SOLUTION: how do i do the 45 45 90 and the 30 60 90 triangles

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Question 255679: how do i do the 45 45 90 and the 30 60 90 triangles
Found 3 solutions by Alan3354, CharlesG2, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
It depends on what you want to do with them.

Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
how do i do the 45 45 90 and the 30 60 90 triangles?
pythagorean theorem is for right triangle
(triangles with a 90 degree angle as one of the 3 angles)
a^2+b^2=c^2 where a and b are 2 of the sides and c is the diagonal (longest side) otherwise known as the hypotenuse
45 45 90
take a rectangle and slice it in two at the diagonal and take one of the halves,
that is this triangle
length ^2 + width ^2 = diagonal ^ 2
30 60 90
that is half of an equalateral triangle (a triangle with 3 equal sides)
the short side will be half the base and will be opposite the 30 degree angle
the height will be opposite the 60 degree angle
the hypotenuse will be opposite the 90 degree angle
(1/2 base)^2 + height ^2 = hypotenuse ^2
you can also use these ratios to come up with the sides if you know the angle
sine
opp/hyp
cosine
adj/hyp
tangent
opp/adj
the hypotenuse will always be opposite to the 90 degree angle
these are some of the ratios for these triangles
sine 45 = sqrt(2)/2
sine 90 = 1
sine 30 = 0.5
sine 60 = sqrt(3)/2

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
how do i do the 45 45 90 and the 30 60 90 triangles

For the 45°-45°-90° right triangle

Start with a 1x1 square,



It has four right, or 90° angles.

Now cut it into two congruent right triangles by drawing
a diagonal from the lower left to the upper right, like this:



That diagonal bisects the right angles on the lower left
and upper right corners, each into two 45° angles. Now
take away the top triangle, leaving only the bottom half:



Now you have a 45°-45°-90° right triangle.



Use the Pythagorean theorem to calculate its hypotenuse:

c%5E2=a%5E2%2Bb%5E2
c%5E2=1%5E2%2B1%5E2
c%5E2=1%2B1
c%5E2=2
c=sqrt%282%29

So the length of the hypotenuse is sqrt%282%29



Now memorize the way this right triangle looks and the 
lengths of the three sides.  For you will need it to find 
the exact values for the sine, cosine, and tangent of 45°:

sin%28%2245%B0%22%29=%28opposite%29%2F%28hypotenuse%29=1%2Fsqrt%282%29=sqrt%282%29%2F2, (after rationalizing the denominator)

cos%28%2245%B0%22%29=%28adjacent%29%2F%28hypotenuse%29=1%2Fsqrt%282%29=sqrt%282%29%2F2, (after rationalizing the denominator)

[Notice that the sine and cosine of 45° are the same!]

tan%28%2245%B0%22%29=%28opposite%29%2F%28adjacent%29=1%2F1=1.

--------------------------------------------------------

For the 30°-60°-90° right triangle

Start with an equilateral triangle, each side of
which has length 2,



It has three 60° angles.



Now cut it into two congruent triangles by drawing a median, which
is also an altitude as well as a bisector of the upper 60°-vertex angle:


  
That bisects the upper 60° angle into two 30° angles. It also
bisects the base of the equilateral into two parts, each with
length 1. Now take away the triangle on the right, leaving only 
the one on the left:




Now you have a 30°-60°-90° right triangle.

Use the Pythagorean theorem to calculate its altitude:

c%5E2=a%5E2%2Bb%5E2
2%5E2=1%5E2%2Bb%5E2
4=1%2Bb%5E2
3=b%5E2
sqrt%283%29=b

So the length of the altitude is sqrt%283%29


  
Now memorize the way this right triangle looks and the lengths 
of the three sides.  For you will need it to find the exact
values for the sine, cosine, and tangent of BOTH 60° AND 30°:

sin%28%2260%B0%22%29=%28opposite%29%2F%28hypotenuse%29=sqrt%283%29%2F2,

cos%28%2260%B0%22%29=%28adjacent%29%2F%28hypotenuse%29=1%2F2, 

tan%28%2260%B0%22%29=%28opposite%29%2F%28adjacent%29=sqrt%283%29%2F1=sqrt%283%29.

sin%28%2230%B0%22%29=%28opposite%29%2F%28hypotenuse%29=1%2F2,

cos%28%2230%B0%22%29=%28adjacent%29%2F%28hypotenuse%29=sqrt%283%29%2F2, 

tan%28%2230%B0%22%29=%28opposite%29%2F%28adjacent%29=+1%2Fsqrt%283%29=sqrt%283%29%2F3,
after rationalizing

[Notice that the sine of 60° and the cosine of 30° are the same,
and also that the sine of 30° and the cosine of 60° are the same!]

Edwin