Show that in any triangle ABC, if AC > AB then angle ABC > angle ACB.
I've started trying to do this... But I'm stuck.. I made a new point B' on AC so that AB = AB', but then I'm not sure what to do next... Help!
2007-02-01 21:11:10 · 7 answers · asked by yogastar02 2 in Science & Mathematics ➔ Mathematics
I don't think using the law of sines is exactly what they have in mind, this is a geometry proof, not a trig proof.
Very good, your first step is a good one. Look at the triangle AB'C. It has 3 angles a, b, c, and you know a + b + c = 180.
You also know, since by construction AB = AB', then angle AB'B = angls ABC. AB'B is supplemental to AB'C, so AB'B + AB'C = 180, and since ABC = AB'B, you know
a + (180 - ABC) + c = 180
a + c = ABC
and angle a (= B'AC) > 0, so c < ABC
2007-02-01 23:20:41 · answer #1 · answered by sofarsogood 5 · 0⤊ 0⤋
Use the Law of Sines.
The way you have labelled your triangle we have:
BC/sinA = AC/sinB = AB/sinC
sinB = (AC/AB)sinC > sinC
Therefore angle ABC > angle ACB.
2007-02-01 21:31:16 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋
By sine rule:
b/sin B = c/sin C and assume b > c
b sin C = c sin B
(b/c) sin C = sin B
But b/c is greater than 1 because b > c
Therefore sin C < sin B
sin B > sin C
angle B > angle C
angle ABC > angle ACB as required
2007-02-01 21:47:40 · answer #3 · answered by Como 7 · 0⤊ 0⤋
I in basic terms hate them while i are not getting them. Kinda accepted, appropriate? thankfully, we had a curriculum and the instructor develop into no longer allowed to coach proofs all 3 hundred and sixty 5 days around and not something. via the way, is Geometry a mandatory direction on your college? because of the fact in mine that is not any longer. So technically I chosen to go through, and that made issues greater effectual.
2016-11-02 02:58:55 · answer #4 · answered by ? 4 · 0⤊ 0⤋
assume the statement "if AC>AB then angle ABC> angle ACB" is true. then sin(ABC-ACB) >0 (positive)
A' is the projection of A to BC
angle ABC and angle ACB are acute angles
sinABC=AA'/AB
sinACB=AA'/AC
since, sin^2 x + cos^2 x = 1
cosABC= sqr(AB^2-AA'^2)/AB
cosACB= sqr(AC^2-AA'^2)/AC
sin(ABC-ACB)
= sinABC cosACB - cosABC sinACB
= (AA'*sqr(AC^2-AA'^2))/(AC*AB) - (sqr(AB^2-AA'^2)*AA')/(AB*AC)
= (AA'*sqr(AC^2-AA'^2) - sqr(AB^2-AA'^2)*AA)/(AB*AC)
since AC>AB, sqr(AC^2-AA'^2)>sqr(AB^2-AA'^2), which means sin(ABC-ACB)>0
therefore, it is proved that angle ABC >angle ACB
2007-02-01 22:06:39 · answer #5 · answered by 99%cocoa 1 · 0⤊ 0⤋
Please read and study theorem which said that
"in a triangle big side of triangle is always in front of big angle of a triangle."
2007-02-01 21:31:37 · answer #6 · answered by Mritunjay 2 · 1⤊ 1⤋
I hated too , but really they are very interesting when u understand them. U need guidance.
Look for someone who can guide you....
2007-02-01 21:37:31 · answer #7 · answered by @rrsu 4 · 0⤊ 0⤋