# a triangle problem of angles

suppose in triangle ABC , angle of BAC is 60 degree. if K is intersection point of [CM] median(for segment[AB] )and [BN] altitude. also suppose |KM|=1 cm and |CK|=6 cm calculate angels of triangle ABC?

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could you show a picture? – yiyi Mar 8 '13 at 6:41

at first: we suppose $BN=x+y$and $AC=z$,$NC=w$, $BM=a$ and angle $ACM=\gamma$

we have in triangle $BKM$ by $\cos$

law: $1=a^2+y^2-\sqrt3ay$ then $$y=\frac{\sqrt3a +(or -)\sqrt{4-a^2}}{2}$$

$y>x$ then $$y=\frac{\sqrt3a + \sqrt{4-a^2}}{2}$$ . in (right angle)triangle $ABN$ :$\sin 60=\frac{y+x}{2a}$ then $y+x=\sqrt3a$ ($IV$) so $x=\frac{\sqrt3a -\sqrt{4-a^2}}{2}$ ($III$)
we have in triangle $AMC$ by $\sin$ law : $$\frac{\sin A}{7}=\frac{\sin \gamma}{a}$$ then $$\sin\gamma=\frac{\sqrt3a}{14}$$ ($I$)
in (right angle) triangle $NKC$: $\sin \gamma=\frac{x}{6}$($II$) then by($II$) and($I$) and ($III$)

we have : $$\frac{a\sqrt3-\sqrt{4-a^2}}{2*6}=\frac{a\sqrt3}{14}$$ so $196-49a^2=3a^2$ then $a=\frac{7}{\sqrt{13}}$($VI$)

we have in triangle $AMC$ by $\cos$ law: $49=a^2+z^2-az$ so $z^2-az+z^2-49=0$ then

$$z=\frac{a+\sqrt{196-3a^2}}{2}$$ ( $$z=\frac{a-\sqrt{196-3a^2}}{2}$$ is not acceptable because $A=60$ in (right angle)triangle $ABN$ then $\cos 60 =\frac{z-w}{2a}$ then $z=a+w$).

so $$w=\frac{\sqrt{196-3a^2}-a}{2}$$($V$).

in (right angle) triangle $BNC$ : $\tan C=\frac{x+y}{w}$ then by($V$)and($IV$) :

$\tan C = \frac{2\sqrt3a}{\sqrt{196-3a^2}-a}$ so by ($VI$) :

$\tan C=\frac{\sqrt3}{3}$ so $C=30$ and $B=90$

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I can't make sense of your answer. It will be much more feasible if you format the math correctly. A tutorial can be found here. – Daryl Dec 12 '12 at 20:47
Thank you for usefull help. Daryl – agustin Dec 13 '12 at 20:56