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November 29th, 2012, 11:05 AM   #1
 
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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.
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November 30th, 2012, 01:38 PM   #2
 
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Re: a problem of triangle

Ok. AM=MB=NA=NM and let them be x . So BN=x.sqrt(3) . Let angle <ACM be ▀. Then <MKB=90-▀ degrees. Use law of sines in triangle MKB and find cos▀ then find CN, it should probably be 3x, just calculated in my mind. So AB is 2x and AC is 4x then find CB using law of cosines,
If you know all sides and one angle use law of sines and find other angles..
I think CK is 2, you mistyped it... Check it again.
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November 30th, 2012, 02:32 PM   #3
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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
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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.
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December 12th, 2012, 12:13 PM   #5
 
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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
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December 12th, 2012, 07:40 PM   #6
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Re: a problem of triangle

C'est quoi le probleme, Augustin?
You knew the answer?
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