h
p..b....f...v....m...w
t...d....th.th..l.....(r)
(t..d)..s...z....n.....r
(t..d)..sh.zh.(n)...(r)
k...g.............ng...E
A syllable is a part of speech an exhalation tapering in time at beginning and end because of obstruction from the mouth. There are five different types of sounds.
The vowels made by the tongue when it is more distant from the roof of the mouth (short a,e,i,o,u, oo in soot, and ou in ought) cannot be placed together in the same syllable. Perhaps this is because when making these sounds the tongue has no reference point, other than itself. These vowels also seem to have the greatest variety of spellings, and differ greatly from accent to accent.
The vowels w, r, and long E, can be placed next to anything. when placed with each other, or the other vowels, they create diphthongs, during which the mouth begins in one vowels position, and then slides into another. Some claim that these diphthongs are each their own sound, to which opinion I point ought that the mouth must change position while it's making them. Also, if the long A, I, and O are considered too be single sounds, the similarly composed aw (shout) and uE (like) would have to be included as well.
There are four consonants in English made by a tone (resonant) escaping the nose while the mouth tries to stop escaping air. These are l, m, n, and ng. For convenience sake I'll call them nasals.
There are eight consonants in English made by the mouth forcing air through a small aperture. They come in pairs - resonant (with a tone) and non-resonant. For convenience sake I'll call them continuants.
There are six English consonants which cannot be held. These also come in resonant and non-resonant pairs. For convenience sake I'll call them stops
These are by no means all the consonants that the mouth can make. The German 'ch' is the most conspicuous because of the gap between g and ng.
There are consonants which, like the vowels, when combined are often mistaken to be one sound. This happens whenever a stop immediately precedes a continuant made with the mouth in the same position (in the same row at the top). This is why various languages have used one letter (or a misleading combination of letters) to signify t-s, t-sh (ch), d-z, and d-zh (j).
It is difficult for the mouth to stop giving consonants tones once it starts, or start giving them tones when it hasn't been. This is why when English words become plural, the pronunciation of the written s depends on the preceding consonant. (In Latin, the z sound seems to have been so unpopular that when s was pared with n {the resonant with the greatest power to change an s, being a nasal in the same position of the mouth} either the s or the n were usually omitted.)
I saved h for last because I didn't know what to do with it. It might be a consonant, but it doesn't require parts of the mouth to touch. It could be a vowel, but it's not said with a tone. I guess it's neither.
Thursday, October 6, 2011
Thursday, July 28, 2011
Law of Tangents
A way of directly proving the law of tangents without using circles.
Given:
1. Triangle ABC, with sides AB, AC, and opposite angles ACB, ABC.
2. Euclid.
3. The tangent of an angle is the ratio of the opposite side over the shorter adjacent side when that angle is part of a right triangle
To prove that the sum of two sides times the tangent of half of the difference between the angles opposite equals the difference between the two sides times the tangent of half of the sum of the angles opposite.
That is, that (AC-AB) x tangent((ABC+ACB)/2)=(AC+AB) x tangent((ABD-ACB)/2)
Take the two sides, such as may be AB and AC.
If the sides are equal
The difference between AB and BC is nothing.
In a triangle angles are to each other as their opposite sides.
Therefore the difference between angles ABC and ACB is nothing.
Nothing times something equals nothing.
According to the terms in the given, nothing equals nothing.
If the sides are not equal
Let AC>AB.
Cut off from AC line AD equal to AB.
Join BD
Cut BD in half at point E.
Draw AE through to point F on line BC.
AD was made equal to AB.
ED was cut equal to EB.
AE equals AE
Therefore traingles ADE and ABE are similar.
Therefore angles AEB and AED are equal.
Therefore line AF cuts line BD at right angles at E.
Any sums of all angles in a triangle are the same.
ABC+ACB+BAC=ABD+ADB+BAD
BAC is BAD
Therefore ABC+ACB=ABD+ADB
ABD=ADB
Therefore ABC+ACB=2ABD
Therefore ABD=(ABC+ACB)/2
ABC-ABD=ABC-(ABC+ACB)/2=(2ABC-(ABC+ACB))/2=(2ABC-ABC-ACB)/2=(ABC-ACB)/2
But ABC-ABD=EBF
Therefore EBF=(ABC-ACB)/2
AEB and BEF are right angles.
Triangles AEB and BEF are right triangles
Therefore tangent(ABE)=AE:EB
And tangent(EBF)=EF:BE
Cut DC in half at point G
Draw HGI parallel to FEA with point H on line BC
Draw AH parallel to BC
AB+AC:AC-AB :: AD+AC:AC-AD :: AD+AD+DC:AD+DC-AD :: 2AD+DC:DC
2AD+DC:DC :: AD+DC/2:DC/2 :: AD+DG:GC :: AG:GC
AI and HC are parallel
Therefore AG:GC :: IG:GH
But DG=half DC
And DE=half DB
Therefore EG is parallel to BC
Therefore AI, EG, FH are parallel
And IH, AF are parallel
Therefore IG:GH=AE:EH
With all these discovered we can say:
Tangent((ABC+ABD)/2):tangent((ABC-ABD)/2)=tangent(ABE):tangent(EBF) :: (AE:EB):(EF:EB) :: AE:EF
And AE:EF=IG:GH :: AG:GC :: AB+AC:AC-AB.
Therefore tangent((ABC+ABD)/2):tangent((ABC-ABD)/2) :: AB+AC:AC-AB
Therefore tangent((ABC+ABD)/2)(AC-AB)=tangent((ABC-ABD)/2)(AB+AC)
Which is what was to be proven. Q.E.D.
Given:
1. Triangle ABC, with sides AB, AC, and opposite angles ACB, ABC.
2. Euclid.
3. The tangent of an angle is the ratio of the opposite side over the shorter adjacent side when that angle is part of a right triangle
To prove that the sum of two sides times the tangent of half of the difference between the angles opposite equals the difference between the two sides times the tangent of half of the sum of the angles opposite.
That is, that (AC-AB) x tangent((ABC+ACB)/2)=(AC+AB) x tangent((ABD-ACB)/2)
Take the two sides, such as may be AB and AC.
If the sides are equal
The difference between AB and BC is nothing.
In a triangle angles are to each other as their opposite sides.
Therefore the difference between angles ABC and ACB is nothing.
Nothing times something equals nothing.
According to the terms in the given, nothing equals nothing.
If the sides are not equal
Let AC>AB.
Cut off from AC line AD equal to AB.
Join BD
Cut BD in half at point E.
Draw AE through to point F on line BC.
AD was made equal to AB.
ED was cut equal to EB.
AE equals AE
Therefore traingles ADE and ABE are similar.
Therefore angles AEB and AED are equal.
Therefore line AF cuts line BD at right angles at E.
Any sums of all angles in a triangle are the same.
ABC+ACB+BAC=ABD+ADB+BAD
BAC is BAD
Therefore ABC+ACB=ABD+ADB
ABD=ADB
Therefore ABC+ACB=2ABD
Therefore ABD=(ABC+ACB)/2
ABC-ABD=ABC-(ABC+ACB)/2=(2ABC-(ABC+ACB))/2=(2ABC-ABC-ACB)/2=(ABC-ACB)/2
But ABC-ABD=EBF
Therefore EBF=(ABC-ACB)/2
AEB and BEF are right angles.
Triangles AEB and BEF are right triangles
Therefore tangent(ABE)=AE:EB
And tangent(EBF)=EF:BE
Cut DC in half at point G
Draw HGI parallel to FEA with point H on line BC
Draw AH parallel to BC
AB+AC:AC-AB :: AD+AC:AC-AD :: AD+AD+DC:AD+DC-AD :: 2AD+DC:DC
2AD+DC:DC :: AD+DC/2:DC/2 :: AD+DG:GC :: AG:GC
AI and HC are parallel
Therefore AG:GC :: IG:GH
But DG=half DC
And DE=half DB
Therefore EG is parallel to BC
Therefore AI, EG, FH are parallel
And IH, AF are parallel
Therefore IG:GH=AE:EH
With all these discovered we can say:
Tangent((ABC+ABD)/2):tangent((ABC-ABD)/2)=tangent(ABE):tangent(EBF) :: (AE:EB):(EF:EB) :: AE:EF
And AE:EF=IG:GH :: AG:GC :: AB+AC:AC-AB.
Therefore tangent((ABC+ABD)/2):tangent((ABC-ABD)/2) :: AB+AC:AC-AB
Therefore tangent((ABC+ABD)/2)(AC-AB)=tangent((ABC-ABD)/2)(AB+AC)
Which is what was to be proven. Q.E.D.
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