Welcome to our community

Be a part of something great, join today!

find angle ACB

Albert

Well-known member
Jan 25, 2013
1,225
$\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

Well-known member
MHB Math Scholar
Mar 10, 2012
834
$\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 $\angle ACB=75$.
 

Albert

Well-known member
Jan 25, 2013
1,225
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 :cool:
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834

Albert

Well-known member
Jan 25, 2013
1,225
Angle ACB.jpg
from the diagram it is easy to see that :
$\angle ACB =30^o +45^o =75^o$
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834