Find angle without trigonometry
up vote
5
down vote
favorite
I solved the following problem using the sine law. Desired value is $angle MAC=10°$. Can you find a geometric solution?
geometry euclidean-geometry geometric-transformation
edited yesterday
greedoid
35.2k114489
asked yesterday
Ghartal
2,86211335
add a comment |
up vote
5
down vote
favorite
I solved the following problem using the sine law. Desired value is $angle MAC=10°$. Can you find a geometric solution?
geometry euclidean-geometry geometric-transformation
edited yesterday
greedoid
35.2k114489
asked yesterday
Ghartal
2,86211335
It seems like it is a least not solvable via algebra hence the given conditions lead to a system of four equation with four unknowns where the determinate of the matrix of coefficients equals zero.
– mrtaurho
yesterday
1
Trigonometry expresses relationships between the ratios of various lengths within a right triangle. Any triangle can be broken up into some combination of right triangles. In principle any statement in trigonometry has a corresponding expression in solely geometric terms.
– TurlocTheRed
yesterday
I thought the Sine Law For Triangles $was$ geometry.
– DanielWainfleet
yesterday
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I solved the following problem using the sine law. Desired value is $angle MAC=10°$. Can you find a geometric solution?
geometry euclidean-geometry geometric-transformation
edited yesterday
greedoid
35.2k114489
asked yesterday
Ghartal
2,86211335
I solved the following problem using the sine law. Desired value is $angle MAC=10°$. Can you find a geometric solution?
geometry euclidean-geometry geometric-transformation
geometry euclidean-geometry geometric-transformation
edited yesterday
greedoid
35.2k114489
asked yesterday
Ghartal
2,86211335
edited yesterday
greedoid
35.2k114489
asked yesterday
Ghartal
2,86211335
edited yesterday
greedoid
35.2k114489
edited yesterday
greedoid
35.2k114489
edited yesterday
greedoid
35.2k114489
35.2k114489
asked yesterday
Ghartal
2,86211335
asked yesterday
Ghartal
2,86211335
asked yesterday
Ghartal
2,86211335
2,86211335
It seems like it is a least not solvable via algebra hence the given conditions lead to a system of four equation with four unknowns where the determinate of the matrix of coefficients equals zero.
– mrtaurho
yesterday
1
Trigonometry expresses relationships between the ratios of various lengths within a right triangle. Any triangle can be broken up into some combination of right triangles. In principle any statement in trigonometry has a corresponding expression in solely geometric terms.
– TurlocTheRed
yesterday
I thought the Sine Law For Triangles $was$ geometry.
– DanielWainfleet
yesterday
add a comment |
It seems like it is a least not solvable via algebra hence the given conditions lead to a system of four equation with four unknowns where the determinate of the matrix of coefficients equals zero.
– mrtaurho
yesterday
1
Trigonometry expresses relationships between the ratios of various lengths within a right triangle. Any triangle can be broken up into some combination of right triangles. In principle any statement in trigonometry has a corresponding expression in solely geometric terms.
– TurlocTheRed
yesterday
I thought the Sine Law For Triangles $was$ geometry.
– DanielWainfleet
yesterday
It seems like it is a least not solvable via algebra hence the given conditions lead to a system of four equation with four unknowns where the determinate of the matrix of coefficients equals zero.
– mrtaurho
yesterday
It seems like it is a least not solvable via algebra hence the given conditions lead to a system of four equation with four unknowns where the determinate of the matrix of coefficients equals zero.
– mrtaurho
yesterday
1
1
Trigonometry expresses relationships between the ratios of various lengths within a right triangle. Any triangle can be broken up into some combination of right triangles. In principle any statement in trigonometry has a corresponding expression in solely geometric terms.
– TurlocTheRed
yesterday
Trigonometry expresses relationships between the ratios of various lengths within a right triangle. Any triangle can be broken up into some combination of right triangles. In principle any statement in trigonometry has a corresponding expression in solely geometric terms.
– TurlocTheRed
yesterday
I thought the Sine Law For Triangles $was$ geometry.
– DanielWainfleet
yesterday
I thought the Sine Law For Triangles $was$ geometry.
– DanielWainfleet
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
Let's go another way around.
Let $M'$ be such that $angle M'CA = angle M'AC = 10^{circ}$ and let $D$ be a reflection of $C$ across line $AM'$. Then $AM' = CM' = DM'$ and a triangle $ABD$ is equlateral (since $AD = AC = AB$ and $angle BAD = 60^{circ}$). So $ABM'$ and $DBM'$ are congruent so $angle DBM' = 30^{circ}$
Now, since $A$ is a center of circle around through $B,D$ and $C$ we have also $$angle DBC = {1over 2}angle DAC = 10^{circ}$$
so $angle CBM' = 20^{circ}$ and thus $M'=M$.
answered yesterday
greedoid
35.2k114489
Did knowing the answer help you find this solution? It's beautiful! I really enjoyed it.
– Ghartal
yesterday
Yes, but you can find this answer easly by drawing.
– greedoid
yesterday
First idea was to rotate $A$ around $B$ for $-60$ to $D$, but didn't get anywhere. Perhaps you can try.
– greedoid
yesterday
Thanks. I'll try it.
– Ghartal
yesterday
add a comment |
up vote
2
down vote
Maybe you can use the following. Consider a regular 9-gon $I_1I_2ldots I_9$ with center $O$, and find your picture by setting $A=O$, $B=I_3$ and $C=I_5$. Then note that $M$ is the intersection of diagonals $I_1I_5$ and $I_3I_6$.
answered yesterday
SMM
2,17349
This looks like a great idea! But I don't see how it reduces the difficulties!
– Ghartal
yesterday
1
@Ghartal I guess you can consider images of $M$ under isometries which preserve the 9-gon: there are 18 of them: 9 rotations and 9 symmetries. This should be a regular 18-gon implying that the central angle is 20, so your angle is 10.
– SMM
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Let's go another way around.
Let $M'$ be such that $angle M'CA = angle M'AC = 10^{circ}$ and let $D$ be a reflection of $C$ across line $AM'$. Then $AM' = CM' = DM'$ and a triangle $ABD$ is equlateral (since $AD = AC = AB$ and $angle BAD = 60^{circ}$). So $ABM'$ and $DBM'$ are congruent so $angle DBM' = 30^{circ}$
Now, since $A$ is a center of circle around through $B,D$ and $C$ we have also $$angle DBC = {1over 2}angle DAC = 10^{circ}$$
so $angle CBM' = 20^{circ}$ and thus $M'=M$.
answered yesterday
greedoid
35.2k114489
Did knowing the answer help you find this solution? It's beautiful! I really enjoyed it.
– Ghartal
yesterday
Yes, but you can find this answer easly by drawing.
– greedoid
yesterday
First idea was to rotate $A$ around $B$ for $-60$ to $D$, but didn't get anywhere. Perhaps you can try.
– greedoid
yesterday
Thanks. I'll try it.
– Ghartal
yesterday
add a comment |
up vote
3
down vote
Let's go another way around.
Let $M'$ be such that $angle M'CA = angle M'AC = 10^{circ}$ and let $D$ be a reflection of $C$ across line $AM'$. Then $AM' = CM' = DM'$ and a triangle $ABD$ is equlateral (since $AD = AC = AB$ and $angle BAD = 60^{circ}$). So $ABM'$ and $DBM'$ are congruent so $angle DBM' = 30^{circ}$
Now, since $A$ is a center of circle around through $B,D$ and $C$ we have also $$angle DBC = {1over 2}angle DAC = 10^{circ}$$
so $angle CBM' = 20^{circ}$ and thus $M'=M$.
answered yesterday
greedoid
35.2k114489
Did knowing the answer help you find this solution? It's beautiful! I really enjoyed it.
– Ghartal
yesterday
Yes, but you can find this answer easly by drawing.
– greedoid
yesterday
First idea was to rotate $A$ around $B$ for $-60$ to $D$, but didn't get anywhere. Perhaps you can try.
– greedoid
yesterday
Thanks. I'll try it.
– Ghartal
yesterday
add a comment |
up vote
3
down vote
up vote
3
down vote
Let's go another way around.
Let $M'$ be such that $angle M'CA = angle M'AC = 10^{circ}$ and let $D$ be a reflection of $C$ across line $AM'$. Then $AM' = CM' = DM'$ and a triangle $ABD$ is equlateral (since $AD = AC = AB$ and $angle BAD = 60^{circ}$). So $ABM'$ and $DBM'$ are congruent so $angle DBM' = 30^{circ}$
Now, since $A$ is a center of circle around through $B,D$ and $C$ we have also $$angle DBC = {1over 2}angle DAC = 10^{circ}$$
so $angle CBM' = 20^{circ}$ and thus $M'=M$.
answered yesterday
greedoid
35.2k114489
Let's go another way around.
Let $M'$ be such that $angle M'CA = angle M'AC = 10^{circ}$ and let $D$ be a reflection of $C$ across line $AM'$. Then $AM' = CM' = DM'$ and a triangle $ABD$ is equlateral (since $AD = AC = AB$ and $angle BAD = 60^{circ}$). So $ABM'$ and $DBM'$ are congruent so $angle DBM' = 30^{circ}$
Now, since $A$ is a center of circle around through $B,D$ and $C$ we have also $$angle DBC = {1over 2}angle DAC = 10^{circ}$$
so $angle CBM' = 20^{circ}$ and thus $M'=M$.
answered yesterday
greedoid
35.2k114489
edited yesterday
answered yesterday
greedoid
35.2k114489
answered yesterday
greedoid
35.2k114489
answered yesterday
greedoid
35.2k114489
35.2k114489
Did knowing the answer help you find this solution? It's beautiful! I really enjoyed it.
– Ghartal
yesterday
Yes, but you can find this answer easly by drawing.
– greedoid
yesterday
First idea was to rotate $A$ around $B$ for $-60$ to $D$, but didn't get anywhere. Perhaps you can try.
– greedoid
yesterday
Thanks. I'll try it.
– Ghartal
yesterday
add a comment |
Did knowing the answer help you find this solution? It's beautiful! I really enjoyed it.
– Ghartal
yesterday
Yes, but you can find this answer easly by drawing.
– greedoid
yesterday
First idea was to rotate $A$ around $B$ for $-60$ to $D$, but didn't get anywhere. Perhaps you can try.
– greedoid
yesterday
Thanks. I'll try it.
– Ghartal
yesterday
Did knowing the answer help you find this solution? It's beautiful! I really enjoyed it.
– Ghartal
yesterday
Did knowing the answer help you find this solution? It's beautiful! I really enjoyed it.
– Ghartal
yesterday
Yes, but you can find this answer easly by drawing.
– greedoid
yesterday
Yes, but you can find this answer easly by drawing.
– greedoid
yesterday
First idea was to rotate $A$ around $B$ for $-60$ to $D$, but didn't get anywhere. Perhaps you can try.
– greedoid
yesterday
First idea was to rotate $A$ around $B$ for $-60$ to $D$, but didn't get anywhere. Perhaps you can try.
– greedoid
yesterday
Thanks. I'll try it.
– Ghartal
yesterday
Thanks. I'll try it.
– Ghartal
yesterday
add a comment |
up vote
2
down vote
Maybe you can use the following. Consider a regular 9-gon $I_1I_2ldots I_9$ with center $O$, and find your picture by setting $A=O$, $B=I_3$ and $C=I_5$. Then note that $M$ is the intersection of diagonals $I_1I_5$ and $I_3I_6$.
answered yesterday
SMM
2,17349
This looks like a great idea! But I don't see how it reduces the difficulties!
– Ghartal
yesterday
1
@Ghartal I guess you can consider images of $M$ under isometries which preserve the 9-gon: there are 18 of them: 9 rotations and 9 symmetries. This should be a regular 18-gon implying that the central angle is 20, so your angle is 10.
– SMM
yesterday
add a comment |
up vote
2
down vote
Maybe you can use the following. Consider a regular 9-gon $I_1I_2ldots I_9$ with center $O$, and find your picture by setting $A=O$, $B=I_3$ and $C=I_5$. Then note that $M$ is the intersection of diagonals $I_1I_5$ and $I_3I_6$.
answered yesterday
SMM
2,17349
This looks like a great idea! But I don't see how it reduces the difficulties!
– Ghartal
yesterday
1
@Ghartal I guess you can consider images of $M$ under isometries which preserve the 9-gon: there are 18 of them: 9 rotations and 9 symmetries. This should be a regular 18-gon implying that the central angle is 20, so your angle is 10.
– SMM
yesterday
add a comment |
up vote
2
down vote
up vote
2
down vote
Maybe you can use the following. Consider a regular 9-gon $I_1I_2ldots I_9$ with center $O$, and find your picture by setting $A=O$, $B=I_3$ and $C=I_5$. Then note that $M$ is the intersection of diagonals $I_1I_5$ and $I_3I_6$.
answered yesterday
SMM
2,17349
Maybe you can use the following. Consider a regular 9-gon $I_1I_2ldots I_9$ with center $O$, and find your picture by setting $A=O$, $B=I_3$ and $C=I_5$. Then note that $M$ is the intersection of diagonals $I_1I_5$ and $I_3I_6$.
answered yesterday
SMM
2,17349
answered yesterday
SMM
2,17349
answered yesterday
SMM
2,17349
answered yesterday
SMM
2,17349
2,17349
This looks like a great idea! But I don't see how it reduces the difficulties!
– Ghartal
yesterday
1
@Ghartal I guess you can consider images of $M$ under isometries which preserve the 9-gon: there are 18 of them: 9 rotations and 9 symmetries. This should be a regular 18-gon implying that the central angle is 20, so your angle is 10.
– SMM
yesterday
add a comment |
This looks like a great idea! But I don't see how it reduces the difficulties!
– Ghartal
yesterday
1
@Ghartal I guess you can consider images of $M$ under isometries which preserve the 9-gon: there are 18 of them: 9 rotations and 9 symmetries. This should be a regular 18-gon implying that the central angle is 20, so your angle is 10.
– SMM
yesterday
This looks like a great idea! But I don't see how it reduces the difficulties!
– Ghartal
yesterday
This looks like a great idea! But I don't see how it reduces the difficulties!
– Ghartal
yesterday
1
1
@Ghartal I guess you can consider images of $M$ under isometries which preserve the 9-gon: there are 18 of them: 9 rotations and 9 symmetries. This should be a regular 18-gon implying that the central angle is 20, so your angle is 10.
– SMM
yesterday
@Ghartal I guess you can consider images of $M$ under isometries which preserve the 9-gon: there are 18 of them: 9 rotations and 9 symmetries. This should be a regular 18-gon implying that the central angle is 20, so your angle is 10.
– SMM
yesterday
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020173%2ffind-angle-without-trigonometry%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
It seems like it is a least not solvable via algebra hence the given conditions lead to a system of four equation with four unknowns where the determinate of the matrix of coefficients equals zero.
– mrtaurho
yesterday
1
Trigonometry expresses relationships between the ratios of various lengths within a right triangle. Any triangle can be broken up into some combination of right triangles. In principle any statement in trigonometry has a corresponding expression in solely geometric terms.
– TurlocTheRed
yesterday
I thought the Sine Law For Triangles $was$ geometry.
– DanielWainfleet
yesterday