Talk:Azimuth

message: When the introduction says "between a reference plane and a point," does it really mean "between a reference plane, a point, and an observer"? There can be an angle between two lines, between a plane and a line, but not between a plane and just one point.
The former commentary is fuck in unsigned but completally true. I thought exactly the same when I read the definition. Hope somebody can improve the present definition ...
Jasón (talk) 20:36, 1 September 2008 (UTC)
thread_title: Where's the angle?

message: Quote: This assumes, however, that the star (upper) culminates in the south ...
What is this opposed to mean? Maikel (talk) 11:30, 8 August 2008 (UTC)
thread_title: Upper culmination?

message: So... does the word "azimuth" derive from Arabic or Hebrew, or both, or neither, or what?  The current claims appear to conflict, and aren't sourced, either. 130.91.112.68 (talk) 04:52, 3 January 2010 (UTC)
thread_title:  Word Origin? 

message: Thank you Darkmeerkat for blue-ifying the azimuth-related lines, however, by misinterpreting the text I was still able to come to the wrong conclusion about which angle was the azimuth. Maybe if someone could also add one of those curved lines on the inside of the angle. 86.136.86.241 (talk) 23:10, 25 March 2010 (UTC)
replies: text: Isn't azimuth going the wrong way for celestial coordinate system which I think is what that diagram in the lead is for? Dmcq (talk) 22:01, 25 March 2012 (UTC)
replies: text: The original author, User:TWCarlson, has kindly reversed the image so that azimuth is now going around the right way.   --Amble (talk) 23:24, 18 June 2012 (UTC)
thread_title:  Image still could be better 

message: The article says azimuth increases in the clockwise direction, whereas I thought it increased in the westward direction, so I am slightly confused. Could someone confirm the azimuth sign reversal in the following scenario when crossing the latitude of the observed celestial object? Suppose the moon is right above Bangkok, at the zenith, and let's ignore earth's rotation so it stays there. Bangkok is at 20° west of Manilla, at the same latitude. Which azimuth values for the moon would be measured by someone flying from the North Pole to the South Pole, along a meridian over Manilla? The clockwise azimuth definition implies that azimuth increases from +20° to +180° while approaching Manilla, then suddenly changes to -180°, and subsequently increases from -180° to -20° while flying away from Manilla.   Ceinturion (talk) 16:12, 25 April 2013 (UTC)
replies: text: Sorry, my question is a mistake. On second thought, the azimuth would start at 200°, be 270° at Manilla without sign reversal, and end at 340°. There is no discontinuity. Ceinturion (talk) 16:57, 25 April 2013 (UTC)
thread_title:  Clockwise, not westward? 

message: In the [Navigation] and [Mapping] sections, the word 'plane' is used to describe a direction.
Particularly in the context of using a direction as a reference plane.
The term `reference vector` or would be more appropriate.  — Preceding unsigned comment added by Wikibls (talk • contribs) 09:29, 14 July 2013 (UTC)
thread_title:  Reference Plane 

message: Triangles are useful when you need to find unknown lengths and heights. Example:
1. Problem: Find length b.
Solution: We see that we know the length
of the hypotenuse and the measure
of angle A. We also see that
we need to find the length of b.
b is adjacent to angle A.
By definition, cosine = adjacent
over hypotenuse.
Therefore, we write an equation using
the cosine.
cos 19o = b/70
Multiply each side by 70 to isolate
the variable b.
70(cos 19o) = b
Use a calculator to find the
cos 19o.
b = 66.19
Before now, we have been able to use the trigonometric functions to solve right triangles. This doesn't do much good because there are many times triangles will not be right triangles. The Law of Sines is one way that we can solve oblique triangles. It is listed below.
a b c
----- = ----- = -----
sin A sin B sin C
When you know any two angles and any side of a triangle, you can use the law of sines to solve the triangle. Example:
1. Problem: In triangle ABC, a = 4.56,
A = 43o, and C = 57o.
Solve the triangle.
Solution: First, sketch the triangle and
include any given information. Here's our idea of a sketch:
Find angle B.
Angle B = 180o - (43o + 57o) = 80o
Now, we use the law of sines to
find the other sides lengths.
c a
----- = -----
sin C sin A
a(sin C)
c = --------
sin A
Plug in any known information.
4.56(sin 570)
c = -------------
sin 43o
Use a calculator to find the sines.
4.56(.8387)
c = -----------
.6820
c = 5.61
Now solve for b.
b a
----- = -----
sin B sin A
a(sin B)
b = --------
sin A
Plug in any known information.
4.56(sin 80o)
b = -------------
.6820
Use a calculator to find the sines.
4.56(.9848)
b = -----------
.6820
b = 6.58
There is another case when you can use the law of sines. When you know two sides and an angle opposite one of the sides, the law of sines can be used. However, with this case, you have to be aware that there might not be a solution, or there may be two! One solution is also possible. Example:
2. Problem: In triangle ABC, a = 15,
b = 25, and angle A = 47o.
Solve the triangle.
Solution: Start out by looking for the measure of angle B.
a b
----- = -----
sin A sin B
b(sin A)
sin B = --------
a
Plug in any known information.
25(sin 47o)
sin B = -----------
15
Use a calculator to find the sine.
25(.7314)
sin B = ---------
15
sin B = 1.219
Since an angle cannot have a sine
greater than 1, there is no solution
for this triangle.
You can also solve triangles that are not right triangles using the Law of Cosines. There are three rules that make up the law of cosines, but you only need to memorize one because the other two can be obtained by changing the letters (put b in place of a, for example). All three rules are listed below. a2 = b2 + c2 - 2bc(cos A) b2 = a2 + c2 - 2ac(cos B) c2 = a2 + b2 - 2ab(cos C) The law of cosines can be used in two different cases. First, if you know the lengths of all three sides of a triangle, you can use the law of cosines. Secondly, you can use the law of cosines when two sides and the included angle are known. Example:
1. Problem: In triangle ABC, a = 24,
c = 32, and angle B = 115o.
Solve the triangle.
Solution: We know two sides. Find the third
using the law of cosines.
b2 = a2 + c2 - 2ac(cos B)
Plug in any known information.
b2 = 242 + 322 - 2 * 24 * 32(-.4226)
b2 = 2249
Solve for b by taking the square
root of each side.
b = SQRT(2249) = 47.4
Using the law of sines,
angle A = 27.32o.
Since we know that triangles are made of
three angles that sum to 180o,
we can find the measure of angle C by
setting up an equation.
Angle C = 180o - 115o - 27.32o = 37.68o  — Preceding unsigned comment added by Hellom2014 (talk • contribs) 18:50, 2 April 2014 (UTC)
thread_title:  Triangles  

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thread_title:  Drafting Technician 

message: The figure does not clearly depict what the zenith stands for and the reference to the zenith page is not helpful (reference to zenith distance instead?).128.178.7.131 (talk) 09:29, 13 May 2014 (UTC)
replies: text: What is not clear about the depiction? Zenith is shown in the depiction. Why not link to it? - Ac44ck (talk) 01:24, 14 May 2014 (UTC)
thread_title:  Zenith and Azimuth

message: This article and its Talk Page are a shame!
They are a clear example of all that can be criticized of Wikipedia: contributors that only contribute their ignorance. The illustration is confusing, the wording is unclear and plagued by inconsistencies and plain lies, and any dictionary does a better job than this article in its current, dismal state.
By no means an insult is intended to anyone, but just read this sentence:
"The star is the point of interest, the reference plane is the horizon or the surface of the sea"
Whoever believes that the surface of the sea is a plane doesn't belong writing articles in this encyclopedia, he belongs back in 2nd-grade Geometry classes.
--AVM (talk) 21:15, 18 November 2014 (UTC)
thread_title:  A SHAME 

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thread_title:  External links modified 

message: While the current page is clearly trying to describe azimuth in a more general way, I was thinking that it might be good to add a section relating specifically to launch_azimuth (the azimuth to launch a rocket, to enter a target plane). I believe simply adding a few lines from https://www.orbiterwiki.org/wiki/Launch_Azimuth might be useful, most specifically the calculation section could use the simple az=asin(cos(inclination)/cos(altitude)) equation, with a short description of what it is, the terms, and that it obviously wont work regardless of the values for inc&alt. And that it won't be perfect for rotating bodies.
As a side-note, I have seen the phrase azimuth used for the launch-window to enter a plane, and although I suspect it is a misnomer (unsure), a mention about it - or possibly even calculations - might improve the page. --83.248.100.73 (talk) 14:42, 9 September 2019 (UTC)
thread_title:  launch_azimuth 

message: If the azimuth is expressed in decimal degrees (gons) the factor 180 should be 200.Madyno (talk) 10:21, 16 August 2020 (UTC)
thread_title:  Cartography