



There are six pieces of information to know about a triangle:










The length of each of the three
sides 


The measure of each of the three angles. 





A good question to ask would be: 


What is the least number of these six things we need to know to find
the others? 





The answer is: 


We need to know 3 of the six things, 


and at least one of the 3 things must be the length of a side. 





We can solve for all six things if we know: 





The length of all 3 sides 


The length of any 2 sides and the measure of any 1 angle 


The length of any 1 side and the measure of any 2 angles 





BUT WE
CAN'T DO A THING 


IF WE ONLY KNOW THE MEASURE OF THE THREE
ANGLES! 





Remember
back on the first page where we had a tiny triangle 


and a great big triangle
that both had the same angles? 





I promised some shortcuts
on this page, so let's get to it. 


First we need a triangle with the sides and angles labeled, like this: 








The angles
are labeled with capital letters (A,B,C) 


and the sides are
labeled with small letters (a,b,c). 


Also, the labeling is
arranged so that a letter used for a side 


is across from the
same letter used for an angle. 





OK, here's the big trick for this triangle or any other triangle: 





Sin
A 
= 
Sin
B 
= 
Sin
C 



a 
b 
c 






What this
puppy says 


is that the ratio of the sine of any angle 


to the
side opposite that angle 


will be the same number for all sides
and angles in the triangle! 





Example: 


On
the last page, we had this triangle for an example. 





Now, to use the
shortcut: 


Call the side on the right "a" which makes the 55° angle angle A. 


Call the angle on the top "B" and the side on the bottom
"b." 


Call the 45° angle at bottom right
"C" and the side across from it "c." 











So, using our shortcut (we don' need all three terms at the same time): 











Substituting the things we know we get: 











Now we
need to solve this for A, 


first multiply both sides by 5a and
cancel where we can: 





5a(Sin
55°) 
= 
5a(Sin
45°) 


a 
5 






5(Sin 55°)
= a(Sin 45°) 





Now divide
both sides by Sin 45 to get the a by itself. 


(Math types say
isolate the a, but saying that might hurt the a's feelings). 


Then we'll cancel where we can. 





5(Sin
55°) 
= 
a(Sin
45°) 


Sin
45° 
Sin
45° 












The stuff
on the left is all just numbers (Math types say constants). 


We
can use a calculator or table to find the values of the sines,
then simplify: 











5.793
= a 





So let's
put that info into our triangle: 











Now we get angle B from the "sum of the angles equals 180°" thing 





180° = 55°
+ 45° + B 


B = 180° 
55°  45° 


B = 80° 











Now we just need to find side
b. We can use the sine ratio trick: 

















Multiply both sides by 5
b and cancel where we can: 





5b(Sin
80°) 
= 
5b(Sin
45°) 


b 
5 






Simplify: 





5(Sin 80°) = b(Sin 45°) 





Divide both sides by Sin 45° and cancel where we can: 





5(Sin
80°) 
= 
b(Sin
45°) 


(Sin
45°) 
(Sin
45°) 






Simplify,
look up the values of the sines, multiply and divide to get b: 











6.964 = b 


Put that
info into the triangle: 








This is exactly
the same answer we got on the last page, 


but this time we didn't have to
cut the triangle 


into two right triangles or anything. 





This little trick with the sine is called the law of sines. 


Yeah, real imaginative those math types, eh? 





Some of you real math fans might be interested in seeing 


where we got this
" law of sines" shortcut trick. 


If not, go have a snack, we'll tell you when to come back. 





OK, if
anyone's still around, here it is: 





We start with
a triangle labeled like before: 








Now we draw the line
cutting our triangle into 2 right triangles 


like we did
on the last page. 


Call the line h. 











From the definition
of the sine, we know that: 





Sine
A 
= 
h 
and 
Sine
B 
= 
h 


b 
a 






Solve each
of these for h: 





h = b x
SineA and h = a x
Sine B 





Since the
h in one of these is the same h as it is in the other one, 


we
can put them together. 





b x
SineA = h = a x
Sine B 





We really
don't even need the h in there. So we can write it as: 





b x
SineA = a x
Sine B 





Now divide
both sides by a x b, cancel and simplify: 





b
x
SineA

= 
a
x
Sine B



a
x
b

a
x
b













There was
nothing special about the side we used when we split the
triangle, 


we could have split it like this: 











Rotate the
triangle so it's easier to see what we have: 











And go
through the same steps as before: 





Sine
C 
= 
h 
and 
Sine
B 
= 
h 


b 
c 






Solve each
of these for h: 





h = b x
Sine C and h = c x
Sine B 





Since the
h in one of these is the same h as it is in the other one, 


we
can put them together. 





c x
Sine B = h = b x
Sine C 





We really
don't even need the h in there. So we can write it as: 





c x
Sine B = b x
Sine C 





Now divide
both sides by b x c, cancel and simplify: 





c
x
Sine B 
= 
b
x
Sine C 


b
x
c

b
x
c 












The Sine
^{B}/b in one of these is the same values as it is in the other one 


so we can put the two equations together. 





Sin
A 
= 
Sin
B 
= 
Sin
C 



a 
b 
c 






And there
you are. 





One last
thought on this. Can you see that since: 





Sin
A 
= 
Sin
B 
= 
Sin
C 



a 
b 
c 






It is also
true that: 





a 
= 
b 
= 
c 



Sin
A 
Sin
B 
Sin
C 






OK,
PROOF IS OVER, YOU CAN COME BACK NOW. 





REMEMBER: 


For this trick to work, we need to know the length of at least one side, 


and he measure of at least two angles. 


If we only know the length of the three sides (and none of the angles), 


or if we only know the measure of one of the angles 


and the length of two of the sides, we're stuck. 





We would
need another trick to solve triangles like these: 











WOW! Does
that sound like a lead in to the next page or what? 





copyright 2008 Bruce Kirkpatrick

