MHB Finding $\angle QCA$ from Altitude $AM$ of Triangle $ABC$

[anemone]

If $Q$ is a point on the altitude $AM$ of triangle $ABC$, and that $\angle QBA=20^{\circ}$, $\angle QBC=40^{\circ}$ and $\angle QCB=30^{\circ}$, find $\angle QCA$.

 

[Albert1]

If $Q$ is a point on the altitude $AM$ of triangle $ABC$, and that $\angle QBA=20^{\circ}$, $\angle QBC=40^{\circ}$ and $\angle QCB=30^{\circ}$, find $\angle QCA$.

Spoiler

Point Q must be the orthocenter of
$\triangle ABC $
we have :$\angle QCA+30+40=90$
$\therefore \angle QCA=20^o$

 

[anemone]

Spoiler

Point Q must be the orthocenter of
$\triangle ABC $
we have :$\angle QCA+30+40=90$
$\therefore \angle QCA=20^o$

Well done, Albert! Well done!(Yes) And thanks for participating!