Special Right Triangles One of the good things about math is that you can recreate it yourself, if you can remember the basics.

So let’s pretend you suddenly have a Special Right Triangles test, but only vaguely remember anything about them.

Special Right Triangles YIKES!

Don’t Panic.

Special Right Triangles Looking carefully, I see there are only two kinds of right triangles here….

Special Right Triangles There is the 30 60-90….

… and the 45 45-90.

Special Right Triangles Let’s deal with this one first...

1 1 And instead of dealing with x, let’s make it easier and have the length of the legs be 1.

Special Right Triangles Ok… so, it’s a right triangle… and the first thing I think of when I see a right triangle is…..

1

THE PYTHAGOREAN THEOREM!

1

Special Right Triangles 1 + 1 = c 2 2 = c 2 c = sqr root 2 …and if I want the hypotenuse, all I have to do is solve 1 2 + 1 2 = c 2 . 1 1

Special Right Triangles …and every triangle that has the same angles as this one will be similar to it… …which means that they will all be dilations of this one… with some zoom factor/ratio that I can call r.

sqr root 2

• r

Special Right Triangles OK… Now I’m ready… bring on the problems.

sqr root 2

• r

Special Right Triangles

sqr root 2

• r This one is 45-45-90.

The length of one leg is 18… which means 18 = 1 • r.

So it’s easy enough to figure out that 18 = r.

And since the hypotenuse is r • sqr root 2…

Special Right Triangles x = 18 • sqr root 2

sqr root 2

• r

Special Right Triangles

sqr root 2

• r NEXT!

This one is also 45-45-90.

In fact, the only difference is that r = 3 • sqr root 2 And since the hypotenuse is r • sqr root 2…

Special Right Triangles x = (3 • sqr root 2) • sqr root 2 x = 3 • (sqr root 2 • sqr root 2) x = 3 • 2 x = 6

sqr root 2

• r

Special Right Triangles

sqr root 2

NEXT!

This one is also 45-45-90.

But we’re given the hypotenuse, instead of a leg!

• r We know the hypotenuse is r • sqr root 2…

Special Right Triangles

sqr root 2

• r 18 = r • sqr root 2 18 • sqr root 2 = r • sqr root 2 • sqr root 2 18 • sqr root 2 = r • 2 9 • sqr root 2 = r … and so does x

Special Right Triangles 60 0 60 0 60 0 Let’s take on the 30-60-90 now.

This one starts off as an equilateral triangle… with all sides equal… and all angles equal to 60 degrees.

Then, we cut it in half.

Special Right Triangles 2 30 0 30 0 2 60 0 60 0 1 1 So now, the two angles at the top are 30 degrees each.

And if the original sides of the equilateral triangle had a length of two, the bottom is cut in half, too!

Special Right Triangles 30 0 2 60 0 1 Now, let’s just look at the half we care about… the 30-60-90 triangle.

Notice that the hypotenuse is twice as long as the side opposite the 30 0 angle.

That’s always going to be true!

Special Right Triangles 1 2 + h 2 = 2 2 1 + h 2 = 4 h 2 = 3 h = sqr root 3

h

30 0 What about the height?

1 2 60 0 This is a job for…..

THE PYTHAGOREAN THEOREM!

a 2 + b 2 = c 2

Special Right Triangles r •

sqr root 3

30 0 2 • r 60 0 1 • r Because every 30-60-90 triangle will be similar to this one… The sides will always be proportional to these sides!

So we are all set to get started.

Special Right Triangles 30 0 r •

sqr root 3

30 0 2 • r 60 0 1 • r The missing angle is 30 0 .

We are given the length of the side opposite that angle, so r = 8.

The hypotenuse, y, is equal to 2r… or 16.

The side across from the 60 0 angle has to be

r • sqr root 3…

so x = 8 • sqr root 3

Special Right Triangles r •

sqr root 3

30 0 2 • r 60 0 1 • r Let’s do another.

Special Right Triangles r •

sqr root 3

30 0 2 • r 60 0 1 • r The hypotenuse, which has to be 2 • r, is equal to 11.

That means r, the side opposite the 30 0 has to be 5.5….

and so x = 5.5.

angle, The side across from the 60 0

sqr root 3…

angle has to be r • so y = 5.5 • sqr root 3

Special Right Triangles r •

sqr root 3

30 0 2 • r 60 0 1 • r Bring on the next one!

Special Right Triangles 30 0 2 • r r •

sqr root 3

30 0 60 0 60 0 1 • r Since this is an isoceles triangle, the other base angle is also 60 0 .

And the half-angle on the right is 30 0 .

And we can focus on just the part we care about!

Special Right Triangles 30 0 2 • r r •

sqr root 3

30 0 60 0 60 0 1 • r The hypotenuse, which has to be 2 • r, is equal to 20.

That means r, the side opposite the 30 0 has to be 10….

and so y = 10.

angle, The side across from the 60 0

sqr root 3…

angle has to be r • so x = 10 • sqr root 3

Special Right Triangles And, finally….

r •

sqr root 3

30 0 2 • r 60 0 1 • r

Special Right Triangles r •

sqr root 3

30 0 2 • r 60 0 1 • r This time, we are given the length of the side opposite the 60 0 angle, which has to be r • sqr

root 3.

If 12 = r • sqr root 3… 12

• sqr root 3

= (r • sqr root 3)

• sqr root 3

12 • sqr root 3 = r • (sqr root 3 • sqr root 3) 12 • sqr root 3 = r • 3 4 • sqr root 3 = r

Special Right Triangles r •

sqr root 3

30 0 2 • r Since r = 4 • sqr root 3… 60 0 1 • r and that is the side opposite the 30 0 angle… x = 4 • sqr root 3

Special Right Triangles r •

sqr root 3

30 0 2 • r 60 0 1 • r And, again, since r = 4 • sqr root 3… and the hypotenuse (y) has to be twice as long… y = 8 • sqr root 3

Special Right Triangles

sqr root 2

30 0 2 • r r •

sqr root 3

1 • r 1 • r 60 0 1 • r So, if you ever have to answer questions about Special Right Triangles, now you know that you can create the “formulas” from scratch, just by using the Pythagorean Theorem.