how do you prove that in a triangle, the greater angle has the greater side opposite to it? i know how to prove that the greater side has the greater angle opposite to it, but can someone please tell me the converse of this theorum?
2006-09-04 02:18:33 · 7 answers · asked by Smithereenian 2 in Science & Mathematics ➔ Mathematics
you can't work the same proof backwords, i already tried that. it all comes down to the same thing that one angle is greater than the other.
2006-09-04 02:29:06 · update #1
suppose the triangle is ABC. A is the tip, BC is the base. we know that angleB is greater than angleC, how would you prove that side AC is greater than side AB?
2006-09-04 02:30:39 · update #2
please give me simple answers, i'm only in ninth grade.
2006-09-04 02:31:21 · update #3
wel, sweetie, this is exactly how i worked out that the greater angle has the greater side opposite to it, thanks a lot for your answer, it's the most logical; but how would you prove that the greater angle has the greater side opposite? by your proof, we can prove that angleB is greater than angleC because AC is greater than AB. but how would you prove the converse? that is not something i can get.
2006-09-04 03:09:04 · update #4
well dear i hope it helps you,
In any triangle the angle opposite the greater side is greater.
Let ABC be a triangle having the side AC greater than AB.
I say that the angle ABC is also greater than the angle BCA.
http://www.largeimagehost.com/image/upload-image/free-image-host/2057/tringle.PNG.html
Since AC is greater than AB, make AD equal to AB, and join BD.
Since the angle ADB is an exterior angle of the triangle BCD, therefore it is greater than the interior and opposite angle DCB.
But the angle ADB equals the angle ABD, since the side AB equals AD, therefore the angle ABD is also greater than the angle ACB. Therefore the angle ABC is much greater than the angle ACB.
Therefore in any triangle the angle opposite the greater side is greater.
Good Luck.
2006-09-04 02:39:52 · answer #1 · answered by sweetie 5 · 2⤊ 0⤋
Well, here's a simple proof I can come up with using the Sine Law.
Proof by contraction: If A is the greatest angle in a triangle, then without any loss of generality, b is the longest side b>a.
a/sinA = b/sinB
bsinA = asinB
From here, there are two cases: A<90 and A>90 (everything is in degrees):
CASE I:
A<90
Since A is the biggest side, B
sinB < sinA
Also, it is given that:
a < b
Since sinB, sinA, a and b are all positive, and b and sinA rest on one side of the equality
asinB = bsinA
We arrive at:
asinB < bsinA
A contradition. Therefore, b cannot be greater than a for the case A<90.
-----
CASE II (A=90 also falls into this category but I didn't include it in the case description because I don't have the approriate key on my keyboard):
A>90
Since A resides in the second quadrant 90
sinA = sin(180-A)
By properties of reference angles.
Since A is greater than 90, 180-A is less than 90, therefore B is acute since B<180-A. Since both B and 180-A reside in the first quadrant [0,90], we can take the sine of both sides:
sinB < sin(180-A)
And substitute in the equation sin(180-A) = A, so:
sinA > sinB
From here, it follows like the first case such that:
b > a
sinA > sinB
asinB = bsinA
but according to the two equalities:
asinB < bsinA
And therein lies the contradiction.
Since, A must be in the range (0,180), we have proven that a>b for all triangles in which A is the greatest angle. Q.E.D.
Hope this helps!
2006-09-04 06:40:12 · answer #2 · answered by Steven X 2 · 1⤊ 0⤋
You can prove it by contradiction.
Say that if A is the greatest angle a is NOT the greatest side.
So one of the other sides must be it, say b.
This implies that B must be the greatest angle, but u already know that A is.
This is a contradiction.
So our assumption is incoorect. So a IS the greatest side
2006-09-04 03:26:35 · answer #3 · answered by the_answerer 2 · 0⤊ 0⤋
well you can solve it backwards.
let's say:
You say we think that 'a' is not the longest, so for example 'b' is.then you can prove that 'B' is the biggest angle that is not acceptable because 'A' is the biggest. when we face this paradox,it means that we thought wrong and 'a' is the longest
(ther is also a second condition that two sides and/or angles are the same,in that case we have a triangle that is much more easier to fight)
2006-09-04 02:32:25 · answer #4 · answered by Farshad Gh 2 · 0⤊ 1⤋
(sinA/a)=(sinB/b)=(sinC/c)
"A" represent the angle, "a" the side opposite the angle.
no sure is this what you want
2006-09-04 02:27:04 · answer #5 · answered by 黑夜 1 · 0⤊ 0⤋
It's basically the same proof, only worked 'backwards' âº
Doug
2006-09-04 02:26:23 · answer #6 · answered by doug_donaghue 7 · 0⤊ 1⤋
it is too lengthy
2006-09-04 02:50:48 · answer #7 · answered by CHIMPU 2 · 0⤊ 2⤋