a problem of triangle

Math Help on Cosine (cos), Sine (sin), Tangent (tan), Cotangent (cot), Cosecant (cosec), Secant (sec), Arccos, Arcsin, Arctan, Hypotenuse, Angles, Formulas, Trigonometric Circle, Unit Circles, Quadrants, Rotations; Triangles, Rectangles, Squares, Parallelograms, Quadrangles, Lozenges, Lines, Perpendicular, Parallel, Perpendicular, Parallel Lines, Bisector, Median, Gravity Center, Circumcenter, Circles, Pythagorean Theorem, Thales, Height, Side, Length, Ruler, Compass, Constructions, Formulas; Quadratic equations (Second degree equations), Absolute Values, Inequalities; Events, Random, Mean, Variance, Expectation, Wins, Losses, Bernoulli, Newton's Binomial Formula, Multinomial Formula, Tests, Samples on My Math Forum.

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.
agustin975
Jack of Clubs

Posts: 30
Joined: Sat Nov 10, 2012 3:30 pm

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=90-ß 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.

etidhor
Queen of Hearts

Posts: 50
Joined: Wed Nov 28, 2012 9:26 am

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: Select all
`           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
I'm just an imagination of your figment...
Denis
Super User

Posts: 2908
Joined: Mon Oct 31, 2011 7:29 pm

etidhor wrote: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.
skipjack
Global Moderator

Posts: 10325
Joined: Sat Dec 23, 2006 2:27 pm

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
agustin975
Jack of Clubs

Posts: 30
Joined: Sat Nov 10, 2012 3:30 pm

Re: a problem of triangle

C'est quoi le probleme, Augustin?
I'm just an imagination of your figment...
Denis
Super User

Posts: 2908
Joined: Mon Oct 31, 2011 7:29 pm