Mixing
Find the degree measure of an arc of a circle
$begingroup$
Can anyone help me with this problem? I think I probably miss some theorems.
In teh figure angle AED = 30, and the minor arc AB, BC, and CD are all equal. The degree measure of the minor arc AD is
a) not enought information, b) 30, c) 105, d) 45, e) none of these.
geometry
$endgroup$
add a comment |
$begingroup$
Can anyone help me with this problem? I think I probably miss some theorems.
In teh figure angle AED = 30, and the minor arc AB, BC, and CD are all equal. The degree measure of the minor arc AD is
a) not enought information, b) 30, c) 105, d) 45, e) none of these.
geometry
$endgroup$
2
$begingroup$
You need to know the theorem that $[m(BC)-m(AD)]/2=30^circ$. I think that with that, you can answer the question.
$endgroup$
– Lubin
Apr 16 '16 at 17:40
$begingroup$
@Lubin Instead of remembering the theorem, I tried to prove it, but failed. Can you give me a hint on or show me how to prove the theorem? Thank you again.
$endgroup$
– user321645
Apr 16 '16 at 18:01
$begingroup$
Do you know the theorem that the if $A$, $B$, and $C$ are three points on the circumference of a circle, then the measure of angle $ACB$ is half the measure of the arc $AB$?
$endgroup$
– Lubin
Apr 16 '16 at 18:04
add a comment |
$begingroup$
Can anyone help me with this problem? I think I probably miss some theorems.
In teh figure angle AED = 30, and the minor arc AB, BC, and CD are all equal. The degree measure of the minor arc AD is
a) not enought information, b) 30, c) 105, d) 45, e) none of these.
geometry
$endgroup$
Can anyone help me with this problem? I think I probably miss some theorems.
In teh figure angle AED = 30, and the minor arc AB, BC, and CD are all equal. The degree measure of the minor arc AD is
a) not enought information, b) 30, c) 105, d) 45, e) none of these.
geometry
geometry
asked Apr 16 '16 at 17:26
user321645
2
$begingroup$
You need to know the theorem that $[m(BC)-m(AD)]/2=30^circ$. I think that with that, you can answer the question.
$endgroup$
– Lubin
Apr 16 '16 at 17:40
$begingroup$
@Lubin Instead of remembering the theorem, I tried to prove it, but failed. Can you give me a hint on or show me how to prove the theorem? Thank you again.
$endgroup$
– user321645
Apr 16 '16 at 18:01
$begingroup$
Do you know the theorem that the if $A$, $B$, and $C$ are three points on the circumference of a circle, then the measure of angle $ACB$ is half the measure of the arc $AB$?
$endgroup$
– Lubin
Apr 16 '16 at 18:04
add a comment |
2
$begingroup$
You need to know the theorem that $[m(BC)-m(AD)]/2=30^circ$. I think that with that, you can answer the question.
$endgroup$
– Lubin
Apr 16 '16 at 17:40
$begingroup$
@Lubin Instead of remembering the theorem, I tried to prove it, but failed. Can you give me a hint on or show me how to prove the theorem? Thank you again.
$endgroup$
– user321645
Apr 16 '16 at 18:01
$begingroup$
Do you know the theorem that the if $A$, $B$, and $C$ are three points on the circumference of a circle, then the measure of angle $ACB$ is half the measure of the arc $AB$?
$endgroup$
– Lubin
Apr 16 '16 at 18:04
2
2
$begingroup$
You need to know the theorem that $[m(BC)-m(AD)]/2=30^circ$. I think that with that, you can answer the question.
$endgroup$
– Lubin
Apr 16 '16 at 17:40
$begingroup$
You need to know the theorem that $[m(BC)-m(AD)]/2=30^circ$. I think that with that, you can answer the question.
$endgroup$
– Lubin
Apr 16 '16 at 17:40
$begingroup$
@Lubin Instead of remembering the theorem, I tried to prove it, but failed. Can you give me a hint on or show me how to prove the theorem? Thank you again.
$endgroup$
– user321645
Apr 16 '16 at 18:01
$begingroup$
@Lubin Instead of remembering the theorem, I tried to prove it, but failed. Can you give me a hint on or show me how to prove the theorem? Thank you again.
$endgroup$
– user321645
Apr 16 '16 at 18:01
$begingroup$
Do you know the theorem that the if $A$, $B$, and $C$ are three points on the circumference of a circle, then the measure of angle $ACB$ is half the measure of the arc $AB$?
$endgroup$
– Lubin
Apr 16 '16 at 18:04
$begingroup$
Do you know the theorem that the if $A$, $B$, and $C$ are three points on the circumference of a circle, then the measure of angle $ACB$ is half the measure of the arc $AB$?
$endgroup$
– Lubin
Apr 16 '16 at 18:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A little bit more on how to prove this is as follows.
An inscribed angle is an angle whose vertex is on the circle and whose sides contain chords of the circle.
In the question angles BAC and ACD are inscribed angles.
If an angle is inscribe in a circle, then its measure is half the measure of its intercepted arc.
From this we conclude that angle BAC = (BC)/2 and angle ACD = (AD)/2.
Now from the exterior angle theorem we have.
Angle AED = abs[ angle BAC - angle ACD ]
Now if we substitute our previous equations into this exterior angle theorem we get.
Angle AED = abs [ (BC)/2 - (AD)/2]
The absolute value is there to make sure the angle is positive.
My final result is
which means that it depends on the radius r of the circle. So I don't think there is enough information to solve.
edited Mar 9 '17 at 17:33
Community♦
1
answered Apr 16 '16 at 20:12
Erock BroxErock Brox
23828
$endgroup$
add a comment |
$begingroup$
From the symmetry that $AB=CD$, we know that $Delta BCE$ is isosceles and that lines $AD$ and $BC$ are parallel. We can add some angles to the original diagram:
The major arc $BD = 2cdot105^circ$, hence $AB=BC=CD=105^circ$ and minor arc $AD=45^circ$.
answered Nov 12 '17 at 10:38
nickgardnickgard
1,9301414
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A little bit more on how to prove this is as follows.
An inscribed angle is an angle whose vertex is on the circle and whose sides contain chords of the circle.
In the question angles BAC and ACD are inscribed angles.
If an angle is inscribe in a circle, then its measure is half the measure of its intercepted arc.
From this we conclude that angle BAC = (BC)/2 and angle ACD = (AD)/2.
Now from the exterior angle theorem we have.
Angle AED = abs[ angle BAC - angle ACD ]
Now if we substitute our previous equations into this exterior angle theorem we get.
Angle AED = abs [ (BC)/2 - (AD)/2]
The absolute value is there to make sure the angle is positive.
My final result is
which means that it depends on the radius r of the circle. So I don't think there is enough information to solve.
edited Mar 9 '17 at 17:33
Community♦
1
answered Apr 16 '16 at 20:12
Erock BroxErock Brox
23828
$endgroup$
add a comment |
$begingroup$
A little bit more on how to prove this is as follows.
An inscribed angle is an angle whose vertex is on the circle and whose sides contain chords of the circle.
In the question angles BAC and ACD are inscribed angles.
If an angle is inscribe in a circle, then its measure is half the measure of its intercepted arc.
From this we conclude that angle BAC = (BC)/2 and angle ACD = (AD)/2.
Now from the exterior angle theorem we have.
Angle AED = abs[ angle BAC - angle ACD ]
Now if we substitute our previous equations into this exterior angle theorem we get.
Angle AED = abs [ (BC)/2 - (AD)/2]
The absolute value is there to make sure the angle is positive.
My final result is
which means that it depends on the radius r of the circle. So I don't think there is enough information to solve.
edited Mar 9 '17 at 17:33
Community♦
1
answered Apr 16 '16 at 20:12
Erock BroxErock Brox
23828
$endgroup$
add a comment |
$begingroup$
A little bit more on how to prove this is as follows.
An inscribed angle is an angle whose vertex is on the circle and whose sides contain chords of the circle.
In the question angles BAC and ACD are inscribed angles.
If an angle is inscribe in a circle, then its measure is half the measure of its intercepted arc.
From this we conclude that angle BAC = (BC)/2 and angle ACD = (AD)/2.
Now from the exterior angle theorem we have.
Angle AED = abs[ angle BAC - angle ACD ]
Now if we substitute our previous equations into this exterior angle theorem we get.
Angle AED = abs [ (BC)/2 - (AD)/2]
The absolute value is there to make sure the angle is positive.
My final result is
which means that it depends on the radius r of the circle. So I don't think there is enough information to solve.
edited Mar 9 '17 at 17:33
Community♦
1
answered Apr 16 '16 at 20:12
Erock BroxErock Brox
23828
$endgroup$
A little bit more on how to prove this is as follows.
An inscribed angle is an angle whose vertex is on the circle and whose sides contain chords of the circle.
In the question angles BAC and ACD are inscribed angles.
If an angle is inscribe in a circle, then its measure is half the measure of its intercepted arc.
From this we conclude that angle BAC = (BC)/2 and angle ACD = (AD)/2.
Now from the exterior angle theorem we have.
Angle AED = abs[ angle BAC - angle ACD ]
Now if we substitute our previous equations into this exterior angle theorem we get.
Angle AED = abs [ (BC)/2 - (AD)/2]
The absolute value is there to make sure the angle is positive.
My final result is
which means that it depends on the radius r of the circle. So I don't think there is enough information to solve.
edited Mar 9 '17 at 17:33
Community♦
1
answered Apr 16 '16 at 20:12
Erock BroxErock Brox
23828
edited Mar 9 '17 at 17:33
Community♦
1
edited Mar 9 '17 at 17:33
Community♦
1
edited Mar 9 '17 at 17:33
Community♦
1
1
answered Apr 16 '16 at 20:12
Erock BroxErock Brox
23828
answered Apr 16 '16 at 20:12
Erock BroxErock Brox
23828
answered Apr 16 '16 at 20:12
Erock BroxErock Brox
23828
23828
add a comment |
add a comment |
$begingroup$
From the symmetry that $AB=CD$, we know that $Delta BCE$ is isosceles and that lines $AD$ and $BC$ are parallel. We can add some angles to the original diagram:
The major arc $BD = 2cdot105^circ$, hence $AB=BC=CD=105^circ$ and minor arc $AD=45^circ$.
answered Nov 12 '17 at 10:38
nickgardnickgard
1,9301414
$endgroup$
add a comment |
$begingroup$
From the symmetry that $AB=CD$, we know that $Delta BCE$ is isosceles and that lines $AD$ and $BC$ are parallel. We can add some angles to the original diagram:
The major arc $BD = 2cdot105^circ$, hence $AB=BC=CD=105^circ$ and minor arc $AD=45^circ$.
answered Nov 12 '17 at 10:38
nickgardnickgard
1,9301414
$endgroup$
add a comment |
$begingroup$
From the symmetry that $AB=CD$, we know that $Delta BCE$ is isosceles and that lines $AD$ and $BC$ are parallel. We can add some angles to the original diagram:
The major arc $BD = 2cdot105^circ$, hence $AB=BC=CD=105^circ$ and minor arc $AD=45^circ$.
answered Nov 12 '17 at 10:38
nickgardnickgard
1,9301414
$endgroup$
From the symmetry that $AB=CD$, we know that $Delta BCE$ is isosceles and that lines $AD$ and $BC$ are parallel. We can add some angles to the original diagram:
The major arc $BD = 2cdot105^circ$, hence $AB=BC=CD=105^circ$ and minor arc $AD=45^circ$.
answered Nov 12 '17 at 10:38
nickgardnickgard
1,9301414
answered Nov 12 '17 at 10:38
nickgardnickgard
1,9301414
answered Nov 12 '17 at 10:38
nickgardnickgard
1,9301414
answered Nov 12 '17 at 10:38
nickgardnickgard
1,9301414
1,9301414
add a comment |
add a comment |
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2
$begingroup$
You need to know the theorem that $[m(BC)-m(AD)]/2=30^circ$. I think that with that, you can answer the question.
$endgroup$
– Lubin
Apr 16 '16 at 17:40
$begingroup$
@Lubin Instead of remembering the theorem, I tried to prove it, but failed. Can you give me a hint on or show me how to prove the theorem? Thank you again.
$endgroup$
– user321645
Apr 16 '16 at 18:01
$begingroup$
Do you know the theorem that the if $A$, $B$, and $C$ are three points on the circumference of a circle, then the measure of angle $ACB$ is half the measure of the arc $AB$?
$endgroup$
– Lubin
Apr 16 '16 at 18:04