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 = 50o, / B = 65o and a = 12.  Solve the triangle.
 
Solution:
Find / C
180 - (65 + 50) = 65
Thus / C = 65o 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 = 30o, / C = 70o and b = 10
 
Solution:
Find / A
180 - (30 + 70) = 80
/ A = 80o
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 = 30o
            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 = 45o
            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 = 60o
            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 = 70o
            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 62o, a  telephone pole tilted at an angle of 7o 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) = 81o  Why?
The vertex angle is 180 - ( 62 + 81) = 180 - 143 = 37o
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 62o and measures the angle from C to B as 55o.  To tenths place, approximate the distance from A to B.
 
Solution:
Again, draw a picture!!
 
First, we need / B.
180 - (62 + 55) = 180 - 117 = 63o
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!