Circle o with tangent sn mau=x, mug=2x+20,mgn=3x-40,mna=4x-20

The complete question inn the attached figure

Part a) Find measure arc UG

we know that
mAU+mUG+mGN+mNA=360°
so
x+(2x+20)+(3x-40)+(4x-20)=360°
10x-40=360°
10x=400°
x=40°

mUG=2x+20-----> 2*40+20----> mUG=100°

the answer part a) is
mUG=100°

Part b) Find m ∠GBN

we know that
The measure of the external angle is the semi difference of the arcs that it covers.
so
 m ∠GBN=(1/2)*[arc GN-arc AU]
arc AU=x----> 40°
arc GN=3x-40----> 3*40-40----> 80°
m ∠GBN=(1/2)*[80-40]-----> 20°

the answer part b) is 
m ∠GBN=20°

Part c) Find the m ∠AGN
we know that
The inscribed angle measures half of the arc it comprises.
so
m ∠AGN=(1/2)*[arc AN]
arc AN=4x-20----> 4*40-20----> 140°
m ∠AGN=(1/2)*[140]----> 70°

the answer Part c) is
m ∠AGN= 70°

Part d) Find m ∠UDG
we know that
The inscribed angle measures half of the arc it comprises.
step 1
find m ∠DUG
m ∠DUG=(1/2)*[arc GN]
arc GN=3x-40----> 3*40-40----> 80°
m ∠DUG=(1/2)*[80]----> 40°

step 2
find m ∠UGD
 m ∠UGD=(1/2)*[arc AU]
arc AU=x----> 40°
m ∠UGD=(1/2)*[40]------> 20°

step 3
find m ∠UDG
m∠UDG=180-(m ∠UGD+m ∠DUG)-----> 180-(40+20)----> 120°

the answer Part c) is 
m∠UDG=120°

Part d)
Find m∠SNG

step 1
we know that
The inscribed angle measures half of the arc it comprises.
so
m∠AGN=(1/2)*[arc AN]
arc AN=4x-20------> 4*40-20-----> 140°
m∠AGN=(1/2)*[140]------> 70°

step 2
∠NGS=180-(m∠UGD+m∠AGN)----> 180-(20+70)----> 90°
that means that NGS is a right triangle

step 3
find m∠GSN
we know that
The measure of the external angle is the semi difference of the arcs that it covers
m∠GSN=(1/2)*[arc AU+arc AN-arc GN]---> (1/2)*[2x+20]---> (1/2)*[100]
m∠GSN=50°

step 4
find m∠SNG
m∠SNG  and m∠GSN are complementary angles
so
m∠SNG=90-m∠GSN-----> 90-50-----> 40°

the answer part d) is
m∠SNG=40°