Finding an angle in a triangle? Announcing the arrival of Valued Associate #679: Cesar...
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Finding an angle in a triangle?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Finding value of an angle in a triangle.Length of angle bisector using Law of SinesFind the angles of triangleFind the value of an angleIn triangle $ABC$ find angle $angle BAC$ given that…Find possible integer values for $x$ using angle-edge relationship in triangle.Find angle $alpha$ in this triangle question.Find an angle of a triangle on a larger triangle which cut through its midpointFind the third angle of the triangle(South Africa 2014) Finding angles in an obtuse triangle
$begingroup$
Given the following picture, how do I find $angle CDA$?
Attempt:
I have found the following angles, which I think may be useful:
$$
angle BAC = arctan(3),quad
angle ABC = arctan(2),quad
angle BCA = arctan(1).
$$
But I don't know how to go on.
trigonometry triangle
edited Mar 17 at 23:31
Jack
27.8k1784206
asked Mar 4 at 12:00
Ruby PaRuby Pa
467
$endgroup$
|
show 1 more comment
$begingroup$
Given the following picture, how do I find $angle CDA$?
Attempt:
I have found the following angles, which I think may be useful:
$$
angle BAC = arctan(3),quad
angle ABC = arctan(2),quad
angle BCA = arctan(1).
$$
But I don't know how to go on.
trigonometry triangle
edited Mar 17 at 23:31
Jack
27.8k1784206
asked Mar 4 at 12:00
Ruby PaRuby Pa
467
$endgroup$
$begingroup$
Hey and Welcome to MSE! Could you please tell us where are you stuck? What have you tried so far?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:08
$begingroup$
So this is not the original problem? In this case you shouid post the whole question. Maybe there are some other information there we could use.
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:14
$begingroup$
@Vinyl_coat_jawa I understand what you're saying, but there is no additional information that I have not already provided.
$endgroup$
– Ruby Pa
Mar 4 at 12:15
$begingroup$
But the how do you know these angles?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:26
1
$begingroup$
Where exactly is point $D$ ?
$endgroup$
– Antinous
Mar 4 at 13:51
|
show 1 more comment
$begingroup$
Given the following picture, how do I find $angle CDA$?
Attempt:
I have found the following angles, which I think may be useful:
$$
angle BAC = arctan(3),quad
angle ABC = arctan(2),quad
angle BCA = arctan(1).
$$
But I don't know how to go on.
trigonometry triangle
edited Mar 17 at 23:31
Jack
27.8k1784206
asked Mar 4 at 12:00
Ruby PaRuby Pa
467
$endgroup$
Given the following picture, how do I find $angle CDA$?
Attempt:
I have found the following angles, which I think may be useful:
$$
angle BAC = arctan(3),quad
angle ABC = arctan(2),quad
angle BCA = arctan(1).
$$
But I don't know how to go on.
trigonometry triangle
trigonometry triangle
edited Mar 17 at 23:31
Jack
27.8k1784206
asked Mar 4 at 12:00
Ruby PaRuby Pa
467
edited Mar 17 at 23:31
Jack
27.8k1784206
asked Mar 4 at 12:00
Ruby PaRuby Pa
467
edited Mar 17 at 23:31
Jack
27.8k1784206
edited Mar 17 at 23:31
Jack
27.8k1784206
edited Mar 17 at 23:31
Jack
27.8k1784206
27.8k1784206
asked Mar 4 at 12:00
Ruby PaRuby Pa
467
asked Mar 4 at 12:00
Ruby PaRuby Pa
467
asked Mar 4 at 12:00
Ruby PaRuby Pa
467
467
$begingroup$
Hey and Welcome to MSE! Could you please tell us where are you stuck? What have you tried so far?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:08
$begingroup$
So this is not the original problem? In this case you shouid post the whole question. Maybe there are some other information there we could use.
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:14
$begingroup$
@Vinyl_coat_jawa I understand what you're saying, but there is no additional information that I have not already provided.
$endgroup$
– Ruby Pa
Mar 4 at 12:15
$begingroup$
But the how do you know these angles?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:26
1
$begingroup$
Where exactly is point $D$ ?
$endgroup$
– Antinous
Mar 4 at 13:51
|
show 1 more comment
$begingroup$
Hey and Welcome to MSE! Could you please tell us where are you stuck? What have you tried so far?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:08
$begingroup$
So this is not the original problem? In this case you shouid post the whole question. Maybe there are some other information there we could use.
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:14
$begingroup$
@Vinyl_coat_jawa I understand what you're saying, but there is no additional information that I have not already provided.
$endgroup$
– Ruby Pa
Mar 4 at 12:15
$begingroup$
But the how do you know these angles?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:26
1
$begingroup$
Where exactly is point $D$ ?
$endgroup$
– Antinous
Mar 4 at 13:51
$begingroup$
Hey and Welcome to MSE! Could you please tell us where are you stuck? What have you tried so far?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:08
$begingroup$
Hey and Welcome to MSE! Could you please tell us where are you stuck? What have you tried so far?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:08
$begingroup$
So this is not the original problem? In this case you shouid post the whole question. Maybe there are some other information there we could use.
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:14
$begingroup$
So this is not the original problem? In this case you shouid post the whole question. Maybe there are some other information there we could use.
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:14
$begingroup$
@Vinyl_coat_jawa I understand what you're saying, but there is no additional information that I have not already provided.
$endgroup$
– Ruby Pa
Mar 4 at 12:15
$begingroup$
@Vinyl_coat_jawa I understand what you're saying, but there is no additional information that I have not already provided.
$endgroup$
– Ruby Pa
Mar 4 at 12:15
$begingroup$
But the how do you know these angles?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:26
$begingroup$
But the how do you know these angles?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:26
1
1
$begingroup$
Where exactly is point $D$ ?
$endgroup$
– Antinous
Mar 4 at 13:51
$begingroup$
Where exactly is point $D$ ?
$endgroup$
– Antinous
Mar 4 at 13:51
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Assuming the points are at the exact grid points where it looks like they are, the line $AD$ is orthogonal to $BC$. The line $BC$ goes $2$ units down for each unit we go to the right, and if we turn that $90^circ$, we get a line which goes $1$ unit up for every $2$ units to the right. That's what $AD$ does.
answered Mar 4 at 12:31
ArthurArthur
123k7122211
$endgroup$
$begingroup$
@RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable.
$endgroup$
– Arthur
Mar 4 at 12:33
$begingroup$
Ahhh yes I see what you mean. Thanks for your help! :)
$endgroup$
– Ruby Pa
Mar 4 at 12:36
add a comment |
$begingroup$
Another approach:
Note that $angle DAB = arctan(dfrac{1}{2})\ $ (two units over four ones). Thus,
$$angle ADC = 180°- (angle DAB+ angle CBA) = 180° - arctan(dfrac{1}{2})-arctan(2) = 90°.$$
answered Mar 4 at 12:48
Alex SilvaAlex Silva
2,79631532
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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oldest
votes
$begingroup$
Assuming the points are at the exact grid points where it looks like they are, the line $AD$ is orthogonal to $BC$. The line $BC$ goes $2$ units down for each unit we go to the right, and if we turn that $90^circ$, we get a line which goes $1$ unit up for every $2$ units to the right. That's what $AD$ does.
answered Mar 4 at 12:31
ArthurArthur
123k7122211
$endgroup$
$begingroup$
@RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable.
$endgroup$
– Arthur
Mar 4 at 12:33
$begingroup$
Ahhh yes I see what you mean. Thanks for your help! :)
$endgroup$
– Ruby Pa
Mar 4 at 12:36
add a comment |
$begingroup$
Assuming the points are at the exact grid points where it looks like they are, the line $AD$ is orthogonal to $BC$. The line $BC$ goes $2$ units down for each unit we go to the right, and if we turn that $90^circ$, we get a line which goes $1$ unit up for every $2$ units to the right. That's what $AD$ does.
answered Mar 4 at 12:31
ArthurArthur
123k7122211
$endgroup$
$begingroup$
@RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable.
$endgroup$
– Arthur
Mar 4 at 12:33
$begingroup$
Ahhh yes I see what you mean. Thanks for your help! :)
$endgroup$
– Ruby Pa
Mar 4 at 12:36
add a comment |
$begingroup$
Assuming the points are at the exact grid points where it looks like they are, the line $AD$ is orthogonal to $BC$. The line $BC$ goes $2$ units down for each unit we go to the right, and if we turn that $90^circ$, we get a line which goes $1$ unit up for every $2$ units to the right. That's what $AD$ does.
answered Mar 4 at 12:31
ArthurArthur
123k7122211
$endgroup$
Assuming the points are at the exact grid points where it looks like they are, the line $AD$ is orthogonal to $BC$. The line $BC$ goes $2$ units down for each unit we go to the right, and if we turn that $90^circ$, we get a line which goes $1$ unit up for every $2$ units to the right. That's what $AD$ does.
answered Mar 4 at 12:31
ArthurArthur
123k7122211
answered Mar 4 at 12:31
ArthurArthur
123k7122211
answered Mar 4 at 12:31
ArthurArthur
123k7122211
answered Mar 4 at 12:31
ArthurArthur
123k7122211
123k7122211
$begingroup$
@RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable.
$endgroup$
– Arthur
Mar 4 at 12:33
$begingroup$
Ahhh yes I see what you mean. Thanks for your help! :)
$endgroup$
– Ruby Pa
Mar 4 at 12:36
add a comment |
$begingroup$
@RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable.
$endgroup$
– Arthur
Mar 4 at 12:33
$begingroup$
Ahhh yes I see what you mean. Thanks for your help! :)
$endgroup$
– Ruby Pa
Mar 4 at 12:36
$begingroup$
@RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable.
$endgroup$
– Arthur
Mar 4 at 12:33
$begingroup$
@RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable.
$endgroup$
– Arthur
Mar 4 at 12:33
$begingroup$
Ahhh yes I see what you mean. Thanks for your help! :)
$endgroup$
– Ruby Pa
Mar 4 at 12:36
$begingroup$
Ahhh yes I see what you mean. Thanks for your help! :)
$endgroup$
– Ruby Pa
Mar 4 at 12:36
add a comment |
$begingroup$
Another approach:
Note that $angle DAB = arctan(dfrac{1}{2})\ $ (two units over four ones). Thus,
$$angle ADC = 180°- (angle DAB+ angle CBA) = 180° - arctan(dfrac{1}{2})-arctan(2) = 90°.$$
answered Mar 4 at 12:48
Alex SilvaAlex Silva
2,79631532
$endgroup$
add a comment |
$begingroup$
Another approach:
Note that $angle DAB = arctan(dfrac{1}{2})\ $ (two units over four ones). Thus,
$$angle ADC = 180°- (angle DAB+ angle CBA) = 180° - arctan(dfrac{1}{2})-arctan(2) = 90°.$$
answered Mar 4 at 12:48
Alex SilvaAlex Silva
2,79631532
$endgroup$
add a comment |
$begingroup$
Another approach:
Note that $angle DAB = arctan(dfrac{1}{2})\ $ (two units over four ones). Thus,
$$angle ADC = 180°- (angle DAB+ angle CBA) = 180° - arctan(dfrac{1}{2})-arctan(2) = 90°.$$
answered Mar 4 at 12:48
Alex SilvaAlex Silva
2,79631532
$endgroup$
Another approach:
Note that $angle DAB = arctan(dfrac{1}{2})\ $ (two units over four ones). Thus,
$$angle ADC = 180°- (angle DAB+ angle CBA) = 180° - arctan(dfrac{1}{2})-arctan(2) = 90°.$$
answered Mar 4 at 12:48
Alex SilvaAlex Silva
2,79631532
edited Mar 29 at 10:56
answered Mar 4 at 12:48
Alex SilvaAlex Silva
2,79631532
answered Mar 4 at 12:48
Alex SilvaAlex Silva
2,79631532
answered Mar 4 at 12:48
Alex SilvaAlex Silva
2,79631532
2,79631532
add a comment |
add a comment |
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$begingroup$
Hey and Welcome to MSE! Could you please tell us where are you stuck? What have you tried so far?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:08
$begingroup$
So this is not the original problem? In this case you shouid post the whole question. Maybe there are some other information there we could use.
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:14
$begingroup$
@Vinyl_coat_jawa I understand what you're saying, but there is no additional information that I have not already provided.
$endgroup$
– Ruby Pa
Mar 4 at 12:15
$begingroup$
But the how do you know these angles?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:26
1
$begingroup$
Where exactly is point $D$ ?
$endgroup$
– Antinous
Mar 4 at 13:51