Answer :
|CD| is parallel to |AB|
[tex]\angle ACD=\angle ACB+\angle DCB[/tex][tex]\begin{gathered} ACBisatriangle.Thesumofanglesinatriangleis180^o. \\ \angle ACB+\angle ABC+\angle BAC=180^o \\ \angle ACB=180^o-35^o-115^o \\ \angle ACB=30^o \end{gathered}[/tex]Also;
[tex]\begin{gathered} S\text{ ince CD and AB are parallel;} \\ \angle DCB=\angle ABC\text{ (because alternate angles are equal)} \\ \angle DCB=35^o \end{gathered}[/tex][tex]\begin{gathered} \angle ACD=\angle ACB+\angle DCB \\ \angle ACD=30^o+35^o \\ \angle ACD=65^o \end{gathered}[/tex]ANSWERS:
(a) The measured angle ACD is 65degrees
(b) The measured angle DCB is 35degrees
(c) The measured angle ACB is 30 degrees
