How do I construct this triangle [closed]
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I was trying to draw the following triangle in latex tikz and I just could not find a way to do it with respect to the given conditions.
- Is it possible to construct it without trigonometry or analytical geometry?
- Is it possible to show that it cannot be done without trigonometry or analytical geometry?
- (Regrettably) I tried to do it via solving some trigonometrical equations, then I realized I am getting nowhere. So, how would I approach with trigonometry?
Note 1: In the actual drawing, $c=6$, but that does not matter, obviously.
Note 2: I am not looking for LaTeX help.
geometry trigonometry triangle
edited Jan 20 at 12:52
Martin Sleziak
44.7k10118272
asked Jan 20 at 10:10
blackenedblackened
332211
$endgroup$
closed as off-topic by Cesareo, TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn Jan 20 at 22:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 1 more comment
$begingroup$
I was trying to draw the following triangle in latex tikz and I just could not find a way to do it with respect to the given conditions.
- Is it possible to construct it without trigonometry or analytical geometry?
- Is it possible to show that it cannot be done without trigonometry or analytical geometry?
- (Regrettably) I tried to do it via solving some trigonometrical equations, then I realized I am getting nowhere. So, how would I approach with trigonometry?
Note 1: In the actual drawing, $c=6$, but that does not matter, obviously.
Note 2: I am not looking for LaTeX help.
geometry trigonometry triangle
edited Jan 20 at 12:52
Martin Sleziak
44.7k10118272
asked Jan 20 at 10:10
blackenedblackened
332211
$endgroup$
closed as off-topic by Cesareo, TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn Jan 20 at 22:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
What concrete is given in the exercise?
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:13
$begingroup$
Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
$endgroup$
– Henry
Jan 20 at 10:15
$begingroup$
And $$angle {CEB}=120^{circ}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:18
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@Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
$endgroup$
– blackened
Jan 20 at 10:20
1
$begingroup$
"figure not to scale"...
$endgroup$
– David Mitra
Jan 20 at 10:21
|
show 1 more comment
$begingroup$
I was trying to draw the following triangle in latex tikz and I just could not find a way to do it with respect to the given conditions.
- Is it possible to construct it without trigonometry or analytical geometry?
- Is it possible to show that it cannot be done without trigonometry or analytical geometry?
- (Regrettably) I tried to do it via solving some trigonometrical equations, then I realized I am getting nowhere. So, how would I approach with trigonometry?
Note 1: In the actual drawing, $c=6$, but that does not matter, obviously.
Note 2: I am not looking for LaTeX help.
geometry trigonometry triangle
edited Jan 20 at 12:52
Martin Sleziak
44.7k10118272
asked Jan 20 at 10:10
blackenedblackened
332211
$endgroup$
I was trying to draw the following triangle in latex tikz and I just could not find a way to do it with respect to the given conditions.
- Is it possible to construct it without trigonometry or analytical geometry?
- Is it possible to show that it cannot be done without trigonometry or analytical geometry?
- (Regrettably) I tried to do it via solving some trigonometrical equations, then I realized I am getting nowhere. So, how would I approach with trigonometry?
Note 1: In the actual drawing, $c=6$, but that does not matter, obviously.
Note 2: I am not looking for LaTeX help.
geometry trigonometry triangle
geometry trigonometry triangle
edited Jan 20 at 12:52
Martin Sleziak
44.7k10118272
asked Jan 20 at 10:10
blackenedblackened
332211
edited Jan 20 at 12:52
Martin Sleziak
44.7k10118272
asked Jan 20 at 10:10
blackenedblackened
332211
edited Jan 20 at 12:52
Martin Sleziak
44.7k10118272
edited Jan 20 at 12:52
Martin Sleziak
44.7k10118272
edited Jan 20 at 12:52
Martin Sleziak
44.7k10118272
44.7k10118272
asked Jan 20 at 10:10
blackenedblackened
332211
asked Jan 20 at 10:10
blackenedblackened
332211
asked Jan 20 at 10:10
blackenedblackened
332211
332211
closed as off-topic by Cesareo, TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn Jan 20 at 22:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Cesareo, TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn Jan 20 at 22:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
What concrete is given in the exercise?
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:13
$begingroup$
Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
$endgroup$
– Henry
Jan 20 at 10:15
$begingroup$
And $$angle {CEB}=120^{circ}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:18
$begingroup$
@Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
$endgroup$
– blackened
Jan 20 at 10:20
1
$begingroup$
"figure not to scale"...
$endgroup$
– David Mitra
Jan 20 at 10:21
|
show 1 more comment
1
$begingroup$
What concrete is given in the exercise?
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:13
$begingroup$
Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
$endgroup$
– Henry
Jan 20 at 10:15
$begingroup$
And $$angle {CEB}=120^{circ}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:18
$begingroup$
@Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
$endgroup$
– blackened
Jan 20 at 10:20
1
$begingroup$
"figure not to scale"...
$endgroup$
– David Mitra
Jan 20 at 10:21
1
1
$begingroup$
What concrete is given in the exercise?
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:13
$begingroup$
What concrete is given in the exercise?
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:13
$begingroup$
Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
$endgroup$
– Henry
Jan 20 at 10:15
$begingroup$
Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
$endgroup$
– Henry
Jan 20 at 10:15
$begingroup$
And $$angle {CEB}=120^{circ}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:18
$begingroup$
And $$angle {CEB}=120^{circ}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:18
$begingroup$
@Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
$endgroup$
– blackened
Jan 20 at 10:20
$begingroup$
@Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
$endgroup$
– blackened
Jan 20 at 10:20
1
1
$begingroup$
"figure not to scale"...
$endgroup$
– David Mitra
Jan 20 at 10:21
$begingroup$
"figure not to scale"...
$endgroup$
– David Mitra
Jan 20 at 10:21
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
As noted in the comments, a little angle-chasing shows that $phi + psi = 60^circ$. As a result, we can find $A^prime$ on $overline{BC}$ such that $triangle ABEcong triangle A^prime BE$, as shown:
Now, recall that, in a triangle, smaller angles are opposite smaller sides.
$$begin{align}
triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
triangle A^prime CE: quad phantom{60^circ-;}phi < 60^circ &quadimpliesquad p < r
end{align}$$
Consequently, $2p < q + r$, which contradicts the desired condition. $square$
answered Jan 20 at 11:17
BlueBlue
48.5k870154
$endgroup$
add a comment |
$begingroup$
This triangle cannot be constructed for any $phi$. If we assume $phi$ can be any value and still have the triangle be constructed, we can see that length $BE = 2 cdot a$ as $phi to 60^circ$. Then applying the sine rule on lengths BE and AE gives: $$frac{2cdot a}{sin{(phi + 60^circ)}}=frac{a}{sin{(60^circ - phi)}}$$
For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt{3} cdot a$ by the sine rule and we are left with a right angled triangle ABC. By Pythagoras' we then have $AB^2+BC^2=AC^2$ which is not true with the given lengths - a contradiction. So there is no way to construct the given triangle.
answered Jan 20 at 10:42
Peter ForemanPeter Foreman
2,16913
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As noted in the comments, a little angle-chasing shows that $phi + psi = 60^circ$. As a result, we can find $A^prime$ on $overline{BC}$ such that $triangle ABEcong triangle A^prime BE$, as shown:
Now, recall that, in a triangle, smaller angles are opposite smaller sides.
$$begin{align}
triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
triangle A^prime CE: quad phantom{60^circ-;}phi < 60^circ &quadimpliesquad p < r
end{align}$$
Consequently, $2p < q + r$, which contradicts the desired condition. $square$
answered Jan 20 at 11:17
BlueBlue
48.5k870154
$endgroup$
add a comment |
$begingroup$
As noted in the comments, a little angle-chasing shows that $phi + psi = 60^circ$. As a result, we can find $A^prime$ on $overline{BC}$ such that $triangle ABEcong triangle A^prime BE$, as shown:
Now, recall that, in a triangle, smaller angles are opposite smaller sides.
$$begin{align}
triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
triangle A^prime CE: quad phantom{60^circ-;}phi < 60^circ &quadimpliesquad p < r
end{align}$$
Consequently, $2p < q + r$, which contradicts the desired condition. $square$
answered Jan 20 at 11:17
BlueBlue
48.5k870154
$endgroup$
add a comment |
$begingroup$
As noted in the comments, a little angle-chasing shows that $phi + psi = 60^circ$. As a result, we can find $A^prime$ on $overline{BC}$ such that $triangle ABEcong triangle A^prime BE$, as shown:
Now, recall that, in a triangle, smaller angles are opposite smaller sides.
$$begin{align}
triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
triangle A^prime CE: quad phantom{60^circ-;}phi < 60^circ &quadimpliesquad p < r
end{align}$$
Consequently, $2p < q + r$, which contradicts the desired condition. $square$
answered Jan 20 at 11:17
BlueBlue
48.5k870154
$endgroup$
As noted in the comments, a little angle-chasing shows that $phi + psi = 60^circ$. As a result, we can find $A^prime$ on $overline{BC}$ such that $triangle ABEcong triangle A^prime BE$, as shown:
Now, recall that, in a triangle, smaller angles are opposite smaller sides.
$$begin{align}
triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
triangle A^prime CE: quad phantom{60^circ-;}phi < 60^circ &quadimpliesquad p < r
end{align}$$
Consequently, $2p < q + r$, which contradicts the desired condition. $square$
answered Jan 20 at 11:17
BlueBlue
48.5k870154
edited Jan 20 at 11:23
answered Jan 20 at 11:17
BlueBlue
48.5k870154
answered Jan 20 at 11:17
BlueBlue
48.5k870154
answered Jan 20 at 11:17
BlueBlue
48.5k870154
48.5k870154
add a comment |
add a comment |
$begingroup$
This triangle cannot be constructed for any $phi$. If we assume $phi$ can be any value and still have the triangle be constructed, we can see that length $BE = 2 cdot a$ as $phi to 60^circ$. Then applying the sine rule on lengths BE and AE gives: $$frac{2cdot a}{sin{(phi + 60^circ)}}=frac{a}{sin{(60^circ - phi)}}$$
For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt{3} cdot a$ by the sine rule and we are left with a right angled triangle ABC. By Pythagoras' we then have $AB^2+BC^2=AC^2$ which is not true with the given lengths - a contradiction. So there is no way to construct the given triangle.
answered Jan 20 at 10:42
Peter ForemanPeter Foreman
2,16913
$endgroup$
add a comment |
$begingroup$
This triangle cannot be constructed for any $phi$. If we assume $phi$ can be any value and still have the triangle be constructed, we can see that length $BE = 2 cdot a$ as $phi to 60^circ$. Then applying the sine rule on lengths BE and AE gives: $$frac{2cdot a}{sin{(phi + 60^circ)}}=frac{a}{sin{(60^circ - phi)}}$$
For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt{3} cdot a$ by the sine rule and we are left with a right angled triangle ABC. By Pythagoras' we then have $AB^2+BC^2=AC^2$ which is not true with the given lengths - a contradiction. So there is no way to construct the given triangle.
answered Jan 20 at 10:42
Peter ForemanPeter Foreman
2,16913
$endgroup$
add a comment |
$begingroup$
This triangle cannot be constructed for any $phi$. If we assume $phi$ can be any value and still have the triangle be constructed, we can see that length $BE = 2 cdot a$ as $phi to 60^circ$. Then applying the sine rule on lengths BE and AE gives: $$frac{2cdot a}{sin{(phi + 60^circ)}}=frac{a}{sin{(60^circ - phi)}}$$
For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt{3} cdot a$ by the sine rule and we are left with a right angled triangle ABC. By Pythagoras' we then have $AB^2+BC^2=AC^2$ which is not true with the given lengths - a contradiction. So there is no way to construct the given triangle.
answered Jan 20 at 10:42
Peter ForemanPeter Foreman
2,16913
$endgroup$
This triangle cannot be constructed for any $phi$. If we assume $phi$ can be any value and still have the triangle be constructed, we can see that length $BE = 2 cdot a$ as $phi to 60^circ$. Then applying the sine rule on lengths BE and AE gives: $$frac{2cdot a}{sin{(phi + 60^circ)}}=frac{a}{sin{(60^circ - phi)}}$$
For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt{3} cdot a$ by the sine rule and we are left with a right angled triangle ABC. By Pythagoras' we then have $AB^2+BC^2=AC^2$ which is not true with the given lengths - a contradiction. So there is no way to construct the given triangle.
answered Jan 20 at 10:42
Peter ForemanPeter Foreman
2,16913
answered Jan 20 at 10:42
Peter ForemanPeter Foreman
2,16913
answered Jan 20 at 10:42
Peter ForemanPeter Foreman
2,16913
answered Jan 20 at 10:42
Peter ForemanPeter Foreman
2,16913
2,16913
add a comment |
add a comment |
1
$begingroup$
What concrete is given in the exercise?
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:13
$begingroup$
Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
$endgroup$
– Henry
Jan 20 at 10:15
$begingroup$
And $$angle {CEB}=120^{circ}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:18
$begingroup$
@Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
$endgroup$
– blackened
Jan 20 at 10:20
1
$begingroup$
"figure not to scale"...
$endgroup$
– David Mitra
Jan 20 at 10:21