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Familiar Triangles


Date: 01/19/99 at 21:56:41
From: Noel
Subject: Integrated math 2b

I need help figuring out special right triangles. I've tried to start 
with a and b but I get confused. I don't know the next step.    
                 
For example one problem is:

Find a and b. Simplify radicals whenever possible.   

     | . 
     |60 .
     |     .
     |       .                     
     |         .14
     |           .
   a |             . 
     |               .
     |____________________. 
          
               b

I need help with 45-45-90 triangles and 30-60-90 triangles. If you have 
any available notes to show me step by step, I'll be very grateful. 
Thank you.


Date: 01/20/99 at 12:35:48
From: Doctor Peterson
Subject: Re: Integrated math 2b

Hi, Noel. Thanks for a carefully explained question!

The two special kinds of triangles are special because two sides are 
related in a simple way. For the 45-45-90 triangle,

    +
    | \
    |45 \
    |     \
   A|       \ C
    |         \
    |           \
    |_            \
    | |           45\
    +-----------------+
             A

the two sides opposite the 45-degree angles are equal. You can get the 
other side from the Pythagorean Theorem:

    C = sqrt(A^2 + A^2) = sqrt(2A^2) = A * sqrt(2)

That is, the hypotenuse is the square root of 2 times the other sides.

For the 30-60-90 triangle, the important thing is that it's exactly 
half of an equilateral triangle:

    +                        +
    |\                      /|\
    | \                    /6|0\
    |30\                  /  |  \
    |   \A              A/   |   \A
   B|    \              /    |    \
    |     \            /     |     \
    |_     \          /      |      \
    | |   60\        /60     |     60\
    +--------+      +--------+--------+
        A/2                  A

That means that the side opposite the 30-degree angle is half the 
length of the hypotenuse. Again, you can get the length of the third 
side by the Pythagorean Theorem:

    B = sqrt(A^2 - (A/2)^2) = sqrt(A^2 - (A^2)/4)
      = sqrt((3/4)A^2) = A * sqrt(3)/2

So in your problem, a is half of 14, or 7 (just imagine completing the 
equilateral triangle by reflecting the triangle in side b if this 
isn't clear.) You can either use the Pythagorean Theorem directly on 
these numbers, or multiply 7 by the square root of 3 from the formula 
above to get the answer.

If you prefer, just try memorizing these pictures:

          +
          | \
          |45 \
          |     \
         1|       \ sqrt(2)
          |         \
          |           \
          |_            \
          | |           45\
          +-----------------+
                   1

          +                        +
          |\                      /|\
          | \                    /6|0\
          |30\                  /  |  \
          |   \ 2            2 /   |   \ 2
   sqrt(3)|    \              /    |    \
          |     \            /     |     \
          |_     \          /      |      \
          | |   60\        /60     |     60\
          +--------+      +--------+--------+
              1                    2


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
Middle School Geometry
Middle School Triangles and Other Polygons

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