Finding the angle b/w two lines in Coordinate Geometry
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In my coaching class I was taught that the tangent of the angle between two lines having slopes $m_1$ and $m_2$ is given by the formula modulus of
$frac{m_1-m_2}{1+m_1m_2}$.
We can then use $tan$-inverse to find the angle.
However, some angles have negative tangent values, which will not be obtained by this formula which uses modulus. But shouldn't these angles also exist between two lines?
analytic-geometry
edited Jan 10 '16 at 5:03
Winther
20.9k33156
asked Jan 10 '16 at 4:55
N.S.JOHNN.S.JOHN
1,187620
$endgroup$
add a comment |
$begingroup$
In my coaching class I was taught that the tangent of the angle between two lines having slopes $m_1$ and $m_2$ is given by the formula modulus of
$frac{m_1-m_2}{1+m_1m_2}$.
We can then use $tan$-inverse to find the angle.
However, some angles have negative tangent values, which will not be obtained by this formula which uses modulus. But shouldn't these angles also exist between two lines?
analytic-geometry
edited Jan 10 '16 at 5:03
Winther
20.9k33156
asked Jan 10 '16 at 4:55
N.S.JOHNN.S.JOHN
1,187620
$endgroup$
$begingroup$
You might consider how many angles are created when two lines intersect.
$endgroup$
– euler1944
Jan 10 '16 at 5:00
$begingroup$
Negative angles might be that they are in clockwise direction thats why we use mod also intersection creates $4$ angles.
$endgroup$
– Archis Welankar
Jan 10 '16 at 5:21
add a comment |
$begingroup$
In my coaching class I was taught that the tangent of the angle between two lines having slopes $m_1$ and $m_2$ is given by the formula modulus of
$frac{m_1-m_2}{1+m_1m_2}$.
We can then use $tan$-inverse to find the angle.
However, some angles have negative tangent values, which will not be obtained by this formula which uses modulus. But shouldn't these angles also exist between two lines?
analytic-geometry
edited Jan 10 '16 at 5:03
Winther
20.9k33156
asked Jan 10 '16 at 4:55
N.S.JOHNN.S.JOHN
1,187620
$endgroup$
In my coaching class I was taught that the tangent of the angle between two lines having slopes $m_1$ and $m_2$ is given by the formula modulus of
$frac{m_1-m_2}{1+m_1m_2}$.
We can then use $tan$-inverse to find the angle.
However, some angles have negative tangent values, which will not be obtained by this formula which uses modulus. But shouldn't these angles also exist between two lines?
analytic-geometry
analytic-geometry
edited Jan 10 '16 at 5:03
Winther
20.9k33156
asked Jan 10 '16 at 4:55
N.S.JOHNN.S.JOHN
1,187620
edited Jan 10 '16 at 5:03
Winther
20.9k33156
asked Jan 10 '16 at 4:55
N.S.JOHNN.S.JOHN
1,187620
edited Jan 10 '16 at 5:03
Winther
20.9k33156
edited Jan 10 '16 at 5:03
Winther
20.9k33156
edited Jan 10 '16 at 5:03
Winther
20.9k33156
20.9k33156
asked Jan 10 '16 at 4:55
N.S.JOHNN.S.JOHN
1,187620
asked Jan 10 '16 at 4:55
N.S.JOHNN.S.JOHN
1,187620
asked Jan 10 '16 at 4:55
N.S.JOHNN.S.JOHN
1,187620
1,187620
$begingroup$
You might consider how many angles are created when two lines intersect.
$endgroup$
– euler1944
Jan 10 '16 at 5:00
$begingroup$
Negative angles might be that they are in clockwise direction thats why we use mod also intersection creates $4$ angles.
$endgroup$
– Archis Welankar
Jan 10 '16 at 5:21
add a comment |
$begingroup$
You might consider how many angles are created when two lines intersect.
$endgroup$
– euler1944
Jan 10 '16 at 5:00
$begingroup$
Negative angles might be that they are in clockwise direction thats why we use mod also intersection creates $4$ angles.
$endgroup$
– Archis Welankar
Jan 10 '16 at 5:21
$begingroup$
You might consider how many angles are created when two lines intersect.
$endgroup$
– euler1944
Jan 10 '16 at 5:00
$begingroup$
You might consider how many angles are created when two lines intersect.
$endgroup$
– euler1944
Jan 10 '16 at 5:00
$begingroup$
Negative angles might be that they are in clockwise direction thats why we use mod also intersection creates $4$ angles.
$endgroup$
– Archis Welankar
Jan 10 '16 at 5:21
$begingroup$
Negative angles might be that they are in clockwise direction thats why we use mod also intersection creates $4$ angles.
$endgroup$
– Archis Welankar
Jan 10 '16 at 5:21
add a comment |
2 Answers
2
active
oldest
votes
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Before you use the formula, you should determine what type of angle you are looking for, specifically, acute or obtuse - when two lines intersect, two pairs of identical angles are formed. To specify which angle you are targeting, use your formula, with $m_{2}$ being the angle's starting line. If you get a negative output after taking inverse tangent, just take the positive of the answer. This results from the fact that inverse tangent is an odd function; specifically, $arctan(-theta) = -arctan(theta).$ Hope this helps!
answered Jan 10 '16 at 5:31
K. JiangK. Jiang
3,0311513
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add a comment |
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My interpretation of the vertical bars in the formula ($tan theta = |dfrac {m_1 – m_2}{1 + m_1m_2}|$) is NOT modulus but absolute value instead.
This means $tan theta = + dfrac {m_1 – m_2}{1 + m_1m_2}$ or $tan theta =–dfrac {m_1 – m_2}{1 + m_1m_2}$.
If one value does not give an acute value of $theta$, the other will.
answered Jan 10 '16 at 5:38
MickMick
12.1k31641
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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votes
$begingroup$
Before you use the formula, you should determine what type of angle you are looking for, specifically, acute or obtuse - when two lines intersect, two pairs of identical angles are formed. To specify which angle you are targeting, use your formula, with $m_{2}$ being the angle's starting line. If you get a negative output after taking inverse tangent, just take the positive of the answer. This results from the fact that inverse tangent is an odd function; specifically, $arctan(-theta) = -arctan(theta).$ Hope this helps!
answered Jan 10 '16 at 5:31
K. JiangK. Jiang
3,0311513
$endgroup$
add a comment |
$begingroup$
Before you use the formula, you should determine what type of angle you are looking for, specifically, acute or obtuse - when two lines intersect, two pairs of identical angles are formed. To specify which angle you are targeting, use your formula, with $m_{2}$ being the angle's starting line. If you get a negative output after taking inverse tangent, just take the positive of the answer. This results from the fact that inverse tangent is an odd function; specifically, $arctan(-theta) = -arctan(theta).$ Hope this helps!
answered Jan 10 '16 at 5:31
K. JiangK. Jiang
3,0311513
$endgroup$
add a comment |
$begingroup$
Before you use the formula, you should determine what type of angle you are looking for, specifically, acute or obtuse - when two lines intersect, two pairs of identical angles are formed. To specify which angle you are targeting, use your formula, with $m_{2}$ being the angle's starting line. If you get a negative output after taking inverse tangent, just take the positive of the answer. This results from the fact that inverse tangent is an odd function; specifically, $arctan(-theta) = -arctan(theta).$ Hope this helps!
answered Jan 10 '16 at 5:31
K. JiangK. Jiang
3,0311513
$endgroup$
Before you use the formula, you should determine what type of angle you are looking for, specifically, acute or obtuse - when two lines intersect, two pairs of identical angles are formed. To specify which angle you are targeting, use your formula, with $m_{2}$ being the angle's starting line. If you get a negative output after taking inverse tangent, just take the positive of the answer. This results from the fact that inverse tangent is an odd function; specifically, $arctan(-theta) = -arctan(theta).$ Hope this helps!
answered Jan 10 '16 at 5:31
K. JiangK. Jiang
3,0311513
answered Jan 10 '16 at 5:31
K. JiangK. Jiang
3,0311513
answered Jan 10 '16 at 5:31
K. JiangK. Jiang
3,0311513
answered Jan 10 '16 at 5:31
K. JiangK. Jiang
3,0311513
3,0311513
add a comment |
add a comment |
$begingroup$
My interpretation of the vertical bars in the formula ($tan theta = |dfrac {m_1 – m_2}{1 + m_1m_2}|$) is NOT modulus but absolute value instead.
This means $tan theta = + dfrac {m_1 – m_2}{1 + m_1m_2}$ or $tan theta =–dfrac {m_1 – m_2}{1 + m_1m_2}$.
If one value does not give an acute value of $theta$, the other will.
answered Jan 10 '16 at 5:38
MickMick
12.1k31641
$endgroup$
add a comment |
$begingroup$
My interpretation of the vertical bars in the formula ($tan theta = |dfrac {m_1 – m_2}{1 + m_1m_2}|$) is NOT modulus but absolute value instead.
This means $tan theta = + dfrac {m_1 – m_2}{1 + m_1m_2}$ or $tan theta =–dfrac {m_1 – m_2}{1 + m_1m_2}$.
If one value does not give an acute value of $theta$, the other will.
answered Jan 10 '16 at 5:38
MickMick
12.1k31641
$endgroup$
add a comment |
$begingroup$
My interpretation of the vertical bars in the formula ($tan theta = |dfrac {m_1 – m_2}{1 + m_1m_2}|$) is NOT modulus but absolute value instead.
This means $tan theta = + dfrac {m_1 – m_2}{1 + m_1m_2}$ or $tan theta =–dfrac {m_1 – m_2}{1 + m_1m_2}$.
If one value does not give an acute value of $theta$, the other will.
answered Jan 10 '16 at 5:38
MickMick
12.1k31641
$endgroup$
My interpretation of the vertical bars in the formula ($tan theta = |dfrac {m_1 – m_2}{1 + m_1m_2}|$) is NOT modulus but absolute value instead.
This means $tan theta = + dfrac {m_1 – m_2}{1 + m_1m_2}$ or $tan theta =–dfrac {m_1 – m_2}{1 + m_1m_2}$.
If one value does not give an acute value of $theta$, the other will.
answered Jan 10 '16 at 5:38
MickMick
12.1k31641
answered Jan 10 '16 at 5:38
MickMick
12.1k31641
answered Jan 10 '16 at 5:38
MickMick
12.1k31641
answered Jan 10 '16 at 5:38
MickMick
12.1k31641
12.1k31641
add a comment |
add a comment |
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$begingroup$
You might consider how many angles are created when two lines intersect.
$endgroup$
– euler1944
Jan 10 '16 at 5:00
$begingroup$
Negative angles might be that they are in clockwise direction thats why we use mod also intersection creates $4$ angles.
$endgroup$
– Archis Welankar
Jan 10 '16 at 5:21