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What is the value of $alpha$ and $beta$ in a triangle?

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3

1

$begingroup$

On triangle $ABC$, with angles $alpha$ over $A$, $beta$ over $B$, and $gamma$ over $C$. Where $gamma$ is $140^circ$.

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 $180^circ$.

geometry triangles

share|cite|improve this question

edited Mar 31 at 2:08

Rócherz

3,0363823

asked Mar 30 at 17:13

Peter ParadaPeter Parada

1559

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What have you done? Have you, for instance, drawn this thing?
  $endgroup$
  – Arthur
  Mar 30 at 17:16
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 30 at 17:18
  
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 30 at 17:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $x+y neq 180$, points $A$, $E$, and $C$ are not collinear.
  $endgroup$
  – MackTuesday
  Mar 30 at 21:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 30 at 22:08
  
  

  
  
  
  

add a comment | 

3

1

$begingroup$

On triangle $ABC$, with angles $alpha$ over $A$, $beta$ over $B$, and $gamma$ over $C$. Where $gamma$ is $140^circ$.

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 $180^circ$.

geometry triangles

share|cite|improve this question

edited Mar 31 at 2:08

Rócherz

3,0363823

asked Mar 30 at 17:13

Peter ParadaPeter Parada

1559

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What have you done? Have you, for instance, drawn this thing?
  $endgroup$
  – Arthur
  Mar 30 at 17:16
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 30 at 17:18
  
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 30 at 17:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $x+y neq 180$, points $A$, $E$, and $C$ are not collinear.
  $endgroup$
  – MackTuesday
  Mar 30 at 21:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 30 at 22:08
  
  

  
  
  
  

add a comment | 

$begingroup$

On triangle $ABC$, with angles $alpha$ over $A$, $beta$ over $B$, and $gamma$ over $C$. Where $gamma$ is $140^circ$.

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 $180^circ$.

geometry triangles

share|cite|improve this question

edited Mar 31 at 2:08

Rócherz

3,0363823

asked Mar 30 at 17:13

Peter ParadaPeter Parada

1559

$endgroup$

share|cite|improve this question

edited Mar 31 at 2:08

Rócherz

3,0363823

asked Mar 30 at 17:13

Peter ParadaPeter Parada

1559

share|cite|improve this question

edited Mar 31 at 2:08

Rócherz

3,0363823

asked Mar 30 at 17:13

Peter ParadaPeter Parada

1559

edited Mar 31 at 2:08

Rócherz

3,0363823

edited Mar 31 at 2:08

Rócherz

3,0363823

asked Mar 30 at 17:13

Peter ParadaPeter Parada

1559

asked Mar 30 at 17:13

Peter ParadaPeter Parada

1559

2 Answers
2

active

oldest

votes

5

$begingroup$

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

answered Mar 30 at 17:32

Peter ForemanPeter Foreman

9,2111321

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice.......[+1]
  $endgroup$
  – Dr. Mathva
  Mar 30 at 17:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @peter-foreman How have you deduced that DEC is 2𝛼?
  $endgroup$
  – Peter Parada
  Mar 30 at 17:45
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 30 at 17:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. Very nice!
  $endgroup$
  – Peter Parada
  Mar 30 at 17:58
  

  
  
  
  

add a comment | 

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

edited Mar 30 at 17:39

answered Mar 30 at 17:32

Dr. MathvaDr. Mathva

4,0641731

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 30 at 17:35
  
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 30 at 17:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Exactly @PeterForeman!
  $endgroup$
  – Dr. Mathva
  Mar 30 at 17:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ok. Got it. Thanks for the picture.
  $endgroup$
  – Peter Parada
  Mar 30 at 17:40
  

  
  
  
  

add a comment | 

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5

$begingroup$

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

answered Mar 30 at 17:32

Peter ForemanPeter Foreman

9,2111321

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice.......[+1]
  $endgroup$
  – Dr. Mathva
  Mar 30 at 17:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @peter-foreman How have you deduced that DEC is 2𝛼?
  $endgroup$
  – Peter Parada
  Mar 30 at 17:45
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 30 at 17:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. Very nice!
  $endgroup$
  – Peter Parada
  Mar 30 at 17:58
  

  
  
  
  

add a comment | 

5

$begingroup$

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

answered Mar 30 at 17:32

Peter ForemanPeter Foreman

9,2111321

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice.......[+1]
  $endgroup$
  – Dr. Mathva
  Mar 30 at 17:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @peter-foreman How have you deduced that DEC is 2𝛼?
  $endgroup$
  – Peter Parada
  Mar 30 at 17:45
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 30 at 17:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. Very nice!
  $endgroup$
  – Peter Parada
  Mar 30 at 17:58
  

  
  
  
  

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

share|cite|improve this answer

answered Mar 30 at 17:32

Peter ForemanPeter Foreman

9,2111321

$endgroup$

share|cite|improve this answer

answered Mar 30 at 17:32

Peter ForemanPeter Foreman

9,2111321

answered Mar 30 at 17:32

Peter ForemanPeter Foreman

9,2111321

answered Mar 30 at 17:32

Peter ForemanPeter Foreman

9,2111321

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

edited Mar 30 at 17:39

answered Mar 30 at 17:32

Dr. MathvaDr. Mathva

4,0641731

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 30 at 17:35
  
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 30 at 17:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Exactly @PeterForeman!
  $endgroup$
  – Dr. Mathva
  Mar 30 at 17:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ok. Got it. Thanks for the picture.
  $endgroup$
  – Peter Parada
  Mar 30 at 17:40
  

  
  
  
  

add a comment | 

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

edited Mar 30 at 17:39

answered Mar 30 at 17:32

Dr. MathvaDr. Mathva

4,0641731

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 30 at 17:35
  
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 30 at 17:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Exactly @PeterForeman!
  $endgroup$
  – Dr. Mathva
  Mar 30 at 17:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ok. Got it. Thanks for the picture.
  $endgroup$
  – Peter Parada
  Mar 30 at 17:40
  

  
  
  
  

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 $$angle ACBnot= angle DCA$$

share|cite|improve this answer

edited Mar 30 at 17:39

answered Mar 30 at 17:32

Dr. MathvaDr. Mathva

4,0641731

$endgroup$

share|cite|improve this answer

edited Mar 30 at 17:39

answered Mar 30 at 17:32

Dr. MathvaDr. Mathva

4,0641731

answered Mar 30 at 17:32

Dr. MathvaDr. Mathva

4,0641731

answered Mar 30 at 17:32

Dr. MathvaDr. Mathva

4,0641731