Finding Angle ACB in Triangle ABC

[Albert1]

$\triangle ABC , \angle ABC=45^o,\,\, point \,\, D\,\, on \,\, \,\overline{BC} $

$and,\,\, 2\overline{BD}=\overline{CD},\,\, \angle DAB=15^o$

$find :\,\, \angle ACB=?$

 

[caffeinemachine]

$\triangle ABC , \angle ABC=45^o,\,\, point \,\, D\,\, on \,\, \,\overline{BC} $

$and,\,\, 2\overline{BD}=\overline{CD},\,\, \angle DAB=15^o$

$find :\,\, \angle ACB=?$

Find a point $M$ between $A$ and $D$ such that $\angle MBD=30$. Note that $MD=DM$ and $AM=MB$. Say $BD=1$. The above leads to $AD=\sqrt 3 + 1$. Further note that $\angle CMD=90$ and hence $\angle MCA=45$. Consequently $\angle ACB=75$.

 

[Albert1]

Find a point $M$ between $A$ and $D$ such that $\angle MBD=30$. Note that $MD=DM$ and $AM=MB$. Say $BD=1$. The above leads to $AD=\sqrt 3 + 1$. Further note that $\angle CMD=90$ and hence $\angle MCA=45$. Consequently $\angle ACB=75$.

caffeinemachine :very good solution

 

[caffeinemachine]

caffeinemachine :very good solution

Thanks. :) If you have a different one then please post it.

 

[Albert1]

View attachment 1209
from the diagram it is easy to see that :
$\angle ACB =30^o +45^o =75^o$

 

[caffeinemachine]

https://www.physicsforums.com/attachments/1209
from the diagram it is easy to see that :
$\angle ACB =30^o +45^o =75^o$

Awesome!