Special right triangles
30-60-90 Triangle Side Ratios Proof Proving the ratios between the sides of a 30-60-90 triangle
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- What i want to do in this video is to discuss
- a special class of triangles called '30-60-90
- triangles'. And I think you know why they're
- called this.
- The measures of its angles are,
- 30 degrees, 60 degrees and 90 degrees.
- And what we're going to prove in this video
- and this tends to be a very useful result
- at least for a lot of what you see in a
- geometry class, and then later on in
- Trigonometry class.
- Is the ratio between the sides of a
- 30-60-90 triangle.
- That if the hypotenuse has length x.
- (so that's, remember the hypotenuse is
- opposite the 90 degree side.)
- If the hypotenuse has length x, what we're going
- to prove is that the shorter side which
- is opposite the 30 degree side has length
- x/2, and that the 60 degree side...
- the 60 degree side, or the side that's opposite
- the 60 degree angle I should say is going
- to be the (the square root of 3 x the shorter side.)
- so the square root of 3 times (x/2)
- that's going to be its length.
- So that's what we're going to prove in this video.
- and in other videos, we're going to apply this.
- We're going to show that this is actually a useful
- result.
- Now lets start with a triangle that we're
- very familiar with... so let me draw ourselves
- an equilateral triangle... so drawing the
- triangles is always the hard part.
- So this is my best shot at an equilateral triangle.
- And lets call this A, B, C
- I'm just going to assume I've constructed an
- equilateral triangle,
- so Triangle ABC is equilateral.
- and if it's equilateral, that means all
- of it's sides are equal, and lets say that it has
- an equilateral with sides of length x.
- so this is going to be x
- this is going to be x, and this is
- going to be x.
- we also know, based on what we seen from
- equilateral triangles before, that the measures
- of all of these angles are going to be 60 degrees.
- So this is going to be 60 degrees
- this is going to be 60 degrees, and then
- this is going to be 60 degrees.
- Now what I'm going to do is I'm going to drop
- an altitude from this top point right over here.
- So I'm going to drop an altitude right down
- and by definition when I'm constructing an
- altitude it is going to intersect the
- base right here... at a right angle.
- So that's going to be right angle, and then
- this is going to be a right angle.
- and it's a pretty straightforward proof to show
- that this is not only an altitude,
- that is perpendicular to this base.
- But it's a pretty straightforward proof to show that it
- bisects the base and you can pause it
- if you like and prove it yourself.
- But it really comes out of the fact that it's easy
- to prove that these two triangles are congruent.
- So let me... prove it for you.
- So lets call this point D right over here.
- So triangles ABD and BDC
- they clearly both share this side, so
- this side is common to both of them right over
- here.
- and then this angle right over here is
- congruent to this angle over there.
- This angle right over here is congruent to
- this angle over here.
- and so, if these two and congruent to each other.
- The third angle has to be congruent to each other.
- So this angle right over here needs to be
- congruent to that angle right over there.
- So these two are congruent.
- And so you can use actually a variety of our
- congruence postulates, we could say
- SAS (side, angle side) congruents
- we could use ASA (angle, side angle)
- any of those to show that triangle ABD
- is congruent to triangle CBD.
- and what that does for us.
- and we could use as I said (Angle,Side,Angle)
- or (Side,Angle,Side)
- whatever we like to use for this.
- what that does for us is that it tells us that
- the corresponding sides of these triangles
- are going to be equal.
- in particular... in particular,
- the length AD is going to be equal to CD
- these are corresponding sides.
- so these are going to be equal to each other.
- and if we know that they're equal to each other.
- that add up to x, remember this was an
- equilateral triangle of length x.
- we know that this side right over here is going
- to be x/2... we know this is going to be x/2
- not only do we know that, but also knew when
- we dropped this altitude we showed that
- this angle has to be congruent to that angle,
- and their measures have to add up to 60.
- So if two things are the same, they add up to 60.
- this is going to be 30 degrees,
- and this is going to be 30 degrees.
- so we already shown one of the interesting parts
- of 30-60-90 triangles.
- That if the hypotenuse. Notice, and I guess I
- didn't point this out that by dropping this
- altitude I've essentially split this equilateral triangle.
- into two 30-60-90 triangles.
- 30-60-90 triangles.
- and so we're already shown that if the side
- opposite the 90 degree side is x.
- That the side opposite the 30 side is going to be
- x/2. That's what we showed right over here.
- Now we just have to come up with a third side,
- the side that is opposite the 60 degree side.
- the side opposite the 60 degree side...
- right over there.
- and lets call that length... well I'll just use
- the letters we already have.
- This is BD.
- now we can just the Pythagorean Theorem
- right here.
- BD^2 + this length right over here ^2
- + (x/2)^2 is going to be equal to the
- hypotenuse squared.
- so we get BD... BD^2 + (x/2)^2
- this is just straight of the Pythagorean Theorem.
- + (x/2)^2 is going to this hypotenuse ^2
- it's going to equal x^2
- and just to be clear... I'm looking at this
- triangle right here...
- I'm looking at this triangle over here in the right.
- and I'm just applying the Pythagorean Theorem.
- this side ^2 + this side ^2 is going to equal
- the hypotenuse ^2.
- The lets solve now for BD
- you get BD^2 + (x^2)/4... (x^2/4)
- is equal to x^2... is equal to x^2.
- You would view this as (4x^2)/4 that's the same
- obviously it's x^2.
- And then if we subtract 1... if you subtract
- (1/4)x^2 from both sides or (x^2)/4
- from both sides. you get
- BD^2 =
- 4x^2/4 - x^2/4 is going to be.
- 3x^2/4
- so its going to be 3x^2/4
- take the principle root of both sides.
- you get BD is equal to... is equal to
- square root(3 times x)... times x
- the principle of 3 is the square root of 3
- the principle root of x^2 is just x
- over the principle root of 4 which is 2
- and BD is a side opposite the 60 degree side
- so we're done, if this hypotenuse is x,
- the side opposite the 30 side is going to be
- x/2
- and the side opposite the 60 degree side
- is going to be the
- square root of (3x/2)
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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