Answer:<CDA≈ 47.5 degrees
Step-by-step explanation:
I note M , so BM _|_ AD
in trg ABM, T Pitagora
AM^2=AB^2-BM^2
AM^2=6.5^2-6^2
AM^2=42.25-36
AM^2=6.25
AM=2.5
I note N , so CN_|_ AD
AD=AM+MN+ND
23=2.5+9+ND
23=11.5+ND
ND=23-11.5
ND=11.5
in trg CND
tg <CDA=CN/ND
=6/11.5
=1.(09)
so <CDA≈ 47.5 degrees
