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November 29th, 2012, 11:05 AM  #1 
Member 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 
Member 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 <ACM be ß. Then <MKB=90ß degrees. Use law of sines in triangle MKB and find cosß 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. 
November 30th, 2012, 02:32 PM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 9,076 Thanks: 622  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 Triangle BCN is a 306090 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  
Global Moderator Joined: Dec 2006 Posts: 16,946 Thanks: 1255  Quote:
 
December 12th, 2012, 12:13 PM  #5 
Member 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^2sqrt(3)ay then y=(sqrt(3)a +(or )sqrt(4a^2))/2 y>x then y=(sqrt(3)a + sqrt(4a^2))/2. in (right angle)triangle ABN :sin60=(y+x)/2a then y+x=sqrt(3)a (****) so x=(sqrt(3)a sqrt(4a^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(4a^2))/2*6=a*sqrt(3)/14 so 19649a^2=3a^2 then a=7/sqrt(13)(******) we have in triangle AMC by cos law: 49=a^2+z^2az so z^2az+z^249=0 then z=(a+sqrt(1963a^2))/2. ( z=(asqrt(1963a^2))/2 is not acceptable because A=60 in (right angle)triangle ABN then cos60 =(zw)/2a then z=a+w). so w=(sqrt(1963a^2)a)/2 (*****). in (right angle) triangle BNC : tanC=(x+y)/w then by(*****)and(****) : 2sqrt(3)a/(sqrt(1963a^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: 9,076 Thanks: 622  Re: a problem of triangle
C'est quoi le probleme, Augustin? You knew the answer? 

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