Calculator for finding the Area of a circle
   
 There are six pieces of information to know about a triangle:
Related Chapters
   
 
 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):
 
Sin A = Sin C


a c
 
 Substituting the things we know we get:
 
Sin 55° = Sin 45°


a 5
 
 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°
 
5(Sin 55°) = a

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 x 0.8192 = a

0.7071

 

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:
 
Sin B = Sin C


b c
 
Sin 80° = Sin 45°


b 5
 
 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:
 
5 x 0.9848 = b

0.7071
 

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
 
SineA = Sine B


a 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
 
Sine B = Sine C


b 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