My Math Forum a problem of triangle

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 November 29th, 2012, 11:05 AM #1 Joined: Nov 2012 Posts: 30 Thanks: 0 a problem of triangle Suppose in triangle ABC : angle of BAC is 60 degrees. If K is intersection point of CM median and BN altitude, also suppose KM=1 cm and CK=6 cm, calculate angles of triangle ABC.
 November 30th, 2012, 01:38 PM #2 Joined: Nov 2012 Posts: 50 Thanks: 0 Re: a problem of triangle Ok. AM=MB=NA=NM and let them be x . So BN=x.sqrt(3) . Let angle
 November 30th, 2012, 02:32 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 3,084 Thanks: 36 Re: a problem of triangle Changing the labelling a bit (I prefer AB > AC > BC): In triangle ABC : angle ABC = 60 degrees. K is intersection point of median AM and altitude CN. KM=1 and AK=6. Calculate angles of triangle ABC. Code: C M K B N A Let a = BC, b = AC, c = AB, m = AM, n = CN. Triangle BCN is a 30-60-90 triangle. Triangle BMN is equilateral (sides = a/2); triangle CMN is isosceles (CM = MN = a/2). Now get to work
December 3rd, 2012, 07:33 AM   #4
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Quote:
 Originally Posted by etidhor I think CK is 2, you mistyped it...
I haven't had time to complete the solution, but I am confident that the problem was stated correctly. I found it (without solution) on a website dated 2003.

 December 12th, 2012, 12:13 PM #5 Joined: Nov 2012 Posts: 30 Thanks: 0 Re: a problem of triangle at first: we suppose BN=x+y,and AC=z,NC=w, BM=a and angle ACM=F we have in triangle BKM by cos law: 1=a^2+y^2-sqrt(3)ay then y=(sqrt(3)a +(or -)sqrt(4-a^2))/2 y>x then y=(sqrt(3)a + sqrt(4-a^2))/2. in (right angle)triangle ABN :sin60=(y+x)/2a then y+x=sqrt(3)a (****) so x=(sqrt(3)a -sqrt(4-a^2))/2 (***) we have in triangle AMC by sin law : sinA/7=sinF/a then sinF=sqrt(3)a/14 (*) in (right angle) triangle NKC: sinF=x/6(**) then by(**) and(*) and (***) we have : (a*sqrt(3)-sqrt(4-a^2))/2*6=a*sqrt(3)/14 so 196-49a^2=3a^2 then a=7/sqrt(13)(******) we have in triangle AMC by cos law: 49=a^2+z^2-az so z^2-az+z^2-49=0 then z=(a+sqrt(196-3a^2))/2. ( z=(a-sqrt(196-3a^2))/2 is not acceptable because A=60 in (right angle)triangle ABN then cos60 =(z-w)/2a then z=a+w). so w=(sqrt(196-3a^2)-a)/2 (*****). in (right angle) triangle BNC : tanC=(x+y)/w then by(*****)and(****) : 2sqrt(3)a/(sqrt(196-3a^2)-a) so by (******) : tanC=sqrt(3)/3 so C=30 and B=90
 December 12th, 2012, 07:40 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 3,084 Thanks: 36 Re: a problem of triangle C'est quoi le probleme, Augustin? You knew the answer?

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