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?
a triangle problem of angles
-
0could you show a picture? – 2013-03-08
1 Answers
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 $B=90$
-
0Thank you for usefull help. Daryl – 2012-12-13