Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

enter image description here

share|improve this question
    
could you show a picture? –  yiyi Mar 8 '13 at 6:41

1 Answer 1

up vote 2 down vote accepted

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 $B=90$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.