<p>If one is concerned with formal geometry, then the spherical trigonometry is the way to go. First, let's try to find the azimuth of intersection of decliation=0 (i..e, the Sun on the equinoxes) with the horizon, at a given latitude <span class="math-container">$\lambda &gt;= 0$</span> :</p>
<p>We shall consider the spherical triangle created by 3 points: the Zenith(Z), the North pole(P) and a point of intersection(M) of dec=0 with the horizon. We already know enough:</p>
<ol>
<li>the length of arc ZP is (<span class="math-container">$90- \lambda$</span>) - that's from the definition of the north pole and the Zenith.</li>
<li>the length of arc ZM is <span class="math-container">$90$</span>. this is also from the definition of the horizon and the Zenith.</li>
<li>the length of the arc PM is <span class="math-container">$90$</span>. This is because the Pole is declination of 90, and the point M is of declination <span class="math-container">$0$</span>.</li>
</ol>
<p>Now, we can employ the <a href="https://en.wikipedia.org/wiki/Spherical_law_of_cosines" rel="nofollow noreferrer">spherical cosine law</a>, let us denote the angle between the arcs ZP and ZM with <span class="math-container">$x$</span> - that's angle is actually the azimuth of point N (*) (defining North pole with azimuth 0):</p>
<p><span class="math-container">$$ \cos(90) = \cos(90- \lambda)*\cos(90) + \sin(90- \lambda)*\sin(90)*\cos(x)  $$</span></p>
<p>We got that <span class="math-container">$\cos(x) = 0$</span> not matter the <span class="math-container">$\lambda$</span>. That means that on the equinox, across all Earth the azimuth of the Sun rise and Set is <span class="math-container">$90$</span> and <span class="math-container">$270$</span> (i.e., East and West).</p>
<p>Now, what happens on a day which other than the equinox where the declination is not necessary 0? Let's set the declination to be <span class="math-container">$\delta$</span>. All above remains the same, but only that the length of the arc PN is <span class="math-container">$90-\delta$</span>; then we have:</p>
<p><span class="math-container">$$ \cos(90-\delta) = \cos(90- \lambda)*\cos(90) + \sin(90- \lambda)*\sin(90)*\cos(x)  $$</span></p>
<p><span class="math-container">$$ =&gt; \sin(\delta) = \cos(\lambda)*\cos(x)$$</span></p>
<p><span class="math-container">$$ =&gt; cos(x) = \dfrac{\sin(\delta)}{\cos(\lambda)}$$</span></p>
<p>Usually this should yield 2 results (one for the Sun rise and one for the Sun set), however, it can also yield no results if, for example, <span class="math-container">$\delta &gt; (90-\lambda$</span>) in which case, there is no rise and no set. either the Sun above or below the horizon all day.</p>
<hr />
<p>(*) An important note: We said that the angle at Z is the difference in Azimuth between the North Pole and our point on the Horizon (M). This is indeed correct, but it is not so trivial to say so. If, for example, we were searching for the angle at the North Pole, we could <strong>not</strong> say that this angle is the difference in RA between the Zenith and our point on the Horizon (M). I think it works in the former case only because both the arcs ZP and ZM are arcs of great circles. (ZM is an arc of a great circle because it is perpendicular to the Horizon and hence an azimuth circle; ZP is an arc of a great circle which is <a href="https://en.wikipedia.org/wiki/Meridian_(astronomy)" rel="nofollow noreferrer">called the Meridan</a> - the circle of RA (and actually also of Azimuth) that goes between the North and South poles that passed through the Zenith); however, PM is not necessarily an arc of a great circle and we could not say that this constitutes an RA circle.</p>

CC BY-SA 4.0
2023-02-02T13:49:12.290Z
2023-02-02T21:27:49.687Z
2023-02-02T21:27:49.687Z
d_e
<p>The angle at which the sun sets is not exactly the same as your latitude, but it can be close. The reason for this is that the sun sets at an angle relative to the horizon that is equal to your latitude on the spring and fall equinoxes, which occur on March 20th/21st and September 22nd/23rd. On these two days, the sun rises due east and sets due west, and its highest point in the sky is directly overhead at the equator. At other times of the year, the angle of the sun at sunset will be different from your latitude due to the tilt of the Earth's axis and the changing position of the Earth in its orbit around the sun.</p>

CC BY-SA 4.0
2023-02-01T17:32:18.557Z
Hi @vishvajit. I do not understand your answer. On what date and what latitude is the azimuth of the rising sun (-90 degrees to make relative to East) approximately the same as the latitude? You mention the equinox. For all latitudes except near the poles, the Sun rises due east (so an angle of 0 degrees relative to east). This is not a good approximation for any latitude other than the equator.
JohnHoltz
CC BY-SA 4.0
2023-02-06T05:59:46.393Z
The point on the horizon where the sun appears is due east. But that does not mean it goes straight up; that only happens at the equator. Instead it follows a sloping path, and the angle of that path matches your latitude because the circle marking your horizon defines a plane at that angle to the earth's axis.
Mark Foskey
CC BY-SA 4.0
2023-02-01T13:05:52.587Z
2023-02-01T13:05:52.587Z
vishvajit
<p>Unfortunately the assertion that your latitude is equal to the Sun rise and set times is not correct.  This is most obvious near the arctic circles where the Sun will rize/set anywhere from <span class="math-container">$0^\circ$</span> to <span class="math-container">$180^\circ$</span>.</p>
<p>The first image below is a <a href="https://en.wikipedia.org/wiki/Solarigraphy" rel="nofollow noreferrer">solarigraph</a> showing the path of the Sun through the sky from one solstice to the next.  You can see that it sets at different azimuths throughout the cycle.  Note the paths are distorted due to how the image was taken, in reality the lines in the center are just as far apart near the edges (for example, <a href="https://www.9news.com/article/weather/weather-colorado/home-made-camera-captures-long-exposure-of-the-sun-angle/73-a64dbc8c-bd9f-4813-9752-adfa78eb7f7b" rel="nofollow noreferrer">here is one facing East</a>).</p>
<p>In the second image, the thin red lines depict where the Sun will rise and set on the summer and winter solstices.</p>
<p><a href="https://i.sstatic.net/kSrJH.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/kSrJH.png" alt="enter image description here" /></a></p>
<p><a href="https://i.sstatic.net/i0HIY.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/i0HIY.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-02-02T06:54:49.130Z
Although the question is not very clear, I believe that it refers to the angle beteen the sun's path and a line perpendicular to the horizon, with it's base at the point of sunset.
Dr Chuck
CC BY-SA 4.0
2023-02-02T13:07:30.287Z
@DrChuck, that's what I thought at first, but the OP clarified it was about azimuth in a comment.
Greg Miller
CC BY-SA 4.0
2023-02-02T16:34:37.877Z
Greg - having read the comment I now have no idea what the question is.
Dr Chuck
CC BY-SA 4.0
2023-02-03T19:01:25.290Z
Hi sorry for the unclear wording. I was having trouble writing down exactly what I was having trouble with. But what @DrChuck said was basically my question.
Susanoo D Ace
CC BY-SA 4.0
2023-02-01T14:37:35.863Z
2023-02-01T14:37:35.863Z
Greg Miller
<p>I learned in class that the sun sets at an angle when it is not an equinox. I also learned that the angle at which it sets is a good approximation of your latitude on earth. I am having trouble drawing a visual model of this geometry.</p>

CC BY-SA 4.0
2023-01-31T22:12:31.803Z
This seems to be missing details: How are you defining the angle at which the sun sets?  The angle between the point of sunset and Due West?  The angle between the path of the Sun at sunset and the horizon? Some other measurement?
notovny
CC BY-SA 4.0
2023-02-01T00:04:29.067Z
The angle between setting straight from east to west, as opposed to setting northeast to northwest or southeast to southwest.
Susanoo D Ace
CC BY-SA 4.0
2023-02-01T11:53:34.733Z
I don't have the time to write up an answer right now, but basically, the premise is false. The angle between due west and the aziumuth of sunset (if it occurs) varies throughout the year for all locations, minimized at equinoxes and maximized at solstices (or whatever date the midnight sun starts for locations inside the polar circles), and for latitudes between roughly 27 degrees and 64 degrees, that angle is always less than the latitude's value.
notovny
CC BY-SA 4.0
2023-02-01T13:21:05.657Z
In summer the sun never sets at higher latitudes inside the polar circle. This alone proves the premise false. Please be more elaborate and descriptive (possibly with a sketch of what angles and what time-of-year you talk about) in your question if you disagree.
planetmaker
CC BY-SA 4.0
2023-02-14T04:30:11.070Z
Draw a circle on a sheet of paper. Call the top of the circle the North Pole. Draw a horizontal line through the middle: the Equator. Draw another horizontal line representing your own latitude. Now, hold the sheet vertically in front of you: this is equivalent to an equinox. Pretend the Sun's on the left, and tilt the sheet *clockwise* by 23.4 degrees: northern winter solstice. Bring it back to vertical: northern vernal equinox. Now tilt it the other way by 23.4 degrees: northern summer solstice. What line would you draw to represent sunset? Should it change as you tilt your sheet? :-)
Chappo Hasn't Forgotten
<observational-astronomy><the-sun>
Why is the angle at which the sun sets the same as your latitude?
Susanoo D Ace
# Why is the angle at which the sun sets the same as your latitude?

(Asked by: username_0 on 2023-01-31)

I learned in class that the sun sets at an angle when it is not an equinox. I also learned that the angle at which it sets is a good approximation of your latitude on earth. I am having trouble drawing a visual model of this geometry.

**Tags:** observational-astronomy, the-sun
> **Comments:**
> **username_1**: This seems to be missing details: How are you defining the angle at which the sun sets?  The angle between the point of sunset and Due West?  The angle between the path of the Sun at sunset and the horizon? Some other measurement?  
> **username_0**: The angle between setting straight from east to west, as opposed to setting northeast to northwest or southeast to southwest.  
> **username_1**: I don't have the time to write up an answer right now, but basically, the premise is false. The angle between due west and the aziumuth of sunset (if it occurs) varies throughout the year for all locations, minimized at equinoxes and maximized at solstices (or whatever date the midnight sun starts for locations inside the polar circles), and for latitudes between roughly 27 degrees and 64 degrees, that angle is always less than the latitude's value.  
> **username_2**: In summer the sun never sets at higher latitudes inside the polar circle. This alone proves the premise false. Please be more elaborate and descriptive (possibly with a sketch of what angles and what time-of-year you talk about) in your question if you disagree.  
> **username_3**: Draw a circle on a sheet of paper. Call the top of the circle the North Pole. Draw a horizontal line through the middle: the Equator. Draw another horizontal line representing your own latitude. Now, hold the sheet vertically in front of you: this is equivalent to an equinox. Pretend the Sun's on the left, and tilt the sheet *clockwise* by 23.4 degrees: northern winter solstice. Bring it back to vertical: northern vernal equinox. Now tilt it the other way by 23.4 degrees: northern summer solstice. What line would you draw to represent sunset? Should it change as you tilt your sheet? :-)  

---

## 3 answers

### Answer by username_4 on 2023-02-02. Score: 4

If one is concerned with formal geometry, then the spherical trigonometry is the way to go. First, let's try to find the azimuth of intersection of decliation=0 (i..e, the Sun on the equinoxes) with the horizon, at a given latitude $\lambda >= 0$ :

We shall consider the spherical triangle created by 3 points: the Zenith(Z), the North pole(P) and a point of intersection(M) of dec=0 with the horizon. We already know enough:

1. the length of arc ZP is ($90- \lambda$) - that's from the definition of the north pole and the Zenith.
2. the length of arc ZM is $90$. this is also from the definition of the horizon and the Zenith.
3. the length of the arc PM is $90$. This is because the Pole is declination of 90, and the point M is of declination $0$.

Now, we can employ the [spherical cosine law](https://en.wikipedia.org/wiki/Spherical_law_of_cosines), let us denote the angle between the arcs ZP and ZM with $x$ - that's angle is actually the azimuth of point N (\*) (defining North pole with azimuth 0):

$$ \cos(90) = \cos(90- \lambda)\*\cos(90) + \sin(90- \lambda)\*\sin(90)\*\cos(x) $$

We got that $\cos(x) = 0$ not matter the $\lambda$. That means that on the equinox, across all Earth the azimuth of the Sun rise and Set is $90$ and $270$ (i.e., East and West).

Now, what happens on a day which other than the equinox where the declination is not necessary 0? Let's set the declination to be $\delta$. All above remains the same, but only that the length of the arc PN is $90-\delta$; then we have:

$$ \cos(90-\delta) = \cos(90- \lambda)\*\cos(90) + \sin(90- \lambda)\*\sin(90)\*\cos(x) $$

$$ => \sin(\delta) = \cos(\lambda)\*\cos(x)$$

$$ => cos(x) = \dfrac{\sin(\delta)}{\cos(\lambda)}$$

Usually this should yield 2 results (one for the Sun rise and one for the Sun set), however, it can also yield no results if, for example, $\delta > (90-\lambda$) in which case, there is no rise and no set. either the Sun above or below the horizon all day.

---

(\*) An important note: We said that the angle at Z is the difference in Azimuth between the North Pole and our point on the Horizon (M). This is indeed correct, but it is not so trivial to say so. If, for example, we were searching for the angle at the North Pole, we could **not** say that this angle is the difference in RA between the Zenith and our point on the Horizon (M). I think it works in the former case only because both the arcs ZP and ZM are arcs of great circles. (ZM is an arc of a great circle because it is perpendicular to the Horizon and hence an azimuth circle; ZP is an arc of a great circle which is [called the Meridan](https://en.wikipedia.org/wiki/Meridian_(astronomy)) - the circle of RA (and actually also of Azimuth) that goes between the North and South poles that passed through the Zenith); however, PM is not necessarily an arc of a great circle and we could not say that this constitutes an RA circle.


---

### Answer by username_5 on 2023-02-01. Score: 3

The angle at which the sun sets is not exactly the same as your latitude, but it can be close. The reason for this is that the sun sets at an angle relative to the horizon that is equal to your latitude on the spring and fall equinoxes, which occur on March 20th/21st and September 22nd/23rd. On these two days, the sun rises due east and sets due west, and its highest point in the sky is directly overhead at the equator. At other times of the year, the angle of the sun at sunset will be different from your latitude due to the tilt of the Earth's axis and the changing position of the Earth in its orbit around the sun.

> **Comments:**
> **username_6**: Hi @vishvajit. I do not understand your answer. On what date and what latitude is the azimuth of the rising sun (-90 degrees to make relative to East) approximately the same as the latitude? You mention the equinox. For all latitudes except near the poles, the Sun rises due east (so an angle of 0 degrees relative to east). This is not a good approximation for any latitude other than the equator.  
> **username_7**: The point on the horizon where the sun appears is due east. But that does not mean it goes straight up; that only happens at the equator. Instead it follows a sloping path, and the angle of that path matches your latitude because the circle marking your horizon defines a plane at that angle to the earth's axis.  

---

### Answer by username_8 on 2023-02-01. Score: 3

Unfortunately the assertion that your latitude is equal to the Sun rise and set times is not correct. This is most obvious near the arctic circles where the Sun will rize/set anywhere from $0^\circ$ to $180^\circ$.

The first image below is a [solarigraph](https://en.wikipedia.org/wiki/Solarigraphy) showing the path of the Sun through the sky from one solstice to the next. You can see that it sets at different azimuths throughout the cycle. Note the paths are distorted due to how the image was taken, in reality the lines in the center are just as far apart near the edges (for example, [here is one facing East](https://www.9news.com/article/weather/weather-colorado/home-made-camera-captures-long-exposure-of-the-sun-angle/73-a64dbc8c-bd9f-4813-9752-adfa78eb7f7b)).

In the second image, the thin red lines depict where the Sun will rise and set on the summer and winter solstices.

[![enter image description here](https://i.sstatic.net/kSrJH.png)](https://i.sstatic.net/kSrJH.png)

[![enter image description here](https://i.sstatic.net/i0HIY.jpg)](https://i.sstatic.net/i0HIY.jpg)

> **Comments:**
> **username_9**: Although the question is not very clear, I believe that it refers to the angle beteen the sun's path and a line perpendicular to the horizon, with it's base at the point of sunset.  
> **username_8**: @DrChuck, that's what I thought at first, but the OP clarified it was about azimuth in a comment.  
> **username_9**: Greg - having read the comment I now have no idea what the question is.  
> **username_0**: Hi sorry for the unclear wording. I was having trouble writing down exactly what I was having trouble with. But what @DrChuck said was basically my question.  

---

