sin
55*
*c = (240 Sin 55)/
Sin
63*
*c = 220.6*
*The distance from A
to
B is 220.6 feet.*
* *
**Let's
head on over and take a look at the Law of Cosines!**