In the given circle, m DE = 105° and m EG = 125°. What is the mZDEG? (3 pts) 1059
A 65°
B 62.5°
C 55°
D 52.5°
125°
O 65
0 62.5°
55°
52.5°

0/5 is 0, 2/0 is undefined, and 0/0 is indeterminate.

step-by-step explanation:

hope my answer has you!

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.