What is the value of α and β in a triangle? The Next CEO of Stack OverflowAngles in an inscribed triangle with one side as diameter, in absolute geometryWhat characteristic of the triangle leads the the existence of the orthocenterConstructing an equilateral triangle from an arbitrary triangle by shifting towards an interior point3-D evaluations of a triangleFind the sum of the lengths of line segments $BD$ and $CE$how to find the other angles of an oblique triangle, given two sides and one angleTriangle bisector length problemWhat additional condition do the angles in this diagram of a triangle satisfy?Find angle given two isosceles triangles.The sum of angles in a triangle is greater than 180?
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What is the value of α and β in a triangle?
The Next CEO of Stack OverflowAngles in an inscribed triangle with one side as diameter, in absolute geometryWhat characteristic of the triangle leads the the existence of the orthocenterConstructing an equilateral triangle from an arbitrary triangle by shifting towards an interior point3-D evaluations of a triangleFind the sum of the lengths of line segments $BD$ and $CE$how to find the other angles of an oblique triangle, given two sides and one angleTriangle bisector length problemWhat additional condition do the angles in this diagram of a triangle satisfy?Find angle given two isosceles triangles.The sum of angles in a triangle is greater than 180?
$begingroup$
On triangle ABC, with angles α over A, β over B, and γ over C. Where γ is 140°.
On $overlineAB$ lies point D (different from A and B).
On $overlineAC$ lies point E (different from A and C).
$overlineAE$, $overlineED$, $overlineDC$, $overlineCB$ are of same length.
What is the value of α and β?
EDIT: Here's my "solution", however the sum of the angles is greater than 180°.
geometry triangles
asked 8 hours ago
Peter ParadaPeter Parada
1256
$endgroup$
add a comment |
$begingroup$
On triangle ABC, with angles α over A, β over B, and γ over C. Where γ is 140°.
On $overlineAB$ lies point D (different from A and B).
On $overlineAC$ lies point E (different from A and C).
$overlineAE$, $overlineED$, $overlineDC$, $overlineCB$ are of same length.
What is the value of α and β?
EDIT: Here's my "solution", however the sum of the angles is greater than 180°.
geometry triangles
asked 8 hours ago
Peter ParadaPeter Parada
1256
$endgroup$
1
$begingroup$
What have you done? Have you, for instance, drawn this thing?
$endgroup$
– Arthur
8 hours ago
1
$begingroup$
I had some solution where α and β were both 40° which was nonsense for me. Will do some drawing for better imagination.
$endgroup$
– Peter Parada
8 hours ago
3
$begingroup$
Your drawing is hopelessly inaccurate. For example, $x$ (which should be $alpha$, by the way) seems to range from $40^circ$ to $110^circ$. Surely you can do better than that?
$endgroup$
– TonyK
7 hours ago
$begingroup$
If $x+y neq 180$, points $A$, $E$, and $C$ are not collinear.
$endgroup$
– MackTuesday
3 hours ago
$begingroup$
It then follows that $AECB$ is a quadrilateral, not a triangle. It turns out $y = 0$. It all stems from the fact that $ED = CD$.
$endgroup$
– MackTuesday
3 hours ago
add a comment |
$begingroup$
On triangle ABC, with angles α over A, β over B, and γ over C. Where γ is 140°.
On $overlineAB$ lies point D (different from A and B).
On $overlineAC$ lies point E (different from A and C).
$overlineAE$, $overlineED$, $overlineDC$, $overlineCB$ are of same length.
What is the value of α and β?
EDIT: Here's my "solution", however the sum of the angles is greater than 180°.
geometry triangles
asked 8 hours ago
Peter ParadaPeter Parada
1256
$endgroup$
On triangle ABC, with angles α over A, β over B, and γ over C. Where γ is 140°.
On $overlineAB$ lies point D (different from A and B).
On $overlineAC$ lies point E (different from A and C).
$overlineAE$, $overlineED$, $overlineDC$, $overlineCB$ are of same length.
What is the value of α and β?
EDIT: Here's my "solution", however the sum of the angles is greater than 180°.
geometry triangles
geometry triangles
asked 8 hours ago
Peter ParadaPeter Parada
1256
asked 8 hours ago
Peter ParadaPeter Parada
1256
edited 7 hours ago
Peter Parada
asked 8 hours ago
Peter ParadaPeter Parada
1256
asked 8 hours ago
Peter ParadaPeter Parada
1256
asked 8 hours ago
Peter ParadaPeter Parada
1256
1256
1
$begingroup$
What have you done? Have you, for instance, drawn this thing?
$endgroup$
– Arthur
8 hours ago
1
$begingroup$
I had some solution where α and β were both 40° which was nonsense for me. Will do some drawing for better imagination.
$endgroup$
– Peter Parada
8 hours ago
3
$begingroup$
Your drawing is hopelessly inaccurate. For example, $x$ (which should be $alpha$, by the way) seems to range from $40^circ$ to $110^circ$. Surely you can do better than that?
$endgroup$
– TonyK
7 hours ago
$begingroup$
If $x+y neq 180$, points $A$, $E$, and $C$ are not collinear.
$endgroup$
– MackTuesday
3 hours ago
$begingroup$
It then follows that $AECB$ is a quadrilateral, not a triangle. It turns out $y = 0$. It all stems from the fact that $ED = CD$.
$endgroup$
– MackTuesday
3 hours ago
add a comment |
1
$begingroup$
What have you done? Have you, for instance, drawn this thing?
$endgroup$
– Arthur
8 hours ago
1
$begingroup$
I had some solution where α and β were both 40° which was nonsense for me. Will do some drawing for better imagination.
$endgroup$
– Peter Parada
8 hours ago
3
$begingroup$
Your drawing is hopelessly inaccurate. For example, $x$ (which should be $alpha$, by the way) seems to range from $40^circ$ to $110^circ$. Surely you can do better than that?
$endgroup$
– TonyK
7 hours ago
$begingroup$
If $x+y neq 180$, points $A$, $E$, and $C$ are not collinear.
$endgroup$
– MackTuesday
3 hours ago
$begingroup$
It then follows that $AECB$ is a quadrilateral, not a triangle. It turns out $y = 0$. It all stems from the fact that $ED = CD$.
$endgroup$
– MackTuesday
3 hours ago
1
1
$begingroup$
What have you done? Have you, for instance, drawn this thing?
$endgroup$
– Arthur
8 hours ago
$begingroup$
What have you done? Have you, for instance, drawn this thing?
$endgroup$
– Arthur
8 hours ago
1
1
$begingroup$
I had some solution where α and β were both 40° which was nonsense for me. Will do some drawing for better imagination.
$endgroup$
– Peter Parada
8 hours ago
$begingroup$
I had some solution where α and β were both 40° which was nonsense for me. Will do some drawing for better imagination.
$endgroup$
– Peter Parada
8 hours ago
3
3
$begingroup$
Your drawing is hopelessly inaccurate. For example, $x$ (which should be $alpha$, by the way) seems to range from $40^circ$ to $110^circ$. Surely you can do better than that?
$endgroup$
– TonyK
7 hours ago
$begingroup$
Your drawing is hopelessly inaccurate. For example, $x$ (which should be $alpha$, by the way) seems to range from $40^circ$ to $110^circ$. Surely you can do better than that?
$endgroup$
– TonyK
7 hours ago
$begingroup$
If $x+y neq 180$, points $A$, $E$, and $C$ are not collinear.
$endgroup$
– MackTuesday
3 hours ago
$begingroup$
If $x+y neq 180$, points $A$, $E$, and $C$ are not collinear.
$endgroup$
– MackTuesday
3 hours ago
$begingroup$
It then follows that $AECB$ is a quadrilateral, not a triangle. It turns out $y = 0$. It all stems from the fact that $ED = CD$.
$endgroup$
– MackTuesday
3 hours ago
$begingroup$
It then follows that $AECB$ is a quadrilateral, not a triangle. It turns out $y = 0$. It all stems from the fact that $ED = CD$.
$endgroup$
– MackTuesday
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I drew a simple diagram from which one can deduce that
$$alpha=10^circ$$
$$beta=40^circ-alpha=30^circ$$
by noticing that angle $DEC=2alpha$ and hence angle $DCB=140^circ-2alpha$ giving the value of the angle $DBC=20^circ+alpha$. But as angle $DBC$ is equal to angle $ABC$ we get that
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$
answered 7 hours ago
Peter ForemanPeter Foreman
4,9501216
$endgroup$
$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
7 hours ago
$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
7 hours ago
1
$begingroup$
$ADE=alpha$ due to the isosceles triangle formed. $AED=180^circ-2alpha$ as angles in a triangle add to $180^circ$. $DEC=180^circ-AED=2alpha$ as a straight line contains $180^circ$.
$endgroup$
– Peter Foreman
7 hours ago
$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
7 hours ago
add a comment |
$begingroup$
The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...
And $$angle ACBnot= angle DCA$$
answered 7 hours ago
Dr. MathvaDr. Mathva
3,120528
$endgroup$
$begingroup$
Not sure if we are on the same page, but I am assuming exacly that. If 2 sides are equal, then 2 angles in triangle are also equal. - "The two angles opposite the legs are equal and are always acute" - en.wikipedia.org/wiki/Isosceles_triangle
$endgroup$
– Peter Parada
7 hours ago
1
$begingroup$
No what he means is that all of the triangles in your diagram have different valued base angles - they are not all simply $x$. Just because two side lengths are the same does not imply all angles of a triangle are the same.
$endgroup$
– Peter Foreman
7 hours ago
$begingroup$
Exactly @PeterForeman!
$endgroup$
– Dr. Mathva
7 hours ago
1
$begingroup$
Ok. Got it. Thanks for the picture.
$endgroup$
– Peter Parada
7 hours ago
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
$begingroup$
I drew a simple diagram from which one can deduce that
$$alpha=10^circ$$
$$beta=40^circ-alpha=30^circ$$
by noticing that angle $DEC=2alpha$ and hence angle $DCB=140^circ-2alpha$ giving the value of the angle $DBC=20^circ+alpha$. But as angle $DBC$ is equal to angle $ABC$ we get that
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$
answered 7 hours ago
Peter ForemanPeter Foreman
4,9501216
$endgroup$
$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
7 hours ago
$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
7 hours ago
1
$begingroup$
$ADE=alpha$ due to the isosceles triangle formed. $AED=180^circ-2alpha$ as angles in a triangle add to $180^circ$. $DEC=180^circ-AED=2alpha$ as a straight line contains $180^circ$.
$endgroup$
– Peter Foreman
7 hours ago
$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
7 hours ago
add a comment |
$begingroup$
I drew a simple diagram from which one can deduce that
$$alpha=10^circ$$
$$beta=40^circ-alpha=30^circ$$
by noticing that angle $DEC=2alpha$ and hence angle $DCB=140^circ-2alpha$ giving the value of the angle $DBC=20^circ+alpha$. But as angle $DBC$ is equal to angle $ABC$ we get that
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$
answered 7 hours ago
Peter ForemanPeter Foreman
4,9501216
$endgroup$
$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
7 hours ago
$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
7 hours ago
1
$begingroup$
$ADE=alpha$ due to the isosceles triangle formed. $AED=180^circ-2alpha$ as angles in a triangle add to $180^circ$. $DEC=180^circ-AED=2alpha$ as a straight line contains $180^circ$.
$endgroup$
– Peter Foreman
7 hours ago
$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
7 hours ago
add a comment |
$begingroup$
I drew a simple diagram from which one can deduce that
$$alpha=10^circ$$
$$beta=40^circ-alpha=30^circ$$
by noticing that angle $DEC=2alpha$ and hence angle $DCB=140^circ-2alpha$ giving the value of the angle $DBC=20^circ+alpha$. But as angle $DBC$ is equal to angle $ABC$ we get that
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$
answered 7 hours ago
Peter ForemanPeter Foreman
4,9501216
$endgroup$
I drew a simple diagram from which one can deduce that
$$alpha=10^circ$$
$$beta=40^circ-alpha=30^circ$$
by noticing that angle $DEC=2alpha$ and hence angle $DCB=140^circ-2alpha$ giving the value of the angle $DBC=20^circ+alpha$. But as angle $DBC$ is equal to angle $ABC$ we get that
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$
answered 7 hours ago
Peter ForemanPeter Foreman
4,9501216
answered 7 hours ago
Peter ForemanPeter Foreman
4,9501216
answered 7 hours ago
Peter ForemanPeter Foreman
4,9501216
answered 7 hours ago
Peter ForemanPeter Foreman
4,9501216
4,9501216
$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
7 hours ago
$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
7 hours ago
1
$begingroup$
$ADE=alpha$ due to the isosceles triangle formed. $AED=180^circ-2alpha$ as angles in a triangle add to $180^circ$. $DEC=180^circ-AED=2alpha$ as a straight line contains $180^circ$.
$endgroup$
– Peter Foreman
7 hours ago
$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
7 hours ago
add a comment |
$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
7 hours ago
$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
7 hours ago
1
$begingroup$
$ADE=alpha$ due to the isosceles triangle formed. $AED=180^circ-2alpha$ as angles in a triangle add to $180^circ$. $DEC=180^circ-AED=2alpha$ as a straight line contains $180^circ$.
$endgroup$
– Peter Foreman
7 hours ago
$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
7 hours ago
$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
7 hours ago
$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
7 hours ago
$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
7 hours ago
$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
7 hours ago
1
1
$begingroup$
$ADE=alpha$ due to the isosceles triangle formed. $AED=180^circ-2alpha$ as angles in a triangle add to $180^circ$. $DEC=180^circ-AED=2alpha$ as a straight line contains $180^circ$.
$endgroup$
– Peter Foreman
7 hours ago
$begingroup$
$ADE=alpha$ due to the isosceles triangle formed. $AED=180^circ-2alpha$ as angles in a triangle add to $180^circ$. $DEC=180^circ-AED=2alpha$ as a straight line contains $180^circ$.
$endgroup$
– Peter Foreman
7 hours ago
$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
7 hours ago
$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
7 hours ago
add a comment |
$begingroup$
The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...
And $$angle ACBnot= angle DCA$$
answered 7 hours ago
Dr. MathvaDr. Mathva
3,120528
$endgroup$
$begingroup$
Not sure if we are on the same page, but I am assuming exacly that. If 2 sides are equal, then 2 angles in triangle are also equal. - "The two angles opposite the legs are equal and are always acute" - en.wikipedia.org/wiki/Isosceles_triangle
$endgroup$
– Peter Parada
7 hours ago
1
$begingroup$
No what he means is that all of the triangles in your diagram have different valued base angles - they are not all simply $x$. Just because two side lengths are the same does not imply all angles of a triangle are the same.
$endgroup$
– Peter Foreman
7 hours ago
$begingroup$
Exactly @PeterForeman!
$endgroup$
– Dr. Mathva
7 hours ago
1
$begingroup$
Ok. Got it. Thanks for the picture.
$endgroup$
– Peter Parada
7 hours ago
add a comment |
$begingroup$
The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...
And $$angle ACBnot= angle DCA$$
answered 7 hours ago
Dr. MathvaDr. Mathva
3,120528
$endgroup$
$begingroup$
Not sure if we are on the same page, but I am assuming exacly that. If 2 sides are equal, then 2 angles in triangle are also equal. - "The two angles opposite the legs are equal and are always acute" - en.wikipedia.org/wiki/Isosceles_triangle
$endgroup$
– Peter Parada
7 hours ago
1
$begingroup$
No what he means is that all of the triangles in your diagram have different valued base angles - they are not all simply $x$. Just because two side lengths are the same does not imply all angles of a triangle are the same.
$endgroup$
– Peter Foreman
7 hours ago
$begingroup$
Exactly @PeterForeman!
$endgroup$
– Dr. Mathva
7 hours ago
1
$begingroup$
Ok. Got it. Thanks for the picture.
$endgroup$
– Peter Parada
7 hours ago
add a comment |
$begingroup$
The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...
And $$angle ACBnot= angle DCA$$
answered 7 hours ago
Dr. MathvaDr. Mathva
3,120528
$endgroup$
The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...
And $$angle ACBnot= angle DCA$$
answered 7 hours ago
Dr. MathvaDr. Mathva
3,120528
edited 7 hours ago
answered 7 hours ago
Dr. MathvaDr. Mathva
3,120528
answered 7 hours ago
Dr. MathvaDr. Mathva
3,120528
answered 7 hours ago
Dr. MathvaDr. Mathva
3,120528
3,120528
$begingroup$
Not sure if we are on the same page, but I am assuming exacly that. If 2 sides are equal, then 2 angles in triangle are also equal. - "The two angles opposite the legs are equal and are always acute" - en.wikipedia.org/wiki/Isosceles_triangle
$endgroup$
– Peter Parada
7 hours ago
1
$begingroup$
No what he means is that all of the triangles in your diagram have different valued base angles - they are not all simply $x$. Just because two side lengths are the same does not imply all angles of a triangle are the same.
$endgroup$
– Peter Foreman
7 hours ago
$begingroup$
Exactly @PeterForeman!
$endgroup$
– Dr. Mathva
7 hours ago
1
$begingroup$
Ok. Got it. Thanks for the picture.
$endgroup$
– Peter Parada
7 hours ago
add a comment |
$begingroup$
Not sure if we are on the same page, but I am assuming exacly that. If 2 sides are equal, then 2 angles in triangle are also equal. - "The two angles opposite the legs are equal and are always acute" - en.wikipedia.org/wiki/Isosceles_triangle
$endgroup$
– Peter Parada
7 hours ago
1
$begingroup$
No what he means is that all of the triangles in your diagram have different valued base angles - they are not all simply $x$. Just because two side lengths are the same does not imply all angles of a triangle are the same.
$endgroup$
– Peter Foreman
7 hours ago
$begingroup$
Exactly @PeterForeman!
$endgroup$
– Dr. Mathva
7 hours ago
1
$begingroup$
Ok. Got it. Thanks for the picture.
$endgroup$
– Peter Parada
7 hours ago
$begingroup$
Not sure if we are on the same page, but I am assuming exacly that. If 2 sides are equal, then 2 angles in triangle are also equal. - "The two angles opposite the legs are equal and are always acute" - en.wikipedia.org/wiki/Isosceles_triangle
$endgroup$
– Peter Parada
7 hours ago
$begingroup$
Not sure if we are on the same page, but I am assuming exacly that. If 2 sides are equal, then 2 angles in triangle are also equal. - "The two angles opposite the legs are equal and are always acute" - en.wikipedia.org/wiki/Isosceles_triangle
$endgroup$
– Peter Parada
7 hours ago
1
1
$begingroup$
No what he means is that all of the triangles in your diagram have different valued base angles - they are not all simply $x$. Just because two side lengths are the same does not imply all angles of a triangle are the same.
$endgroup$
– Peter Foreman
7 hours ago
$begingroup$
No what he means is that all of the triangles in your diagram have different valued base angles - they are not all simply $x$. Just because two side lengths are the same does not imply all angles of a triangle are the same.
$endgroup$
– Peter Foreman
7 hours ago
$begingroup$
Exactly @PeterForeman!
$endgroup$
– Dr. Mathva
7 hours ago
$begingroup$
Exactly @PeterForeman!
$endgroup$
– Dr. Mathva
7 hours ago
1
1
$begingroup$
Ok. Got it. Thanks for the picture.
$endgroup$
– Peter Parada
7 hours ago
$begingroup$
Ok. Got it. Thanks for the picture.
$endgroup$
– Peter Parada
7 hours ago
add a comment |
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1
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What have you done? Have you, for instance, drawn this thing?
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– Arthur
8 hours ago
1
$begingroup$
I had some solution where α and β were both 40° which was nonsense for me. Will do some drawing for better imagination.
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– Peter Parada
8 hours ago
3
$begingroup$
Your drawing is hopelessly inaccurate. For example, $x$ (which should be $alpha$, by the way) seems to range from $40^circ$ to $110^circ$. Surely you can do better than that?
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– TonyK
7 hours ago
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If $x+y neq 180$, points $A$, $E$, and $C$ are not collinear.
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– MackTuesday
3 hours ago
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It then follows that $AECB$ is a quadrilateral, not a triangle. It turns out $y = 0$. It all stems from the fact that $ED = CD$.
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– MackTuesday
3 hours ago