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30-60-90 and 45-45-90 Triangles

Date: 03/15/99 at 21:02:01
From: Kristina
Subject: 30-60-90 Triangles

Please help me. If I have a triangle that is 30-60-90, or a triangle 
that is 45-45-90, how do I find all the sides when given only one side? 
For example, if I'm given the base of 9, how do I find the hypotenuse 
and adjacent side?

Date: 03/16/99 at 11:13:33
From: Doctor Rick
Subject: Re: 30-60-90 Triangles

Hi, Kristina, thanks for your question!

The 45-45-90 triangle is the easiest to see. Because 2 angles are 
equal, it is isosceles. Both legs have the same length. If you know one 
leg (a), you know both legs, and then you can use the Pythagorean 
Theorem to find the hypotenuse c.

              /  |
         c  /    |
          /      | a
        /        |
      /          |

  c = sqrt(a^2 + a^2)
    = sqrt(2*a^2)
    = sqrt(2)*a

So these are the proportions of the sides of a 45-45-90 triangle:

              /  |
   sqrt(2)  /    |
          /      | 1
        /        |
      /          |

In other words, the ratio of the hypotenuse to either leg is sqrt(2):1. 
If you know the length of either the legs or the hypotenuse, c, you can 
use the proportions above to calculate the other. For instance (as you 
asked), if a = 9, then

  c/a = sqrt(2)

  c = a*sqrt(2)
    = 9*sqrt(2)

Or, if c = 3, then

  a/c = 1/sqrt(2)

  a = c/sqrt(2)
    = 3/sqrt(2)
    = (3/2)sqrt(2)

If that last line confuses you, you'll be doing this sort of thing a 
lot, so let's go over it. We like to move square roots to the 
numerator. The way to do this is to multiply both numerator and 
denominator by the square root:

     1         sqrt(2)        sqrt(2)
  ------- = --------------- = -------
  sqrt(2)   sqrt(2)*sqrt(2)      2

since if we square the square root of a number, we get the number back.

Before we move on to the next triangle, I want to take the time to 
introduce trigonometry to you. If you don't understand it right now, 
that's okay. But keep in mind that you'll be seeing more of it in the 
coming years! I start first with a diagram of a right triangle:

     |  \
     |    \  c
   a |      \
     |        \
     |        A \

I've labeled the sides a, b, and c, and the angle across from side a 
as angle A. We use trigonometry as a way to express the ratios of the 
sides. Let's rename the sides in relation to angle A. For example, 
side a is opposite angle A, so we'll call it opposite. Then we have:

            |  \
            |    \   hypotenuse
   opposite |      \
            |        \
            |        A \

Then we call the ratio of opposite/hypotenuse the sine of A, sin(A), 
the ratio of adjacent/hypotenuse the cosine of A, cos(A), and the ratio 
of opposite/adjacent the tangent of A, tan(A). Remember when we talked 
about ratios above? We could have just said that:

   sin(45) = opposite/hypotenuse = a/c = 1/sqrt(2)
   cos(45) = adjacent/hypotenuse = a/c = 1/sqrt(2)

Now for the 30-60-90 triangle. It isn't isosceles like the 45-45-90
triangle, but it is half of an equilateral triangle:

          / | \
       c /  |  \ c
        /   |   \
       /   a|    \
      /     |     \
        b      b

The angles of the equilateral triangle are 60 degrees. The top angle is 
bisected, giving the 30 degree angle. And since the sides of the 
equilateral triangle are equal, 2b = c. That's the key! The side 
opposite the 30 degree angle is half the hypotenuse.

If we know b, then we know c = 2b, and we can fill in the Pythagorean

  (2b)^2 = a^2 + b^2

  a = sqrt((2b)^2 - b^2)
    = sqrt(3b^2)
    = sqrt(3)*b

Now we have the proportions of the sides of a 30-60-90 triangle:

          / |
         /  |
      2 /   |
       /    | sqrt(3)
      /     |

If you know any side, you can find the others by ratios. For instance, 
if you know b = 9, then

  a/b = sqrt(3)/1

    a = b*sqrt(3)
      = 9*sqrt(3)

  c/b = 2

    c = 2b
      = 18

If you know a = 4, then

  b/a = 1/sqrt(3)

    b = a/sqrt(3)
      = 4/sqrt(3)
      = (4/3)*sqrt(2)

  c/a = 2/sqrt(3)

    c = 2a/sqrt(3)
      = 8/sqrt(3)
      = (8/3)*sqrt(3)

Can you express these ratios in the trigonometric form introduced 

The main thing to get out of this is to memorize those two proportion
figures (or, if you're like me, remember how to derive them from
Pythagoras). Then if you are comfortable with the use of ratios, you 
can find all you need to know about either of these special triangles.

- Doctor Rick, The Math Forum   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry
Middle School Geometry
Middle School Triangles and Other Polygons

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