calculate angle of line with negative slope
$begingroup$
I want to use the formula
$$
tan(alpha)=m
$$
for negative slopes but always get negative degrees. For instance, say the slope of a line $g$ is $-1$. Using the formula above (arctan$(-1)=alpha$), I get $-45$ degrees instead of $135$ degrees.
Why exactly does this formula not returning correct angles for negative slopes?
angle slope
asked Dec 18 '18 at 10:07
lagunalaguna
767
$endgroup$
add a comment |
$begingroup$
I want to use the formula
$$
tan(alpha)=m
$$
for negative slopes but always get negative degrees. For instance, say the slope of a line $g$ is $-1$. Using the formula above (arctan$(-1)=alpha$), I get $-45$ degrees instead of $135$ degrees.
Why exactly does this formula not returning correct angles for negative slopes?
angle slope
asked Dec 18 '18 at 10:07
lagunalaguna
767
$endgroup$
$begingroup$
It all depends on how you define $arctan$. It is usually defined as the inverse function of the restriction of $tan$ on $left( -frac{pi}{2}, frac{pi}{2}right)$.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 10:12
add a comment |
$begingroup$
I want to use the formula
$$
tan(alpha)=m
$$
for negative slopes but always get negative degrees. For instance, say the slope of a line $g$ is $-1$. Using the formula above (arctan$(-1)=alpha$), I get $-45$ degrees instead of $135$ degrees.
Why exactly does this formula not returning correct angles for negative slopes?
angle slope
asked Dec 18 '18 at 10:07
lagunalaguna
767
$endgroup$
I want to use the formula
$$
tan(alpha)=m
$$
for negative slopes but always get negative degrees. For instance, say the slope of a line $g$ is $-1$. Using the formula above (arctan$(-1)=alpha$), I get $-45$ degrees instead of $135$ degrees.
Why exactly does this formula not returning correct angles for negative slopes?
angle slope
angle slope
asked Dec 18 '18 at 10:07
lagunalaguna
767
asked Dec 18 '18 at 10:07
lagunalaguna
767
asked Dec 18 '18 at 10:07
lagunalaguna
767
asked Dec 18 '18 at 10:07
lagunalaguna
767
asked Dec 18 '18 at 10:07
lagunalaguna
767
767
$begingroup$
It all depends on how you define $arctan$. It is usually defined as the inverse function of the restriction of $tan$ on $left( -frac{pi}{2}, frac{pi}{2}right)$.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 10:12
add a comment |
$begingroup$
It all depends on how you define $arctan$. It is usually defined as the inverse function of the restriction of $tan$ on $left( -frac{pi}{2}, frac{pi}{2}right)$.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 10:12
$begingroup$
It all depends on how you define $arctan$. It is usually defined as the inverse function of the restriction of $tan$ on $left( -frac{pi}{2}, frac{pi}{2}right)$.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 10:12
$begingroup$
It all depends on how you define $arctan$. It is usually defined as the inverse function of the restriction of $tan$ on $left( -frac{pi}{2}, frac{pi}{2}right)$.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 10:12
add a comment |
1 Answer
1
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$begingroup$
Use this formula: $alpha=begin{cases}tan^{-1}m,&mge0\pi+tan^{-1}m,&m<0end{cases}$
answered Dec 18 '18 at 10:11
Shubham JohriShubham Johri
5,192717
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Use this formula: $alpha=begin{cases}tan^{-1}m,&mge0\pi+tan^{-1}m,&m<0end{cases}$
answered Dec 18 '18 at 10:11
Shubham JohriShubham Johri
5,192717
$endgroup$
add a comment |
$begingroup$
Use this formula: $alpha=begin{cases}tan^{-1}m,&mge0\pi+tan^{-1}m,&m<0end{cases}$
answered Dec 18 '18 at 10:11
Shubham JohriShubham Johri
5,192717
$endgroup$
add a comment |
$begingroup$
Use this formula: $alpha=begin{cases}tan^{-1}m,&mge0\pi+tan^{-1}m,&m<0end{cases}$
answered Dec 18 '18 at 10:11
Shubham JohriShubham Johri
5,192717
$endgroup$
Use this formula: $alpha=begin{cases}tan^{-1}m,&mge0\pi+tan^{-1}m,&m<0end{cases}$
answered Dec 18 '18 at 10:11
Shubham JohriShubham Johri
5,192717
answered Dec 18 '18 at 10:11
Shubham JohriShubham Johri
5,192717
answered Dec 18 '18 at 10:11
Shubham JohriShubham Johri
5,192717
answered Dec 18 '18 at 10:11
Shubham JohriShubham Johri
5,192717
5,192717
add a comment |
add a comment |
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$begingroup$
It all depends on how you define $arctan$. It is usually defined as the inverse function of the restriction of $tan$ on $left( -frac{pi}{2}, frac{pi}{2}right)$.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 10:12