What is the value of α and β in a triangle?












3












$begingroup$


On triangle ABC, with angles α over A, β over B, and γ over C. Where γ is 140°.



On $overline{AB}$ lies point D (different from A and B).



On $overline{AC}$ lies point E (different from A and C).



$overline{AE}$, $overline{ED}$, $overline{DC}$, $overline{CB}$ 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°.



enter image description here






geometry triangles







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What have you done? Have you, for instance, drawn this thing?
    $endgroup$
    – Arthur
    1 hour 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
    1 hour ago








  • 2




    $begingroup$
    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
    51 mins ago


















3












$begingroup$


On triangle ABC, with angles α over A, β over B, and γ over C. Where γ is 140°.



On $overline{AB}$ lies point D (different from A and B).



On $overline{AC}$ lies point E (different from A and C).



$overline{AE}$, $overline{ED}$, $overline{DC}$, $overline{CB}$ 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°.



enter image description here






geometry triangles







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What have you done? Have you, for instance, drawn this thing?
    $endgroup$
    – Arthur
    1 hour 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
    1 hour ago








  • 2




    $begingroup$
    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
    51 mins ago
















3












3








3





$begingroup$


On triangle ABC, with angles α over A, β over B, and γ over C. Where γ is 140°.



On $overline{AB}$ lies point D (different from A and B).



On $overline{AC}$ lies point E (different from A and C).



$overline{AE}$, $overline{ED}$, $overline{DC}$, $overline{CB}$ 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°.



enter image description here






geometry triangles







share|cite|improve this question











$endgroup$




On triangle ABC, with angles α over A, β over B, and γ over C. Where γ is 140°.



On $overline{AB}$ lies point D (different from A and B).



On $overline{AC}$ lies point E (different from A and C).



$overline{AE}$, $overline{ED}$, $overline{DC}$, $overline{CB}$ 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°.



enter image description here






geometry triangles




geometry triangles






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 59 mins ago







Peter Parada

















asked 1 hour ago









Peter ParadaPeter Parada

1256




1256








  • 1




    $begingroup$
    What have you done? Have you, for instance, drawn this thing?
    $endgroup$
    – Arthur
    1 hour 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
    1 hour ago








  • 2




    $begingroup$
    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
    51 mins ago
















  • 1




    $begingroup$
    What have you done? Have you, for instance, drawn this thing?
    $endgroup$
    – Arthur
    1 hour 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
    1 hour ago








  • 2




    $begingroup$
    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
    51 mins ago










1




1




$begingroup$
What have you done? Have you, for instance, drawn this thing?
$endgroup$
– Arthur
1 hour ago




$begingroup$
What have you done? Have you, for instance, drawn this thing?
$endgroup$
– Arthur
1 hour 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
1 hour 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
1 hour ago






2




2




$begingroup$
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
51 mins 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
51 mins ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

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
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice.......[+1]
    $endgroup$
    – Dr. Mathva
    49 mins ago










  • $begingroup$
    @peter-foreman How have you deduced that DEC is 2𝛼?
    $endgroup$
    – Peter Parada
    44 mins 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
    35 mins ago










  • $begingroup$
    Thank you. Very nice!
    $endgroup$
    – Peter Parada
    30 mins ago



















3












$begingroup$

The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...





And $$angle ACBnot= angle DCA$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    54 mins 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
    50 mins ago












  • $begingroup$
    Exactly @PeterForeman!
    $endgroup$
    – Dr. Mathva
    49 mins ago






  • 1




    $begingroup$
    Ok. Got it. Thanks for the picture.
    $endgroup$
    – Peter Parada
    48 mins ago












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice.......[+1]
    $endgroup$
    – Dr. Mathva
    49 mins ago










  • $begingroup$
    @peter-foreman How have you deduced that DEC is 2𝛼?
    $endgroup$
    – Peter Parada
    44 mins 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
    35 mins ago










  • $begingroup$
    Thank you. Very nice!
    $endgroup$
    – Peter Parada
    30 mins ago
















4












$begingroup$

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
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice.......[+1]
    $endgroup$
    – Dr. Mathva
    49 mins ago










  • $begingroup$
    @peter-foreman How have you deduced that DEC is 2𝛼?
    $endgroup$
    – Peter Parada
    44 mins 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
    35 mins ago










  • $begingroup$
    Thank you. Very nice!
    $endgroup$
    – Peter Parada
    30 mins ago














4












4








4





$begingroup$

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
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$






share|cite|improve this answer









$endgroup$



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
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 56 mins ago









Peter ForemanPeter Foreman

4,9051216




4,9051216












  • $begingroup$
    Nice.......[+1]
    $endgroup$
    – Dr. Mathva
    49 mins ago










  • $begingroup$
    @peter-foreman How have you deduced that DEC is 2𝛼?
    $endgroup$
    – Peter Parada
    44 mins 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
    35 mins ago










  • $begingroup$
    Thank you. Very nice!
    $endgroup$
    – Peter Parada
    30 mins ago


















  • $begingroup$
    Nice.......[+1]
    $endgroup$
    – Dr. Mathva
    49 mins ago










  • $begingroup$
    @peter-foreman How have you deduced that DEC is 2𝛼?
    $endgroup$
    – Peter Parada
    44 mins 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
    35 mins ago










  • $begingroup$
    Thank you. Very nice!
    $endgroup$
    – Peter Parada
    30 mins ago
















$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
49 mins ago




$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
49 mins ago












$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
44 mins ago




$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
44 mins 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
35 mins 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
35 mins ago












$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
30 mins ago




$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
30 mins ago











3












$begingroup$

The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...





And $$angle ACBnot= angle DCA$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    54 mins 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
    50 mins ago












  • $begingroup$
    Exactly @PeterForeman!
    $endgroup$
    – Dr. Mathva
    49 mins ago






  • 1




    $begingroup$
    Ok. Got it. Thanks for the picture.
    $endgroup$
    – Peter Parada
    48 mins ago
















3












$begingroup$

The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...





And $$angle ACBnot= angle DCA$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    54 mins 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
    50 mins ago












  • $begingroup$
    Exactly @PeterForeman!
    $endgroup$
    – Dr. Mathva
    49 mins ago






  • 1




    $begingroup$
    Ok. Got it. Thanks for the picture.
    $endgroup$
    – Peter Parada
    48 mins ago














3












3








3





$begingroup$

The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...





And $$angle ACBnot= angle DCA$$






share|cite|improve this answer











$endgroup$



The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...





And $$angle ACBnot= angle DCA$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 50 mins ago

























answered 57 mins ago









Dr. MathvaDr. Mathva

3,130528




3,130528












  • $begingroup$
    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
    54 mins 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
    50 mins ago












  • $begingroup$
    Exactly @PeterForeman!
    $endgroup$
    – Dr. Mathva
    49 mins ago






  • 1




    $begingroup$
    Ok. Got it. Thanks for the picture.
    $endgroup$
    – Peter Parada
    48 mins 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
    54 mins 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
    50 mins ago












  • $begingroup$
    Exactly @PeterForeman!
    $endgroup$
    – Dr. Mathva
    49 mins ago






  • 1




    $begingroup$
    Ok. Got it. Thanks for the picture.
    $endgroup$
    – Peter Parada
    48 mins 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
54 mins 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
54 mins 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
50 mins 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
50 mins ago














$begingroup$
Exactly @PeterForeman!
$endgroup$
– Dr. Mathva
49 mins ago




$begingroup$
Exactly @PeterForeman!
$endgroup$
– Dr. Mathva
49 mins ago




1




1




$begingroup$
Ok. Got it. Thanks for the picture.
$endgroup$
– Peter Parada
48 mins ago




$begingroup$
Ok. Got it. Thanks for the picture.
$endgroup$
– Peter Parada
48 mins ago


















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Sidney Franklin

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Questa voce sull'argomento registi statunitensi è solo un abbozzo . Contribuisci a migliorarla secondo le convenzioni di Wikipedia. Sidney Franklin, foto di Carl Van Vechten Oscar alla memoria Irving G. Thalberg 1943 Sidney Franklin , all'anagrafe Sidney Arnold Franklin , (San Francisco, 21 marzo 1893 – Santa Monica, 18 maggio 1972), è stato un regista, produttore cinematografico e attore statunitense. Il 20 giugno 1963, si sposò con l'attrice australiana Enid Bennett. Il matrimonio durò fino alla morte dell'attrice, il 14 maggio 1969 [1] Indice 1 Filmografia 1.1 Regista 1.2 Aiuto regia (parziale) 1.3 Produttore 1.4 Attore (parziale) 2 Note 3 Altri progetti 4 Collegamenti esterni Filmografia | Regista | The Baby , co-regia Chester M. Franklin (1915) The Rivals , co-regia Chester M. Franklin (1915) Little Dick's First Case , co-regia Chester M. Franklin (1915) Her Filmland Hero , co-...
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