- #1
Deviousfred
- 19
- 1
I was asked by one of my professors to prove for the class why the largest side of the tringle corresponds to the largest angle of that triangle.
I was thinking of using law of sines to do so. Please let me know if I am wrong so that I do not look like a fool in front of class.
My triangle is ABC with vertices/angles a,b, and c respectively.
I am saying that angle a is the largest in that triangle, so;
Quick notes: L is my representation for angle.
mLa > mLb > mLc.
Law of Sines:
sin(a)/BC = sin(b)/AC = sin(c)/AB
therefore;
AC*sin(a)=BC*sin(b),
AC*sin(c)=AB*sin(b),
AB*sin(a)=BC*sin(c).
so;
AC/BC = sin(b)/sin(a),
AC/AB = sin(b)/sin(c),
AB/BC = sin(c)/sin(a).
once again;
mLa > mLb > mLc.
therefore;
sin(a) > sin(b) > sin(c)
so if;
AC/BC = sin(b)/sin(a),
AC/AB = sin(b)/sin(c),
AB/BC = sin(c)/sin(a).
then;
AC < BC
AC > AB
AB < BC
so;
BC > AC > AB
I don't know if I need to write anything more down from here. Take it easy on my, I threw this together last night while ignoring my mother-in-law's boring story.
I was thinking of using law of sines to do so. Please let me know if I am wrong so that I do not look like a fool in front of class.
My triangle is ABC with vertices/angles a,b, and c respectively.
I am saying that angle a is the largest in that triangle, so;
Quick notes: L is my representation for angle.
mLa > mLb > mLc.
Law of Sines:
sin(a)/BC = sin(b)/AC = sin(c)/AB
therefore;
AC*sin(a)=BC*sin(b),
AC*sin(c)=AB*sin(b),
AB*sin(a)=BC*sin(c).
so;
AC/BC = sin(b)/sin(a),
AC/AB = sin(b)/sin(c),
AB/BC = sin(c)/sin(a).
once again;
mLa > mLb > mLc.
therefore;
sin(a) > sin(b) > sin(c)
so if;
AC/BC = sin(b)/sin(a),
AC/AB = sin(b)/sin(c),
AB/BC = sin(c)/sin(a).
then;
AC < BC
AC > AB
AB < BC
so;
BC > AC > AB
I don't know if I need to write anything more down from here. Take it easy on my, I threw this together last night while ignoring my mother-in-law's boring story.