Somebody me with this geometry work

m∠EGF = 65°  and m∠CGF = 115°

Step-by-step explanation:

Given;

∠EFG = 50°

EF = FG

Solution,

In ΔEFG m∠EFG = 50° and EF = FG.

Since triangle is an isosceles triangle hence their base angles are always equal.

∴

Let the measure of ∠EGF  be x.

∴

Now by angle Sum property which states "The sum of all the angles of a triangle is 180°."

m∠EFG + m∠FEG + ∠EGF = 180

Hence

m∠EGF = 65°  

Also 'The sum of angles that are formed on a straight line is equal to 180°."

m∠EGF + m∠CGF = 180°

65° + m∠CGF = 180°

m∠CGF = 180° - 65° = 115°

Hence m∠EGF = 65°  m∠CGF = 115°

b

step-by-step explanation:

the word superfluous means unnecessary

the radius is 2 times the diameter.

step-by-step explanation: