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Circle o with tangent SN mAU=x, mUG=2x+20,mGN=3x-40,mNA=4x-20
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Mar 14, 2019
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Circle o with tangent SN
mAU=x, mUG=2x+20,mGN=3x-40,mNA=4x-20
Mathematics
high-school
Galaxywatcher
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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°
Emilie Zawadzki
answered
Mar 19, 2019
by
Emilie Zawadzki
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