
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
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: 6,797 Thanks: 474  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: 15,354 Thanks: 1020  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: 6,797 Thanks: 474  Re: a problem of triangle
C'est quoi le probleme, Augustin? You knew the answer? 

Tags 
problem, triangle 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Triangle problem  jiasyuen  Algebra  1  December 19th, 2013 03:12 AM 
Triangle Problem  Jakarta  Algebra  6  April 3rd, 2013 07:26 AM 
Right Triangle problem  ro_ch  Algebra  3  March 9th, 2013 06:11 PM 
Help with a triangle problem  JohnA  Algebra  12  March 16th, 2012 05:50 AM 
triangle problem  Kath  Algebra  4  February 28th, 2009 03:25 PM 