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°
