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Triangle geometry nastiness

  • Thread starter steppenwolf
  • Start date

steppenwolf

i might just be stupid and blind to a really obvious answer, but this just stopped me dead in our end of unit geometry test today:

prove that in a triangle with two unequal sides the angle opposite the shorter side will be smaller then angle opposite the larger side.

help! so geometry isn't my forté but what don't i see? even just a little hint would be welcome, i'm sure this won't be a challenge for most of you!
 

HallsofIvy

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Here's the best I could come up with off the top of my head- it seems to me to be a bit awkward.


Call the angles (and vertices) A, B, C and suppose the side opposite A is shorter than the side opposite B. Striking an arc with center at C and radius the length of side BC strikes side AC inside the triangle ABC (BECAUSE BC is shorter than AC). Call this point D. Connecting BD gives isosceles triangle BCD. Call the base angles (in other words not angle C) of BCD "theta". The line BD divides angle B into two angles, one of which is theta. Call the other angle at B, "gamma". Then we have theta= A+ gamma and B= theta + gamma.
Those give B= A+ 2gamma. Since the measures of the angles are all positive, B> A.
 
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Let a < b, then we must prove [alpha] < [beta].
Let h be the height of the triangle with respect to c.
Then, h = b * sin [alpha] = a * sin [beta].
So, a/b = sin [alpha] / sin [beta].
If both [alpha] and [beta] are <=90°, then we're done.
If not, then it's obvious that only one of them can be >90°. I think it's easy to prove that this can only be true for [beta].
 

steppenwolf

hallsofivy that is just beautiful! thankyou so much, even if i lost 12% of my mark i have still learnt something. thanks also arcnet, but unfortunately i don't think we were permitted to use trig at all, only axioms and very elementary theorems. i am frustrated as i used the external angle theorem to get through the vast majority of questions but didn't see its use here. thanks again!
 

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