How to get the angle of the figure? [duplicate]

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This question already has an answer here:

• 
  Finding an angle within an 80-80-20 isosceles triangle
  
  4 answers
  
  

I have this problem, that can be solved with elemental knowledge.

In order to challenge, I can't draw extra segments to solved it.

This is the problem and i need to get the measure of $anglex$

My current development is:

Well, the sides with equal colour are congruent, and I have completed all the angles that I could, in red colour.

Then, how i can get the $anglex$ in a elementary form, without draw extra segments? Is possible ?

geometry euclidean-geometry triangle angle

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edited Aug 6 at 17:11

Michael Rozenberg

88.2k1579180

asked Aug 6 at 15:34

Mattiu

780316

marked as duplicate by timon92, José Carlos Santos, Dylan, Xander Henderson, amWhy Aug 7 at 0:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Introducing another variable for one of the angles you haven't solved for may help. Try to write two angles in two different ways using your variables. This will give you a system of two linear equations in two variables.
  – Dzoooks
  Aug 6 at 15:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzoooks i already tried that
  – Mattiu
  Aug 6 at 17:04
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

1

This question already has an answer here:

• 
  Finding an angle within an 80-80-20 isosceles triangle
  
  4 answers
  
  

I have this problem, that can be solved with elemental knowledge.

In order to challenge, I can't draw extra segments to solved it.

This is the problem and i need to get the measure of $anglex$

My current development is:

Well, the sides with equal colour are congruent, and I have completed all the angles that I could, in red colour.

Then, how i can get the $anglex$ in a elementary form, without draw extra segments? Is possible ?

geometry euclidean-geometry triangle angle

share|cite|improve this question

edited Aug 6 at 17:11

Michael Rozenberg

88.2k1579180

asked Aug 6 at 15:34

Mattiu

780316

marked as duplicate by timon92, José Carlos Santos, Dylan, Xander Henderson, amWhy Aug 7 at 0:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Introducing another variable for one of the angles you haven't solved for may help. Try to write two angles in two different ways using your variables. This will give you a system of two linear equations in two variables.
  – Dzoooks
  Aug 6 at 15:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzoooks i already tried that
  – Mattiu
  Aug 6 at 17:04
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

This question already has an answer here:

• 
  Finding an angle within an 80-80-20 isosceles triangle
  
  4 answers
  
  

I have this problem, that can be solved with elemental knowledge.

In order to challenge, I can't draw extra segments to solved it.

This is the problem and i need to get the measure of $anglex$

My current development is:

Well, the sides with equal colour are congruent, and I have completed all the angles that I could, in red colour.

Then, how i can get the $anglex$ in a elementary form, without draw extra segments? Is possible ?

geometry euclidean-geometry triangle angle

share|cite|improve this question

edited Aug 6 at 17:11

Michael Rozenberg

88.2k1579180

asked Aug 6 at 15:34

Mattiu

780316

This question already has an answer here:

• 
  Finding an angle within an 80-80-20 isosceles triangle
  
  4 answers
  
  

I have this problem, that can be solved with elemental knowledge.

In order to challenge, I can't draw extra segments to solved it.

This is the problem and i need to get the measure of $anglex$

My current development is:

Well, the sides with equal colour are congruent, and I have completed all the angles that I could, in red colour.

Then, how i can get the $anglex$ in a elementary form, without draw extra segments? Is possible ?

This question already has an answer here:

• 
  Finding an angle within an 80-80-20 isosceles triangle
  
  4 answers
  
  

geometry euclidean-geometry triangle angle

share|cite|improve this question

edited Aug 6 at 17:11

Michael Rozenberg

88.2k1579180

asked Aug 6 at 15:34

Mattiu

780316

share|cite|improve this question

share|cite|improve this question

edited Aug 6 at 17:11

Michael Rozenberg

88.2k1579180

edited Aug 6 at 17:11

Michael Rozenberg

88.2k1579180

edited Aug 6 at 17:11

Michael Rozenberg

88.2k1579180

88.2k1579180

asked Aug 6 at 15:34

Mattiu

780316

asked Aug 6 at 15:34

Mattiu

780316

asked Aug 6 at 15:34

Mattiu

780316

780316

marked as duplicate by timon92, José Carlos Santos, Dylan, Xander Henderson, amWhy Aug 7 at 0:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by timon92, José Carlos Santos, Dylan, Xander Henderson, amWhy Aug 7 at 0:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Introducing another variable for one of the angles you haven't solved for may help. Try to write two angles in two different ways using your variables. This will give you a system of two linear equations in two variables.
  – Dzoooks
  Aug 6 at 15:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzoooks i already tried that
  – Mattiu
  Aug 6 at 17:04
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Introducing another variable for one of the angles you haven't solved for may help. Try to write two angles in two different ways using your variables. This will give you a system of two linear equations in two variables.
  – Dzoooks
  Aug 6 at 15:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzoooks i already tried that
  – Mattiu
  Aug 6 at 17:04
  

  
  
  
  

1

1

Introducing another variable for one of the angles you haven't solved for may help. Try to write two angles in two different ways using your variables. This will give you a system of two linear equations in two variables.
– Dzoooks
Aug 6 at 15:49

Introducing another variable for one of the angles you haven't solved for may help. Try to write two angles in two different ways using your variables. This will give you a system of two linear equations in two variables.
– Dzoooks
Aug 6 at 15:49

@Dzoooks i already tried that
– Mattiu
Aug 6 at 17:04

@Dzoooks i already tried that
– Mattiu
Aug 6 at 17:04

add a comment | 

3 Answers
3

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votes

up vote
1
down vote

By the sine rule
$$green/red = sin80/sin40 = 2cos40$$
and also
$$green/red = sin(160 - x)/sin x = sin(20 + x)/sin x = sin20cot x + cos 20$$
Therefore
$$sin20cot x = 2cos40 - cos20 = cos40 - 2sin30sin10 = sin50 - sin10 = 2sin20cos30$$
so $cot x = 2cos30 = sqrt3$ and $x = 30$ degrees.

share|cite|improve this answer

answered Aug 6 at 16:13

Michael Behrend

89225

add a comment | 

up vote
1
down vote

I think, we can get a solution without drawing of an extra segments by trigonometry only.

The geometric solution:

Let $Din FA$ such that $DE||BC$ and $DCcap BE=G$.

Thus, $Delta BGC$ and $Delta DGE$ they are equilaterals and since $$measuredangle BFC=measuredangle FCB=50^circ,$$ we obtain:
$$FB=BC=BG,$$ which says that $$measuredangle FGB=frac180^circ-20^circ2=80^circ,$$
which gives
$$measuredangle FGE=100^circ.$$
But also $measuredangle FDE=100^circ$ and $DE=EG$.

Thus, $Delta DEFcongDelta GEF,$ which says
$$measuredangle FEB=frac12measuredangle DEG=30^circ.$$

share|cite|improve this answer

edited Aug 6 at 16:36

answered Aug 6 at 16:31

Michael Rozenberg

88.2k1579180

add a comment | 

up vote
1
down vote

Here are some solutions to this problem:

share|cite|improve this answer

answered Aug 6 at 19:33

Seyed

5,71231221

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

By the sine rule
$$green/red = sin80/sin40 = 2cos40$$
and also
$$green/red = sin(160 - x)/sin x = sin(20 + x)/sin x = sin20cot x + cos 20$$
Therefore
$$sin20cot x = 2cos40 - cos20 = cos40 - 2sin30sin10 = sin50 - sin10 = 2sin20cos30$$
so $cot x = 2cos30 = sqrt3$ and $x = 30$ degrees.

share|cite|improve this answer

answered Aug 6 at 16:13

Michael Behrend

89225

add a comment | 

up vote
1
down vote

By the sine rule
$$green/red = sin80/sin40 = 2cos40$$
and also
$$green/red = sin(160 - x)/sin x = sin(20 + x)/sin x = sin20cot x + cos 20$$
Therefore
$$sin20cot x = 2cos40 - cos20 = cos40 - 2sin30sin10 = sin50 - sin10 = 2sin20cos30$$
so $cot x = 2cos30 = sqrt3$ and $x = 30$ degrees.

share|cite|improve this answer

answered Aug 6 at 16:13

Michael Behrend

89225

add a comment | 

up vote
1
down vote

up vote
1
down vote

By the sine rule
$$green/red = sin80/sin40 = 2cos40$$
and also
$$green/red = sin(160 - x)/sin x = sin(20 + x)/sin x = sin20cot x + cos 20$$
Therefore
$$sin20cot x = 2cos40 - cos20 = cos40 - 2sin30sin10 = sin50 - sin10 = 2sin20cos30$$
so $cot x = 2cos30 = sqrt3$ and $x = 30$ degrees.

share|cite|improve this answer

answered Aug 6 at 16:13

Michael Behrend

89225

By the sine rule
$$green/red = sin80/sin40 = 2cos40$$
and also
$$green/red = sin(160 - x)/sin x = sin(20 + x)/sin x = sin20cot x + cos 20$$
Therefore
$$sin20cot x = 2cos40 - cos20 = cos40 - 2sin30sin10 = sin50 - sin10 = 2sin20cos30$$
so $cot x = 2cos30 = sqrt3$ and $x = 30$ degrees.

share|cite|improve this answer

answered Aug 6 at 16:13

Michael Behrend

89225

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 16:13

Michael Behrend

89225

answered Aug 6 at 16:13

Michael Behrend

89225

answered Aug 6 at 16:13

Michael Behrend

89225

89225

add a comment | 

add a comment | 

up vote
1
down vote

I think, we can get a solution without drawing of an extra segments by trigonometry only.

The geometric solution:

Let $Din FA$ such that $DE||BC$ and $DCcap BE=G$.

Thus, $Delta BGC$ and $Delta DGE$ they are equilaterals and since $$measuredangle BFC=measuredangle FCB=50^circ,$$ we obtain:
$$FB=BC=BG,$$ which says that $$measuredangle FGB=frac180^circ-20^circ2=80^circ,$$
which gives
$$measuredangle FGE=100^circ.$$
But also $measuredangle FDE=100^circ$ and $DE=EG$.

Thus, $Delta DEFcongDelta GEF,$ which says
$$measuredangle FEB=frac12measuredangle DEG=30^circ.$$

share|cite|improve this answer

edited Aug 6 at 16:36

answered Aug 6 at 16:31

Michael Rozenberg

88.2k1579180

add a comment | 

up vote
1
down vote

I think, we can get a solution without drawing of an extra segments by trigonometry only.

The geometric solution:

Let $Din FA$ such that $DE||BC$ and $DCcap BE=G$.

Thus, $Delta BGC$ and $Delta DGE$ they are equilaterals and since $$measuredangle BFC=measuredangle FCB=50^circ,$$ we obtain:
$$FB=BC=BG,$$ which says that $$measuredangle FGB=frac180^circ-20^circ2=80^circ,$$
which gives
$$measuredangle FGE=100^circ.$$
But also $measuredangle FDE=100^circ$ and $DE=EG$.

Thus, $Delta DEFcongDelta GEF,$ which says
$$measuredangle FEB=frac12measuredangle DEG=30^circ.$$

share|cite|improve this answer

edited Aug 6 at 16:36

answered Aug 6 at 16:31

Michael Rozenberg

88.2k1579180

add a comment | 

up vote
1
down vote

up vote
1
down vote

I think, we can get a solution without drawing of an extra segments by trigonometry only.

The geometric solution:

Let $Din FA$ such that $DE||BC$ and $DCcap BE=G$.

Thus, $Delta BGC$ and $Delta DGE$ they are equilaterals and since $$measuredangle BFC=measuredangle FCB=50^circ,$$ we obtain:
$$FB=BC=BG,$$ which says that $$measuredangle FGB=frac180^circ-20^circ2=80^circ,$$
which gives
$$measuredangle FGE=100^circ.$$
But also $measuredangle FDE=100^circ$ and $DE=EG$.

Thus, $Delta DEFcongDelta GEF,$ which says
$$measuredangle FEB=frac12measuredangle DEG=30^circ.$$

share|cite|improve this answer

edited Aug 6 at 16:36

answered Aug 6 at 16:31

Michael Rozenberg

88.2k1579180

I think, we can get a solution without drawing of an extra segments by trigonometry only.

The geometric solution:

Let $Din FA$ such that $DE||BC$ and $DCcap BE=G$.

Thus, $Delta BGC$ and $Delta DGE$ they are equilaterals and since $$measuredangle BFC=measuredangle FCB=50^circ,$$ we obtain:
$$FB=BC=BG,$$ which says that $$measuredangle FGB=frac180^circ-20^circ2=80^circ,$$
which gives
$$measuredangle FGE=100^circ.$$
But also $measuredangle FDE=100^circ$ and $DE=EG$.

Thus, $Delta DEFcongDelta GEF,$ which says
$$measuredangle FEB=frac12measuredangle DEG=30^circ.$$

share|cite|improve this answer

edited Aug 6 at 16:36

answered Aug 6 at 16:31

Michael Rozenberg

88.2k1579180

share|cite|improve this answer

share|cite|improve this answer

edited Aug 6 at 16:36

edited Aug 6 at 16:36

edited Aug 6 at 16:36

answered Aug 6 at 16:31

Michael Rozenberg

88.2k1579180

answered Aug 6 at 16:31

Michael Rozenberg

88.2k1579180

answered Aug 6 at 16:31

Michael Rozenberg

88.2k1579180

88.2k1579180

add a comment | 

add a comment | 

up vote
1
down vote

Here are some solutions to this problem:

share|cite|improve this answer

answered Aug 6 at 19:33

Seyed

5,71231221

add a comment | 

up vote
1
down vote

Here are some solutions to this problem:

share|cite|improve this answer

answered Aug 6 at 19:33

Seyed

5,71231221

add a comment | 

up vote
1
down vote

up vote
1
down vote

Here are some solutions to this problem:

share|cite|improve this answer

answered Aug 6 at 19:33

Seyed

5,71231221

Here are some solutions to this problem:

share|cite|improve this answer

answered Aug 6 at 19:33

Seyed

5,71231221

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 19:33

Seyed

5,71231221

answered Aug 6 at 19:33

Seyed

5,71231221

answered Aug 6 at 19:33

Seyed

5,71231221

5,71231221

add a comment | 

add a comment |