Value of the angle in isosceles triangle. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why are there isosceles triangles?Proving that the volume of a pyramid is one-third that of its corresponding prism.How to find coordinates of 3rd vertex of a right angled triangle when everything else is known?Calculate the angle from the given points coordinates.Geometry: Finding the sides of the triangle with base and altitude givenArea of an isosceles triangle where the tangents of some angles are in geometric progressionCalculate height of triangle given angle and baseHow to calculate the height of a triangle without using vector cross productThree circles in isosceles trianglecyclic quadrilateral inside an isosceles right triangle
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Value of the angle in isosceles triangle.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why are there isosceles triangles?Proving that the volume of a pyramid is one-third that of its corresponding prism.How to find coordinates of 3rd vertex of a right angled triangle when everything else is known?Calculate the angle from the given points coordinates.Geometry: Finding the sides of the triangle with base and altitude givenArea of an isosceles triangle where the tangents of some angles are in geometric progressionCalculate height of triangle given angle and baseHow to calculate the height of a triangle without using vector cross productThree circles in isosceles trianglecyclic quadrilateral inside an isosceles right triangle
$begingroup$
I try to find a way to calculate value of one of the isosceles triangle angles when I have given values of its height h = 200 and base x = 200. Values of those can vary depend on the condition, so I need an universal solution. I'll be very grateful for your help.
trigonometry triangles
asked Nov 16 '14 at 0:58
bluevoxelbluevoxel
150110
$endgroup$
add a comment |
$begingroup$
I try to find a way to calculate value of one of the isosceles triangle angles when I have given values of its height h = 200 and base x = 200. Values of those can vary depend on the condition, so I need an universal solution. I'll be very grateful for your help.
trigonometry triangles
asked Nov 16 '14 at 0:58
bluevoxelbluevoxel
150110
$endgroup$
add a comment |
$begingroup$
I try to find a way to calculate value of one of the isosceles triangle angles when I have given values of its height h = 200 and base x = 200. Values of those can vary depend on the condition, so I need an universal solution. I'll be very grateful for your help.
trigonometry triangles
asked Nov 16 '14 at 0:58
bluevoxelbluevoxel
150110
$endgroup$
I try to find a way to calculate value of one of the isosceles triangle angles when I have given values of its height h = 200 and base x = 200. Values of those can vary depend on the condition, so I need an universal solution. I'll be very grateful for your help.
trigonometry triangles
trigonometry triangles
asked Nov 16 '14 at 0:58
bluevoxelbluevoxel
150110
asked Nov 16 '14 at 0:58
bluevoxelbluevoxel
150110
asked Nov 16 '14 at 0:58
bluevoxelbluevoxel
150110
asked Nov 16 '14 at 0:58
bluevoxelbluevoxel
150110
asked Nov 16 '14 at 0:58
bluevoxelbluevoxel
150110
150110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you have a computer, you can say that
a = atan2(x/2, h)
Alternatively, you can use the simpler form
a = arctan(x/(2h))
If you type this into google, for particular values of $x$ and $h$, you'll get the answer you want. For instance, searching for
atan(100/200) in degrees
returned the answer
26.5650512 degrees
The query
atan(200/(2*200)) in degrees
produced the same thing.
A bit more detail.
Computing sine, cosine, and tangent isn't as easy as square root, but it's not insanely hard. It turns out that if you know $tan(a)$ and $tan(b)$ you can use "addition formulas" to compute $tan(a pm b)$ and $tan(a/2)$. Starting from one known value, like
$tan(45^deg) = 1$, you can compute many other values, enough to build a pretty complete table, and then fill in by interpolating. This takes time and energy, but that's life. Once you have a table of tangents, you can swap the columns to give you "inverse tangent" -- a function that says "what angle has this tangent?" That's called "arctan".
In fact, you can build an arctan table using addition rules as well, which is a bit more direct. Proving the addition laws? That's what trigonometry is all about.
It turns out that $arctan(x)$ can be expressed as a polynomial in $x$...but one with infinitely many terms. Fortunately, for small $x$, most of the terms are very small, so you can get away with evaluating just the first few (i.e., the larger) terms without making much error. This is in fact what the Java Math library's authors do, albeit in a somewhat more sophisticated way. The proof that arctan can be written as a polynomial comes up in calculus, under the general heading of "Approximation by Polynomials", or more specifically, "Taylor Series". Textbooks have multiple chapters about these, so I can't explain it all here, of course. Wish I could, but...
answered Nov 16 '14 at 1:01
John HughesJohn Hughes
65.5k24292
$endgroup$
$begingroup$
To be honest I asked this question in the context of Java, but I would like to know how this function is expressed mathematically. I would like to understand its basis.
$endgroup$
– bluevoxel
Nov 16 '14 at 1:10
$begingroup$
See details added in question.
$endgroup$
– John Hughes
Nov 16 '14 at 1:12
$begingroup$
Cool! Thank you very much. Time to go back to the maths' basics :)
$endgroup$
– bluevoxel
Nov 16 '14 at 1:28
add a comment |
$begingroup$
Using trigonometry, $a=arctan(100/200)$.
answered Nov 16 '14 at 1:01
Suzu HiroseSuzu Hirose
4,18021228
$endgroup$
$begingroup$
I believe this should be $arctan(100/200)$.
$endgroup$
– John Hughes
Nov 16 '14 at 1:02
$begingroup$
@JohnHughes - thanks. I've undeleted this since your post only discusses computing the number.
$endgroup$
– Suzu Hirose
Nov 16 '14 at 1:22
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
oldest
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oldest
votes
$begingroup$
If you have a computer, you can say that
a = atan2(x/2, h)
Alternatively, you can use the simpler form
a = arctan(x/(2h))
If you type this into google, for particular values of $x$ and $h$, you'll get the answer you want. For instance, searching for
atan(100/200) in degrees
returned the answer
26.5650512 degrees
The query
atan(200/(2*200)) in degrees
produced the same thing.
A bit more detail.
Computing sine, cosine, and tangent isn't as easy as square root, but it's not insanely hard. It turns out that if you know $tan(a)$ and $tan(b)$ you can use "addition formulas" to compute $tan(a pm b)$ and $tan(a/2)$. Starting from one known value, like
$tan(45^deg) = 1$, you can compute many other values, enough to build a pretty complete table, and then fill in by interpolating. This takes time and energy, but that's life. Once you have a table of tangents, you can swap the columns to give you "inverse tangent" -- a function that says "what angle has this tangent?" That's called "arctan".
In fact, you can build an arctan table using addition rules as well, which is a bit more direct. Proving the addition laws? That's what trigonometry is all about.
It turns out that $arctan(x)$ can be expressed as a polynomial in $x$...but one with infinitely many terms. Fortunately, for small $x$, most of the terms are very small, so you can get away with evaluating just the first few (i.e., the larger) terms without making much error. This is in fact what the Java Math library's authors do, albeit in a somewhat more sophisticated way. The proof that arctan can be written as a polynomial comes up in calculus, under the general heading of "Approximation by Polynomials", or more specifically, "Taylor Series". Textbooks have multiple chapters about these, so I can't explain it all here, of course. Wish I could, but...
answered Nov 16 '14 at 1:01
John HughesJohn Hughes
65.5k24292
$endgroup$
$begingroup$
To be honest I asked this question in the context of Java, but I would like to know how this function is expressed mathematically. I would like to understand its basis.
$endgroup$
– bluevoxel
Nov 16 '14 at 1:10
$begingroup$
See details added in question.
$endgroup$
– John Hughes
Nov 16 '14 at 1:12
$begingroup$
Cool! Thank you very much. Time to go back to the maths' basics :)
$endgroup$
– bluevoxel
Nov 16 '14 at 1:28
add a comment |
$begingroup$
If you have a computer, you can say that
a = atan2(x/2, h)
Alternatively, you can use the simpler form
a = arctan(x/(2h))
If you type this into google, for particular values of $x$ and $h$, you'll get the answer you want. For instance, searching for
atan(100/200) in degrees
returned the answer
26.5650512 degrees
The query
atan(200/(2*200)) in degrees
produced the same thing.
A bit more detail.
Computing sine, cosine, and tangent isn't as easy as square root, but it's not insanely hard. It turns out that if you know $tan(a)$ and $tan(b)$ you can use "addition formulas" to compute $tan(a pm b)$ and $tan(a/2)$. Starting from one known value, like
$tan(45^deg) = 1$, you can compute many other values, enough to build a pretty complete table, and then fill in by interpolating. This takes time and energy, but that's life. Once you have a table of tangents, you can swap the columns to give you "inverse tangent" -- a function that says "what angle has this tangent?" That's called "arctan".
In fact, you can build an arctan table using addition rules as well, which is a bit more direct. Proving the addition laws? That's what trigonometry is all about.
It turns out that $arctan(x)$ can be expressed as a polynomial in $x$...but one with infinitely many terms. Fortunately, for small $x$, most of the terms are very small, so you can get away with evaluating just the first few (i.e., the larger) terms without making much error. This is in fact what the Java Math library's authors do, albeit in a somewhat more sophisticated way. The proof that arctan can be written as a polynomial comes up in calculus, under the general heading of "Approximation by Polynomials", or more specifically, "Taylor Series". Textbooks have multiple chapters about these, so I can't explain it all here, of course. Wish I could, but...
answered Nov 16 '14 at 1:01
John HughesJohn Hughes
65.5k24292
$endgroup$
$begingroup$
To be honest I asked this question in the context of Java, but I would like to know how this function is expressed mathematically. I would like to understand its basis.
$endgroup$
– bluevoxel
Nov 16 '14 at 1:10
$begingroup$
See details added in question.
$endgroup$
– John Hughes
Nov 16 '14 at 1:12
$begingroup$
Cool! Thank you very much. Time to go back to the maths' basics :)
$endgroup$
– bluevoxel
Nov 16 '14 at 1:28
add a comment |
$begingroup$
If you have a computer, you can say that
a = atan2(x/2, h)
Alternatively, you can use the simpler form
a = arctan(x/(2h))
If you type this into google, for particular values of $x$ and $h$, you'll get the answer you want. For instance, searching for
atan(100/200) in degrees
returned the answer
26.5650512 degrees
The query
atan(200/(2*200)) in degrees
produced the same thing.
A bit more detail.
Computing sine, cosine, and tangent isn't as easy as square root, but it's not insanely hard. It turns out that if you know $tan(a)$ and $tan(b)$ you can use "addition formulas" to compute $tan(a pm b)$ and $tan(a/2)$. Starting from one known value, like
$tan(45^deg) = 1$, you can compute many other values, enough to build a pretty complete table, and then fill in by interpolating. This takes time and energy, but that's life. Once you have a table of tangents, you can swap the columns to give you "inverse tangent" -- a function that says "what angle has this tangent?" That's called "arctan".
In fact, you can build an arctan table using addition rules as well, which is a bit more direct. Proving the addition laws? That's what trigonometry is all about.
It turns out that $arctan(x)$ can be expressed as a polynomial in $x$...but one with infinitely many terms. Fortunately, for small $x$, most of the terms are very small, so you can get away with evaluating just the first few (i.e., the larger) terms without making much error. This is in fact what the Java Math library's authors do, albeit in a somewhat more sophisticated way. The proof that arctan can be written as a polynomial comes up in calculus, under the general heading of "Approximation by Polynomials", or more specifically, "Taylor Series". Textbooks have multiple chapters about these, so I can't explain it all here, of course. Wish I could, but...
answered Nov 16 '14 at 1:01
John HughesJohn Hughes
65.5k24292
$endgroup$
If you have a computer, you can say that
a = atan2(x/2, h)
Alternatively, you can use the simpler form
a = arctan(x/(2h))
If you type this into google, for particular values of $x$ and $h$, you'll get the answer you want. For instance, searching for
atan(100/200) in degrees
returned the answer
26.5650512 degrees
The query
atan(200/(2*200)) in degrees
produced the same thing.
A bit more detail.
Computing sine, cosine, and tangent isn't as easy as square root, but it's not insanely hard. It turns out that if you know $tan(a)$ and $tan(b)$ you can use "addition formulas" to compute $tan(a pm b)$ and $tan(a/2)$. Starting from one known value, like
$tan(45^deg) = 1$, you can compute many other values, enough to build a pretty complete table, and then fill in by interpolating. This takes time and energy, but that's life. Once you have a table of tangents, you can swap the columns to give you "inverse tangent" -- a function that says "what angle has this tangent?" That's called "arctan".
In fact, you can build an arctan table using addition rules as well, which is a bit more direct. Proving the addition laws? That's what trigonometry is all about.
It turns out that $arctan(x)$ can be expressed as a polynomial in $x$...but one with infinitely many terms. Fortunately, for small $x$, most of the terms are very small, so you can get away with evaluating just the first few (i.e., the larger) terms without making much error. This is in fact what the Java Math library's authors do, albeit in a somewhat more sophisticated way. The proof that arctan can be written as a polynomial comes up in calculus, under the general heading of "Approximation by Polynomials", or more specifically, "Taylor Series". Textbooks have multiple chapters about these, so I can't explain it all here, of course. Wish I could, but...
answered Nov 16 '14 at 1:01
John HughesJohn Hughes
65.5k24292
edited Nov 16 '14 at 1:19
answered Nov 16 '14 at 1:01
John HughesJohn Hughes
65.5k24292
answered Nov 16 '14 at 1:01
John HughesJohn Hughes
65.5k24292
answered Nov 16 '14 at 1:01
John HughesJohn Hughes
65.5k24292
65.5k24292
$begingroup$
To be honest I asked this question in the context of Java, but I would like to know how this function is expressed mathematically. I would like to understand its basis.
$endgroup$
– bluevoxel
Nov 16 '14 at 1:10
$begingroup$
See details added in question.
$endgroup$
– John Hughes
Nov 16 '14 at 1:12
$begingroup$
Cool! Thank you very much. Time to go back to the maths' basics :)
$endgroup$
– bluevoxel
Nov 16 '14 at 1:28
add a comment |
$begingroup$
To be honest I asked this question in the context of Java, but I would like to know how this function is expressed mathematically. I would like to understand its basis.
$endgroup$
– bluevoxel
Nov 16 '14 at 1:10
$begingroup$
See details added in question.
$endgroup$
– John Hughes
Nov 16 '14 at 1:12
$begingroup$
Cool! Thank you very much. Time to go back to the maths' basics :)
$endgroup$
– bluevoxel
Nov 16 '14 at 1:28
$begingroup$
To be honest I asked this question in the context of Java, but I would like to know how this function is expressed mathematically. I would like to understand its basis.
$endgroup$
– bluevoxel
Nov 16 '14 at 1:10
$begingroup$
To be honest I asked this question in the context of Java, but I would like to know how this function is expressed mathematically. I would like to understand its basis.
$endgroup$
– bluevoxel
Nov 16 '14 at 1:10
$begingroup$
See details added in question.
$endgroup$
– John Hughes
Nov 16 '14 at 1:12
$begingroup$
See details added in question.
$endgroup$
– John Hughes
Nov 16 '14 at 1:12
$begingroup$
Cool! Thank you very much. Time to go back to the maths' basics :)
$endgroup$
– bluevoxel
Nov 16 '14 at 1:28
$begingroup$
Cool! Thank you very much. Time to go back to the maths' basics :)
$endgroup$
– bluevoxel
Nov 16 '14 at 1:28
add a comment |
$begingroup$
Using trigonometry, $a=arctan(100/200)$.
answered Nov 16 '14 at 1:01
Suzu HiroseSuzu Hirose
4,18021228
$endgroup$
$begingroup$
I believe this should be $arctan(100/200)$.
$endgroup$
– John Hughes
Nov 16 '14 at 1:02
$begingroup$
@JohnHughes - thanks. I've undeleted this since your post only discusses computing the number.
$endgroup$
– Suzu Hirose
Nov 16 '14 at 1:22
add a comment |
$begingroup$
Using trigonometry, $a=arctan(100/200)$.
answered Nov 16 '14 at 1:01
Suzu HiroseSuzu Hirose
4,18021228
$endgroup$
$begingroup$
I believe this should be $arctan(100/200)$.
$endgroup$
– John Hughes
Nov 16 '14 at 1:02
$begingroup$
@JohnHughes - thanks. I've undeleted this since your post only discusses computing the number.
$endgroup$
– Suzu Hirose
Nov 16 '14 at 1:22
add a comment |
$begingroup$
Using trigonometry, $a=arctan(100/200)$.
answered Nov 16 '14 at 1:01
Suzu HiroseSuzu Hirose
4,18021228
$endgroup$
Using trigonometry, $a=arctan(100/200)$.
answered Nov 16 '14 at 1:01
Suzu HiroseSuzu Hirose
4,18021228
edited Nov 16 '14 at 1:21
answered Nov 16 '14 at 1:01
Suzu HiroseSuzu Hirose
4,18021228
answered Nov 16 '14 at 1:01
Suzu HiroseSuzu Hirose
4,18021228
answered Nov 16 '14 at 1:01
Suzu HiroseSuzu Hirose
4,18021228
4,18021228
$begingroup$
I believe this should be $arctan(100/200)$.
$endgroup$
– John Hughes
Nov 16 '14 at 1:02
$begingroup$
@JohnHughes - thanks. I've undeleted this since your post only discusses computing the number.
$endgroup$
– Suzu Hirose
Nov 16 '14 at 1:22
add a comment |
$begingroup$
I believe this should be $arctan(100/200)$.
$endgroup$
– John Hughes
Nov 16 '14 at 1:02
$begingroup$
@JohnHughes - thanks. I've undeleted this since your post only discusses computing the number.
$endgroup$
– Suzu Hirose
Nov 16 '14 at 1:22
$begingroup$
I believe this should be $arctan(100/200)$.
$endgroup$
– John Hughes
Nov 16 '14 at 1:02
$begingroup$
I believe this should be $arctan(100/200)$.
$endgroup$
– John Hughes
Nov 16 '14 at 1:02
$begingroup$
@JohnHughes - thanks. I've undeleted this since your post only discusses computing the number.
$endgroup$
– Suzu Hirose
Nov 16 '14 at 1:22
$begingroup$
@JohnHughes - thanks. I've undeleted this since your post only discusses computing the number.
$endgroup$
– Suzu Hirose
Nov 16 '14 at 1:22
add a comment |
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