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Prove the following: if, in a triangle ABC, BA =CB
O in atriangle
< BAO = < OAC = 20°,
< ACO = 10°,
and < OCB = 30°,
then< OBC = 80°.
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This was trickier than I thought.
Last edited by JaneFairfax (2007-08-06 22:48:14)
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What? ......
That tough?
X'(y-Xβ)=0
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There may be a simpler way; if there is, then I havent been able to spot it. (For example, it turns out that OA = OB. Maybe there is a straightforward way to deduce this but Im totally blind at the moment. )
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I know where you got this problem ^^
It's really nice, of course with a synthetic solution.
Not hard to find.
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