Find x angle in triangleSine Law or No?In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on...
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Find x angle in triangle
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Find x angle in triangle
Sine Law or No?In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on $AC$, with $BC=AD$. Find $angle DBC$. Where's my error?How do I find the base angles without a vertex angle in a isosceles triangle?Prove triangle made from two altitudes and midpoint is isoscelesHas triangle an angle?Geometry: Find angle x in triangleAlternative proof for the equality of two angles in an isosceles triangle.formula for calculating any angle of an isosceles triangleFind unknown angles of a triangle!!!Find the value of an angleEquation of A line in an Isosceles TriangleKiselev's geometry Problem 67: In an isosceles triangle, two medians/bisectors/altitudes are congruent
$begingroup$
I need to find angle x in this isosceles triangle(20-80-80), by using pure geometry, if i can say so. If my calculations are correct (i tried another approach) answer should be 30, but there should be 'easy' way to find this.
Also i found many Langley’s Adventitious Angles exercises which are very similar to mine but yet different.
geometry euclidean-geometry triangle
edited 1 hour ago
Michael Rozenberg
106k1893198
asked 2 hours ago
Andriy KhrystyanovichAndriy Khrystyanovich
1134
New contributor
Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 1 more comment
$begingroup$
I need to find angle x in this isosceles triangle(20-80-80), by using pure geometry, if i can say so. If my calculations are correct (i tried another approach) answer should be 30, but there should be 'easy' way to find this.
Also i found many Langley’s Adventitious Angles exercises which are very similar to mine but yet different.
geometry euclidean-geometry triangle
edited 1 hour ago
Michael Rozenberg
106k1893198
asked 2 hours ago
Andriy KhrystyanovichAndriy Khrystyanovich
1134
New contributor
Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
thats trigonometry solution. not what im looking for
$endgroup$
– Andriy Khrystyanovich
1 hour ago
$begingroup$
@Blue The solution in the linked topic by trigonometry only, because the topic starter looked for trigonometric solution only. I think we need to open this topic.
$endgroup$
– Michael Rozenberg
1 hour ago
$begingroup$
Note: A trigonometric solution is offered in this question.
$endgroup$
– Blue
1 hour ago
$begingroup$
math.stackexchange.com/a/3126628/480425. Here I propose a couple of nice solutions using simple Euclidean geometry.
$endgroup$
– Matteo
1 hour ago
$begingroup$
Possible duplicate of In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on $AC$, with $BC=AD$. Find $angle DBC$. Where's my error?
$endgroup$
– Aretino
1 hour ago
|
show 1 more comment
$begingroup$
I need to find angle x in this isosceles triangle(20-80-80), by using pure geometry, if i can say so. If my calculations are correct (i tried another approach) answer should be 30, but there should be 'easy' way to find this.
Also i found many Langley’s Adventitious Angles exercises which are very similar to mine but yet different.
geometry euclidean-geometry triangle
edited 1 hour ago
Michael Rozenberg
106k1893198
asked 2 hours ago
Andriy KhrystyanovichAndriy Khrystyanovich
1134
New contributor
Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I need to find angle x in this isosceles triangle(20-80-80), by using pure geometry, if i can say so. If my calculations are correct (i tried another approach) answer should be 30, but there should be 'easy' way to find this.
Also i found many Langley’s Adventitious Angles exercises which are very similar to mine but yet different.
geometry euclidean-geometry triangle
geometry euclidean-geometry triangle
edited 1 hour ago
Michael Rozenberg
106k1893198
asked 2 hours ago
Andriy KhrystyanovichAndriy Khrystyanovich
1134
New contributor
Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Michael Rozenberg
106k1893198
asked 2 hours ago
Andriy KhrystyanovichAndriy Khrystyanovich
1134
New contributor
Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Michael Rozenberg
106k1893198
edited 1 hour ago
Michael Rozenberg
106k1893198
edited 1 hour ago
Michael Rozenberg
106k1893198
106k1893198
asked 2 hours ago
Andriy KhrystyanovichAndriy Khrystyanovich
1134
New contributor
Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Andriy KhrystyanovichAndriy Khrystyanovich
1134
asked 2 hours ago
Andriy KhrystyanovichAndriy Khrystyanovich
1134
1134
New contributor
Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
thats trigonometry solution. not what im looking for
$endgroup$
– Andriy Khrystyanovich
1 hour ago
$begingroup$
@Blue The solution in the linked topic by trigonometry only, because the topic starter looked for trigonometric solution only. I think we need to open this topic.
$endgroup$
– Michael Rozenberg
1 hour ago
$begingroup$
Note: A trigonometric solution is offered in this question.
$endgroup$
– Blue
1 hour ago
$begingroup$
math.stackexchange.com/a/3126628/480425. Here I propose a couple of nice solutions using simple Euclidean geometry.
$endgroup$
– Matteo
1 hour ago
$begingroup$
Possible duplicate of In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on $AC$, with $BC=AD$. Find $angle DBC$. Where's my error?
$endgroup$
– Aretino
1 hour ago
|
show 1 more comment
1
$begingroup$
thats trigonometry solution. not what im looking for
$endgroup$
– Andriy Khrystyanovich
1 hour ago
$begingroup$
@Blue The solution in the linked topic by trigonometry only, because the topic starter looked for trigonometric solution only. I think we need to open this topic.
$endgroup$
– Michael Rozenberg
1 hour ago
$begingroup$
Note: A trigonometric solution is offered in this question.
$endgroup$
– Blue
1 hour ago
$begingroup$
math.stackexchange.com/a/3126628/480425. Here I propose a couple of nice solutions using simple Euclidean geometry.
$endgroup$
– Matteo
1 hour ago
$begingroup$
Possible duplicate of In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on $AC$, with $BC=AD$. Find $angle DBC$. Where's my error?
$endgroup$
– Aretino
1 hour ago
1
1
$begingroup$
thats trigonometry solution. not what im looking for
$endgroup$
– Andriy Khrystyanovich
1 hour ago
$begingroup$
thats trigonometry solution. not what im looking for
$endgroup$
– Andriy Khrystyanovich
1 hour ago
$begingroup$
@Blue The solution in the linked topic by trigonometry only, because the topic starter looked for trigonometric solution only. I think we need to open this topic.
$endgroup$
– Michael Rozenberg
1 hour ago
$begingroup$
@Blue The solution in the linked topic by trigonometry only, because the topic starter looked for trigonometric solution only. I think we need to open this topic.
$endgroup$
– Michael Rozenberg
1 hour ago
$begingroup$
Note: A trigonometric solution is offered in this question.
$endgroup$
– Blue
1 hour ago
$begingroup$
Note: A trigonometric solution is offered in this question.
$endgroup$
– Blue
1 hour ago
$begingroup$
math.stackexchange.com/a/3126628/480425. Here I propose a couple of nice solutions using simple Euclidean geometry.
$endgroup$
– Matteo
1 hour ago
$begingroup$
math.stackexchange.com/a/3126628/480425. Here I propose a couple of nice solutions using simple Euclidean geometry.
$endgroup$
– Matteo
1 hour ago
$begingroup$
Possible duplicate of In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on $AC$, with $BC=AD$. Find $angle DBC$. Where's my error?
$endgroup$
– Aretino
1 hour ago
$begingroup$
Possible duplicate of In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on $AC$, with $BC=AD$. Find $angle DBC$. Where's my error?
$endgroup$
– Aretino
1 hour ago
|
show 1 more comment
4 Answers
4
active
oldest
votes
$begingroup$
Construct an equilateral triangle which it's sides are equal to the base of the main triangle.
answered 1 hour ago
SeyedSeyed
6,92341424
$endgroup$
$begingroup$
Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer.
$endgroup$
– greedoid
1 hour ago
$begingroup$
@greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules.
$endgroup$
– Seyed
1 hour ago
$begingroup$
Yes, that is correct if you want another upvote.
$endgroup$
– greedoid
1 hour ago
1
$begingroup$
@greedoid, Thanks for your advise.
$endgroup$
– Seyed
1 hour ago
add a comment |
$begingroup$
Let in $Delta ABC$ we have $AB=AC$, $measuredangle A=20^{circ}$ and $measuredangle ADC=x$ as on your picture.
Let $Min AB$ such that $AD=MD$ and $Kin DC$ such that $MK=AD$.
Also, let $B'in MB$ such that $MB'=AD$ and $C'in KC$ such that $B'C'||BC.$
Thus, $$measuredangle MKA=measuredangle MDK=2cdot20^{circ}=40^{circ}$$ and from here
$$measuredangle B'MK=40^{circ}+20^{circ}=60^{circ},$$ which says $$B'K=MB'=AD=BC.$$
But $$measuredangle B'KC'=60^{circ}+20^{circ}=80^{circ}=measuredangle BCA=measuredangle B'C'A.$$
Thus, $$B'C'=B'K=AD=BC,$$ which says that
$$Bequiv B'$$ and $$Cequiv C'.$$
Id est,
$$measuredangle BDC=10^{circ}+20^{circ}=30^{circ}.$$
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
$begingroup$
Nice as always +1
$endgroup$
– greedoid
1 hour ago
add a comment |
$begingroup$
construct triangle BCE congruent to ADB.
so AB = BE, angle ABE = 80° - 20° = 60°
Thus triangle ABE is equilateral.
AB = AE = AC, since angle CAE = 60° - 20° =40°
angle AEC = (180° - 40° )/2 = 70°
so x = 20° + angle ABD = 20° + angle CBE = 20° + (70° - 60° ) = 30°
answered 1 hour ago
qsmyqsmy
19018
$endgroup$
$begingroup$
Very nice +1......
$endgroup$
– greedoid
1 hour ago
add a comment |
$begingroup$
And my second solution is as follow:
answered 1 hour ago
SeyedSeyed
6,92341424
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Construct an equilateral triangle which it's sides are equal to the base of the main triangle.
answered 1 hour ago
SeyedSeyed
6,92341424
$endgroup$
$begingroup$
Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer.
$endgroup$
– greedoid
1 hour ago
$begingroup$
@greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules.
$endgroup$
– Seyed
1 hour ago
$begingroup$
Yes, that is correct if you want another upvote.
$endgroup$
– greedoid
1 hour ago
1
$begingroup$
@greedoid, Thanks for your advise.
$endgroup$
– Seyed
1 hour ago
add a comment |
$begingroup$
Construct an equilateral triangle which it's sides are equal to the base of the main triangle.
answered 1 hour ago
SeyedSeyed
6,92341424
$endgroup$
$begingroup$
Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer.
$endgroup$
– greedoid
1 hour ago
$begingroup$
@greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules.
$endgroup$
– Seyed
1 hour ago
$begingroup$
Yes, that is correct if you want another upvote.
$endgroup$
– greedoid
1 hour ago
1
$begingroup$
@greedoid, Thanks for your advise.
$endgroup$
– Seyed
1 hour ago
add a comment |
$begingroup$
Construct an equilateral triangle which it's sides are equal to the base of the main triangle.
answered 1 hour ago
SeyedSeyed
6,92341424
$endgroup$
Construct an equilateral triangle which it's sides are equal to the base of the main triangle.
answered 1 hour ago
SeyedSeyed
6,92341424
edited 1 hour ago
answered 1 hour ago
SeyedSeyed
6,92341424
answered 1 hour ago
SeyedSeyed
6,92341424
answered 1 hour ago
SeyedSeyed
6,92341424
6,92341424
$begingroup$
Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer.
$endgroup$
– greedoid
1 hour ago
$begingroup$
@greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules.
$endgroup$
– Seyed
1 hour ago
$begingroup$
Yes, that is correct if you want another upvote.
$endgroup$
– greedoid
1 hour ago
1
$begingroup$
@greedoid, Thanks for your advise.
$endgroup$
– Seyed
1 hour ago
add a comment |
$begingroup$
Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer.
$endgroup$
– greedoid
1 hour ago
$begingroup$
@greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules.
$endgroup$
– Seyed
1 hour ago
$begingroup$
Yes, that is correct if you want another upvote.
$endgroup$
– greedoid
1 hour ago
1
$begingroup$
@greedoid, Thanks for your advise.
$endgroup$
– Seyed
1 hour ago
$begingroup$
Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer.
$endgroup$
– greedoid
1 hour ago
$begingroup$
Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer.
$endgroup$
– greedoid
1 hour ago
$begingroup$
@greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules.
$endgroup$
– Seyed
1 hour ago
$begingroup$
@greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules.
$endgroup$
– Seyed
1 hour ago
$begingroup$
Yes, that is correct if you want another upvote.
$endgroup$
– greedoid
1 hour ago
$begingroup$
Yes, that is correct if you want another upvote.
$endgroup$
– greedoid
1 hour ago
1
1
$begingroup$
@greedoid, Thanks for your advise.
$endgroup$
– Seyed
1 hour ago
$begingroup$
@greedoid, Thanks for your advise.
$endgroup$
– Seyed
1 hour ago
add a comment |
$begingroup$
Let in $Delta ABC$ we have $AB=AC$, $measuredangle A=20^{circ}$ and $measuredangle ADC=x$ as on your picture.
Let $Min AB$ such that $AD=MD$ and $Kin DC$ such that $MK=AD$.
Also, let $B'in MB$ such that $MB'=AD$ and $C'in KC$ such that $B'C'||BC.$
Thus, $$measuredangle MKA=measuredangle MDK=2cdot20^{circ}=40^{circ}$$ and from here
$$measuredangle B'MK=40^{circ}+20^{circ}=60^{circ},$$ which says $$B'K=MB'=AD=BC.$$
But $$measuredangle B'KC'=60^{circ}+20^{circ}=80^{circ}=measuredangle BCA=measuredangle B'C'A.$$
Thus, $$B'C'=B'K=AD=BC,$$ which says that
$$Bequiv B'$$ and $$Cequiv C'.$$
Id est,
$$measuredangle BDC=10^{circ}+20^{circ}=30^{circ}.$$
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
$begingroup$
Nice as always +1
$endgroup$
– greedoid
1 hour ago
add a comment |
$begingroup$
Let in $Delta ABC$ we have $AB=AC$, $measuredangle A=20^{circ}$ and $measuredangle ADC=x$ as on your picture.
Let $Min AB$ such that $AD=MD$ and $Kin DC$ such that $MK=AD$.
Also, let $B'in MB$ such that $MB'=AD$ and $C'in KC$ such that $B'C'||BC.$
Thus, $$measuredangle MKA=measuredangle MDK=2cdot20^{circ}=40^{circ}$$ and from here
$$measuredangle B'MK=40^{circ}+20^{circ}=60^{circ},$$ which says $$B'K=MB'=AD=BC.$$
But $$measuredangle B'KC'=60^{circ}+20^{circ}=80^{circ}=measuredangle BCA=measuredangle B'C'A.$$
Thus, $$B'C'=B'K=AD=BC,$$ which says that
$$Bequiv B'$$ and $$Cequiv C'.$$
Id est,
$$measuredangle BDC=10^{circ}+20^{circ}=30^{circ}.$$
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
$begingroup$
Nice as always +1
$endgroup$
– greedoid
1 hour ago
add a comment |
$begingroup$
Let in $Delta ABC$ we have $AB=AC$, $measuredangle A=20^{circ}$ and $measuredangle ADC=x$ as on your picture.
Let $Min AB$ such that $AD=MD$ and $Kin DC$ such that $MK=AD$.
Also, let $B'in MB$ such that $MB'=AD$ and $C'in KC$ such that $B'C'||BC.$
Thus, $$measuredangle MKA=measuredangle MDK=2cdot20^{circ}=40^{circ}$$ and from here
$$measuredangle B'MK=40^{circ}+20^{circ}=60^{circ},$$ which says $$B'K=MB'=AD=BC.$$
But $$measuredangle B'KC'=60^{circ}+20^{circ}=80^{circ}=measuredangle BCA=measuredangle B'C'A.$$
Thus, $$B'C'=B'K=AD=BC,$$ which says that
$$Bequiv B'$$ and $$Cequiv C'.$$
Id est,
$$measuredangle BDC=10^{circ}+20^{circ}=30^{circ}.$$
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
Let in $Delta ABC$ we have $AB=AC$, $measuredangle A=20^{circ}$ and $measuredangle ADC=x$ as on your picture.
Let $Min AB$ such that $AD=MD$ and $Kin DC$ such that $MK=AD$.
Also, let $B'in MB$ such that $MB'=AD$ and $C'in KC$ such that $B'C'||BC.$
Thus, $$measuredangle MKA=measuredangle MDK=2cdot20^{circ}=40^{circ}$$ and from here
$$measuredangle B'MK=40^{circ}+20^{circ}=60^{circ},$$ which says $$B'K=MB'=AD=BC.$$
But $$measuredangle B'KC'=60^{circ}+20^{circ}=80^{circ}=measuredangle BCA=measuredangle B'C'A.$$
Thus, $$B'C'=B'K=AD=BC,$$ which says that
$$Bequiv B'$$ and $$Cequiv C'.$$
Id est,
$$measuredangle BDC=10^{circ}+20^{circ}=30^{circ}.$$
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
edited 1 hour ago
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
106k1893198
$begingroup$
Nice as always +1
$endgroup$
– greedoid
1 hour ago
add a comment |
$begingroup$
Nice as always +1
$endgroup$
– greedoid
1 hour ago
$begingroup$
Nice as always +1
$endgroup$
– greedoid
1 hour ago
$begingroup$
Nice as always +1
$endgroup$
– greedoid
1 hour ago
add a comment |
$begingroup$
construct triangle BCE congruent to ADB.
so AB = BE, angle ABE = 80° - 20° = 60°
Thus triangle ABE is equilateral.
AB = AE = AC, since angle CAE = 60° - 20° =40°
angle AEC = (180° - 40° )/2 = 70°
so x = 20° + angle ABD = 20° + angle CBE = 20° + (70° - 60° ) = 30°
answered 1 hour ago
qsmyqsmy
19018
$endgroup$
$begingroup$
Very nice +1......
$endgroup$
– greedoid
1 hour ago
add a comment |
$begingroup$
construct triangle BCE congruent to ADB.
so AB = BE, angle ABE = 80° - 20° = 60°
Thus triangle ABE is equilateral.
AB = AE = AC, since angle CAE = 60° - 20° =40°
angle AEC = (180° - 40° )/2 = 70°
so x = 20° + angle ABD = 20° + angle CBE = 20° + (70° - 60° ) = 30°
answered 1 hour ago
qsmyqsmy
19018
$endgroup$
$begingroup$
Very nice +1......
$endgroup$
– greedoid
1 hour ago
add a comment |
$begingroup$
construct triangle BCE congruent to ADB.
so AB = BE, angle ABE = 80° - 20° = 60°
Thus triangle ABE is equilateral.
AB = AE = AC, since angle CAE = 60° - 20° =40°
angle AEC = (180° - 40° )/2 = 70°
so x = 20° + angle ABD = 20° + angle CBE = 20° + (70° - 60° ) = 30°
answered 1 hour ago
qsmyqsmy
19018
$endgroup$
construct triangle BCE congruent to ADB.
so AB = BE, angle ABE = 80° - 20° = 60°
Thus triangle ABE is equilateral.
AB = AE = AC, since angle CAE = 60° - 20° =40°
angle AEC = (180° - 40° )/2 = 70°
so x = 20° + angle ABD = 20° + angle CBE = 20° + (70° - 60° ) = 30°
answered 1 hour ago
qsmyqsmy
19018
answered 1 hour ago
qsmyqsmy
19018
answered 1 hour ago
qsmyqsmy
19018
answered 1 hour ago
qsmyqsmy
19018
19018
$begingroup$
Very nice +1......
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– greedoid
1 hour ago
add a comment |
$begingroup$
Very nice +1......
$endgroup$
– greedoid
1 hour ago
$begingroup$
Very nice +1......
$endgroup$
– greedoid
1 hour ago
$begingroup$
Very nice +1......
$endgroup$
– greedoid
1 hour ago
add a comment |
$begingroup$
And my second solution is as follow:
answered 1 hour ago
SeyedSeyed
6,92341424
$endgroup$
add a comment |
$begingroup$
And my second solution is as follow:
answered 1 hour ago
SeyedSeyed
6,92341424
$endgroup$
add a comment |
$begingroup$
And my second solution is as follow:
answered 1 hour ago
SeyedSeyed
6,92341424
$endgroup$
And my second solution is as follow:
answered 1 hour ago
SeyedSeyed
6,92341424
answered 1 hour ago
SeyedSeyed
6,92341424
answered 1 hour ago
SeyedSeyed
6,92341424
answered 1 hour ago
SeyedSeyed
6,92341424
6,92341424
add a comment |
add a comment |
Andriy Khrystyanovich is a new contributor. Be nice, and check out our Code of Conduct.
Andriy Khrystyanovich is a new contributor. Be nice, and check out our Code of Conduct.
Andriy Khrystyanovich is a new contributor. Be nice, and check out our Code of Conduct.
Andriy Khrystyanovich is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
thats trigonometry solution. not what im looking for
$endgroup$
– Andriy Khrystyanovich
1 hour ago
$begingroup$
@Blue The solution in the linked topic by trigonometry only, because the topic starter looked for trigonometric solution only. I think we need to open this topic.
$endgroup$
– Michael Rozenberg
1 hour ago
$begingroup$
Note: A trigonometric solution is offered in this question.
$endgroup$
– Blue
1 hour ago
$begingroup$
math.stackexchange.com/a/3126628/480425. Here I propose a couple of nice solutions using simple Euclidean geometry.
$endgroup$
– Matteo
1 hour ago
$begingroup$
Possible duplicate of In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on $AC$, with $BC=AD$. Find $angle DBC$. Where's my error?
$endgroup$
– Aretino
1 hour ago