Using functions like sine, cosine and tangent to calculate coordinates in Tikz












2















I am learning TikZ with the pgfmanual. The example given is as follows:



enter image description here



I have achieved the following so far:



enter image description here



The code is as follows:



documentclass{article}
usepackage{tikz}

tikzset{help lines/.style=very thin}
tikzset{My Grid/.style={help lines,color=blue!50}}

begin{document}
begin{tikzpicture}
draw[My Grid] (-4,-4) grid (4,4);
draw (-5,0) node[left] {$(-5,0)$} -- (5,0) node[right] {$(5,0)$};
draw (0,-5) node[below] {$(0,-5)$} -- (0,5) node[above] {$(0,5)$};
draw (0,0) circle [radius=3cm];
shadedraw[left color=gray, right color=green, draw=green!50!black] (0,0) -- (0.75,0) arc [start angle=0, end angle=30, radius=1cm] -- cycle;
draw[red, very thick] (30:3cm) -- (2.6,0);
draw [very thick,orange] (3,0) -- (3,1.7);
end{tikzpicture}
end{document}


To achieve the intersection of the slope and tangent, pgfmanual uses the concept of path and intersections library which is very confusing.



Is there an easier way to tell the system to draw a line from point A to sin(30) as point B and to draw tangent of the angle proportionately instead of using direct numbers.



Please suggest alternative apart from intersections, polar coordinates and paths. Because the pgfmanual already uses them which is hard to understand.










share|improve this question

























  • May be If you specify exactly what is hard for you to understand someone will try to explain it to you

    – Hafid Boukhoulda
    Feb 23 at 8:14











  • @hafid The slope lines and connecting the tangent for the time being

    – subham soni
    Feb 23 at 8:16











  • Do you need to use cos(30) and tan(30) to indicate point coordinates instead of (2.6,0) and (3,1.7)

    – Hafid Boukhoulda
    Feb 23 at 8:23













  • Yes @hafid. That is exactly the requirement.

    – subham soni
    Feb 23 at 8:27











  • I think your approach is simpler than working with explicit sin and cos functions. E.g. the red line can be drawn like draw[red, very thick] (30:3cm) -- (0,0-|30:3cm);, where the last coordinate is the projection of the first one on y=0.

    – marmot
    Feb 23 at 14:11
















2















I am learning TikZ with the pgfmanual. The example given is as follows:



enter image description here



I have achieved the following so far:



enter image description here



The code is as follows:



documentclass{article}
usepackage{tikz}

tikzset{help lines/.style=very thin}
tikzset{My Grid/.style={help lines,color=blue!50}}

begin{document}
begin{tikzpicture}
draw[My Grid] (-4,-4) grid (4,4);
draw (-5,0) node[left] {$(-5,0)$} -- (5,0) node[right] {$(5,0)$};
draw (0,-5) node[below] {$(0,-5)$} -- (0,5) node[above] {$(0,5)$};
draw (0,0) circle [radius=3cm];
shadedraw[left color=gray, right color=green, draw=green!50!black] (0,0) -- (0.75,0) arc [start angle=0, end angle=30, radius=1cm] -- cycle;
draw[red, very thick] (30:3cm) -- (2.6,0);
draw [very thick,orange] (3,0) -- (3,1.7);
end{tikzpicture}
end{document}


To achieve the intersection of the slope and tangent, pgfmanual uses the concept of path and intersections library which is very confusing.



Is there an easier way to tell the system to draw a line from point A to sin(30) as point B and to draw tangent of the angle proportionately instead of using direct numbers.



Please suggest alternative apart from intersections, polar coordinates and paths. Because the pgfmanual already uses them which is hard to understand.










share|improve this question

























  • May be If you specify exactly what is hard for you to understand someone will try to explain it to you

    – Hafid Boukhoulda
    Feb 23 at 8:14











  • @hafid The slope lines and connecting the tangent for the time being

    – subham soni
    Feb 23 at 8:16











  • Do you need to use cos(30) and tan(30) to indicate point coordinates instead of (2.6,0) and (3,1.7)

    – Hafid Boukhoulda
    Feb 23 at 8:23













  • Yes @hafid. That is exactly the requirement.

    – subham soni
    Feb 23 at 8:27











  • I think your approach is simpler than working with explicit sin and cos functions. E.g. the red line can be drawn like draw[red, very thick] (30:3cm) -- (0,0-|30:3cm);, where the last coordinate is the projection of the first one on y=0.

    – marmot
    Feb 23 at 14:11














2












2








2








I am learning TikZ with the pgfmanual. The example given is as follows:



enter image description here



I have achieved the following so far:



enter image description here



The code is as follows:



documentclass{article}
usepackage{tikz}

tikzset{help lines/.style=very thin}
tikzset{My Grid/.style={help lines,color=blue!50}}

begin{document}
begin{tikzpicture}
draw[My Grid] (-4,-4) grid (4,4);
draw (-5,0) node[left] {$(-5,0)$} -- (5,0) node[right] {$(5,0)$};
draw (0,-5) node[below] {$(0,-5)$} -- (0,5) node[above] {$(0,5)$};
draw (0,0) circle [radius=3cm];
shadedraw[left color=gray, right color=green, draw=green!50!black] (0,0) -- (0.75,0) arc [start angle=0, end angle=30, radius=1cm] -- cycle;
draw[red, very thick] (30:3cm) -- (2.6,0);
draw [very thick,orange] (3,0) -- (3,1.7);
end{tikzpicture}
end{document}


To achieve the intersection of the slope and tangent, pgfmanual uses the concept of path and intersections library which is very confusing.



Is there an easier way to tell the system to draw a line from point A to sin(30) as point B and to draw tangent of the angle proportionately instead of using direct numbers.



Please suggest alternative apart from intersections, polar coordinates and paths. Because the pgfmanual already uses them which is hard to understand.










share|improve this question
















I am learning TikZ with the pgfmanual. The example given is as follows:



enter image description here



I have achieved the following so far:



enter image description here



The code is as follows:



documentclass{article}
usepackage{tikz}

tikzset{help lines/.style=very thin}
tikzset{My Grid/.style={help lines,color=blue!50}}

begin{document}
begin{tikzpicture}
draw[My Grid] (-4,-4) grid (4,4);
draw (-5,0) node[left] {$(-5,0)$} -- (5,0) node[right] {$(5,0)$};
draw (0,-5) node[below] {$(0,-5)$} -- (0,5) node[above] {$(0,5)$};
draw (0,0) circle [radius=3cm];
shadedraw[left color=gray, right color=green, draw=green!50!black] (0,0) -- (0.75,0) arc [start angle=0, end angle=30, radius=1cm] -- cycle;
draw[red, very thick] (30:3cm) -- (2.6,0);
draw [very thick,orange] (3,0) -- (3,1.7);
end{tikzpicture}
end{document}


To achieve the intersection of the slope and tangent, pgfmanual uses the concept of path and intersections library which is very confusing.



Is there an easier way to tell the system to draw a line from point A to sin(30) as point B and to draw tangent of the angle proportionately instead of using direct numbers.



Please suggest alternative apart from intersections, polar coordinates and paths. Because the pgfmanual already uses them which is hard to understand.







tikz-pgf intersections






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Feb 23 at 9:22









Hafid Boukhoulda

4,6341625




4,6341625










asked Feb 23 at 8:04









subham sonisubham soni

4,61983185




4,61983185













  • May be If you specify exactly what is hard for you to understand someone will try to explain it to you

    – Hafid Boukhoulda
    Feb 23 at 8:14











  • @hafid The slope lines and connecting the tangent for the time being

    – subham soni
    Feb 23 at 8:16











  • Do you need to use cos(30) and tan(30) to indicate point coordinates instead of (2.6,0) and (3,1.7)

    – Hafid Boukhoulda
    Feb 23 at 8:23













  • Yes @hafid. That is exactly the requirement.

    – subham soni
    Feb 23 at 8:27











  • I think your approach is simpler than working with explicit sin and cos functions. E.g. the red line can be drawn like draw[red, very thick] (30:3cm) -- (0,0-|30:3cm);, where the last coordinate is the projection of the first one on y=0.

    – marmot
    Feb 23 at 14:11



















  • May be If you specify exactly what is hard for you to understand someone will try to explain it to you

    – Hafid Boukhoulda
    Feb 23 at 8:14











  • @hafid The slope lines and connecting the tangent for the time being

    – subham soni
    Feb 23 at 8:16











  • Do you need to use cos(30) and tan(30) to indicate point coordinates instead of (2.6,0) and (3,1.7)

    – Hafid Boukhoulda
    Feb 23 at 8:23













  • Yes @hafid. That is exactly the requirement.

    – subham soni
    Feb 23 at 8:27











  • I think your approach is simpler than working with explicit sin and cos functions. E.g. the red line can be drawn like draw[red, very thick] (30:3cm) -- (0,0-|30:3cm);, where the last coordinate is the projection of the first one on y=0.

    – marmot
    Feb 23 at 14:11

















May be If you specify exactly what is hard for you to understand someone will try to explain it to you

– Hafid Boukhoulda
Feb 23 at 8:14





May be If you specify exactly what is hard for you to understand someone will try to explain it to you

– Hafid Boukhoulda
Feb 23 at 8:14













@hafid The slope lines and connecting the tangent for the time being

– subham soni
Feb 23 at 8:16





@hafid The slope lines and connecting the tangent for the time being

– subham soni
Feb 23 at 8:16













Do you need to use cos(30) and tan(30) to indicate point coordinates instead of (2.6,0) and (3,1.7)

– Hafid Boukhoulda
Feb 23 at 8:23







Do you need to use cos(30) and tan(30) to indicate point coordinates instead of (2.6,0) and (3,1.7)

– Hafid Boukhoulda
Feb 23 at 8:23















Yes @hafid. That is exactly the requirement.

– subham soni
Feb 23 at 8:27





Yes @hafid. That is exactly the requirement.

– subham soni
Feb 23 at 8:27













I think your approach is simpler than working with explicit sin and cos functions. E.g. the red line can be drawn like draw[red, very thick] (30:3cm) -- (0,0-|30:3cm);, where the last coordinate is the projection of the first one on y=0.

– marmot
Feb 23 at 14:11





I think your approach is simpler than working with explicit sin and cos functions. E.g. the red line can be drawn like draw[red, very thick] (30:3cm) -- (0,0-|30:3cm);, where the last coordinate is the projection of the first one on y=0.

– marmot
Feb 23 at 14:11










2 Answers
2






active

oldest

votes


















8














Since in cos(30) and tan(30) there are parentheses you must put these functions inside curly brackets {}



documentclass{article}
usepackage{tikz}
usetikzlibrary{angles,quotes}

tikzset{help lines/.style=very thin}
tikzset{My Grid/.style={help lines,color=blue!50}}

begin{document}
begin{tikzpicture}
draw[My Grid] (-4,-4) grid (4,4);
draw (-5,0) node[left] {$(-5,0)$} -- (5,0) node[right] {$(5,0)$};
draw (0,-5) node[below] {$(0,-5)$} -- (0,5) node[above] {$(0,5)$};
draw (0,0) circle [radius=3cm];

% shadedraw[left color=gray, right color=green, draw=green!50!black]
% (0,0) -- (0.75,0) arc [start angle=0, end angle=30, radius=0.75cm] -- cycle;

coordinate(O)at(0,0);
draw[red, very thick] (30:3cm)coordinate(A)
--({3*cos(30)},0)coordinate(B);

draw [very thick,orange] (3,0) -- (3,{3*tan(30)})coordinate(C);

pic[fill=green!50!black,
angle radius=0.75cm,
angle eccentricity=1.2,
"(alpha)"] {angle=B--O--A};

draw (O)--(C);

end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • Will this ensure that the length is actual value when computed as well or plotted manually with pencil and graph?

    – subham soni
    Feb 23 at 8:32











  • @subhamsoni I notice that the green angle is greeter than 30° I suppose you needed to be 30°

    – Hafid Boukhoulda
    Feb 23 at 8:38













  • I have mentioned end angle as 30. Is there anything I am missing

    – subham soni
    Feb 23 at 8:39











  • @subhamsoni Yes you need the specify the correct radius 0.75 cm and not 1cm since the start point is (0.75,0). Also in Tikz there is another manner to draw angles.

    – Hafid Boukhoulda
    Feb 23 at 8:43













  • it would be great if you can share the other way because precision is always better.

    – subham soni
    Feb 23 at 8:46



















3














Here is an alternative to Hafid's nice answer, going further into the direction you started. You do not need to use any trigonometric function. Just using polar coordinates and projections is sufficient.



documentclass{article}
usepackage{tikz}
usetikzlibrary{calc,angles,quotes}
tikzset{help lines/.style=very thin}
tikzset{My Grid/.style={help lines,color=blue!50}}

begin{document}
begin{tikzpicture}
draw[My Grid] (-4,-4) grid (4,4);
draw (-5,0) node[left] {$(-5,0)$} -- (5,0) node[right] {$(5,0)$};
draw (0,-5) node[below] {$(0,-5)$} -- (0,5) node[above] {$(0,5)$};
draw (0,0) coordinate (O) circle [radius=3cm];
draw[red, very thick] (30:3cm) coordinate (A)
% (30:3cm) is a polar coordinate with angle 30 (degrees) and radius 3cm
-- (0,0-|A) coordinate(Ax)
% (0,0-|30:3cm) is a point that has the x coordinate of A and y=0
% see https://tex.stackexchange.com/a/401429/121799 for more details
node[midway,left]{$sinalpha$};
draw [very thick,orange] (3,0) -- (intersection cs:
first line={(O)--(A)},second line={(3,0)--(3,3)}) coordinate(A')
% (A') is at the intersections of the lines OA and the vertical line through (3,0)
node[midway,right]{$tanalpha$};
pic[fill=green!50,angle radius=1cm,
angle eccentricity=0.6, "$alpha$"] {angle=Ax--O--A};
% that's almost a 1-1 copy of what you can find on p. 560 of the manual
draw (O) -- (A');
draw[very thick,blue] (O) -- (Ax) node[midway,below]{$cosalpha$};
end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • Thanks marmot . Can you please add comments to each line of the tikz code which will help me understand better. Especially the relationship between polar coordinates and projection

    – subham soni
    Feb 23 at 15:12






  • 1





    @subhamsoni I added explanations in the comments in the code.

    – marmot
    Feb 23 at 15:24











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2 Answers
2






active

oldest

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









8














Since in cos(30) and tan(30) there are parentheses you must put these functions inside curly brackets {}



documentclass{article}
usepackage{tikz}
usetikzlibrary{angles,quotes}

tikzset{help lines/.style=very thin}
tikzset{My Grid/.style={help lines,color=blue!50}}

begin{document}
begin{tikzpicture}
draw[My Grid] (-4,-4) grid (4,4);
draw (-5,0) node[left] {$(-5,0)$} -- (5,0) node[right] {$(5,0)$};
draw (0,-5) node[below] {$(0,-5)$} -- (0,5) node[above] {$(0,5)$};
draw (0,0) circle [radius=3cm];

% shadedraw[left color=gray, right color=green, draw=green!50!black]
% (0,0) -- (0.75,0) arc [start angle=0, end angle=30, radius=0.75cm] -- cycle;

coordinate(O)at(0,0);
draw[red, very thick] (30:3cm)coordinate(A)
--({3*cos(30)},0)coordinate(B);

draw [very thick,orange] (3,0) -- (3,{3*tan(30)})coordinate(C);

pic[fill=green!50!black,
angle radius=0.75cm,
angle eccentricity=1.2,
"(alpha)"] {angle=B--O--A};

draw (O)--(C);

end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • Will this ensure that the length is actual value when computed as well or plotted manually with pencil and graph?

    – subham soni
    Feb 23 at 8:32











  • @subhamsoni I notice that the green angle is greeter than 30° I suppose you needed to be 30°

    – Hafid Boukhoulda
    Feb 23 at 8:38













  • I have mentioned end angle as 30. Is there anything I am missing

    – subham soni
    Feb 23 at 8:39











  • @subhamsoni Yes you need the specify the correct radius 0.75 cm and not 1cm since the start point is (0.75,0). Also in Tikz there is another manner to draw angles.

    – Hafid Boukhoulda
    Feb 23 at 8:43













  • it would be great if you can share the other way because precision is always better.

    – subham soni
    Feb 23 at 8:46
















8














Since in cos(30) and tan(30) there are parentheses you must put these functions inside curly brackets {}



documentclass{article}
usepackage{tikz}
usetikzlibrary{angles,quotes}

tikzset{help lines/.style=very thin}
tikzset{My Grid/.style={help lines,color=blue!50}}

begin{document}
begin{tikzpicture}
draw[My Grid] (-4,-4) grid (4,4);
draw (-5,0) node[left] {$(-5,0)$} -- (5,0) node[right] {$(5,0)$};
draw (0,-5) node[below] {$(0,-5)$} -- (0,5) node[above] {$(0,5)$};
draw (0,0) circle [radius=3cm];

% shadedraw[left color=gray, right color=green, draw=green!50!black]
% (0,0) -- (0.75,0) arc [start angle=0, end angle=30, radius=0.75cm] -- cycle;

coordinate(O)at(0,0);
draw[red, very thick] (30:3cm)coordinate(A)
--({3*cos(30)},0)coordinate(B);

draw [very thick,orange] (3,0) -- (3,{3*tan(30)})coordinate(C);

pic[fill=green!50!black,
angle radius=0.75cm,
angle eccentricity=1.2,
"(alpha)"] {angle=B--O--A};

draw (O)--(C);

end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • Will this ensure that the length is actual value when computed as well or plotted manually with pencil and graph?

    – subham soni
    Feb 23 at 8:32











  • @subhamsoni I notice that the green angle is greeter than 30° I suppose you needed to be 30°

    – Hafid Boukhoulda
    Feb 23 at 8:38













  • I have mentioned end angle as 30. Is there anything I am missing

    – subham soni
    Feb 23 at 8:39











  • @subhamsoni Yes you need the specify the correct radius 0.75 cm and not 1cm since the start point is (0.75,0). Also in Tikz there is another manner to draw angles.

    – Hafid Boukhoulda
    Feb 23 at 8:43













  • it would be great if you can share the other way because precision is always better.

    – subham soni
    Feb 23 at 8:46














8












8








8







Since in cos(30) and tan(30) there are parentheses you must put these functions inside curly brackets {}



documentclass{article}
usepackage{tikz}
usetikzlibrary{angles,quotes}

tikzset{help lines/.style=very thin}
tikzset{My Grid/.style={help lines,color=blue!50}}

begin{document}
begin{tikzpicture}
draw[My Grid] (-4,-4) grid (4,4);
draw (-5,0) node[left] {$(-5,0)$} -- (5,0) node[right] {$(5,0)$};
draw (0,-5) node[below] {$(0,-5)$} -- (0,5) node[above] {$(0,5)$};
draw (0,0) circle [radius=3cm];

% shadedraw[left color=gray, right color=green, draw=green!50!black]
% (0,0) -- (0.75,0) arc [start angle=0, end angle=30, radius=0.75cm] -- cycle;

coordinate(O)at(0,0);
draw[red, very thick] (30:3cm)coordinate(A)
--({3*cos(30)},0)coordinate(B);

draw [very thick,orange] (3,0) -- (3,{3*tan(30)})coordinate(C);

pic[fill=green!50!black,
angle radius=0.75cm,
angle eccentricity=1.2,
"(alpha)"] {angle=B--O--A};

draw (O)--(C);

end{tikzpicture}
end{document}


enter image description here






share|improve this answer















Since in cos(30) and tan(30) there are parentheses you must put these functions inside curly brackets {}



documentclass{article}
usepackage{tikz}
usetikzlibrary{angles,quotes}

tikzset{help lines/.style=very thin}
tikzset{My Grid/.style={help lines,color=blue!50}}

begin{document}
begin{tikzpicture}
draw[My Grid] (-4,-4) grid (4,4);
draw (-5,0) node[left] {$(-5,0)$} -- (5,0) node[right] {$(5,0)$};
draw (0,-5) node[below] {$(0,-5)$} -- (0,5) node[above] {$(0,5)$};
draw (0,0) circle [radius=3cm];

% shadedraw[left color=gray, right color=green, draw=green!50!black]
% (0,0) -- (0.75,0) arc [start angle=0, end angle=30, radius=0.75cm] -- cycle;

coordinate(O)at(0,0);
draw[red, very thick] (30:3cm)coordinate(A)
--({3*cos(30)},0)coordinate(B);

draw [very thick,orange] (3,0) -- (3,{3*tan(30)})coordinate(C);

pic[fill=green!50!black,
angle radius=0.75cm,
angle eccentricity=1.2,
"(alpha)"] {angle=B--O--A};

draw (O)--(C);

end{tikzpicture}
end{document}


enter image description here







share|improve this answer














share|improve this answer



share|improve this answer








edited Feb 23 at 10:54

























answered Feb 23 at 8:27









Hafid BoukhouldaHafid Boukhoulda

4,6341625




4,6341625













  • Will this ensure that the length is actual value when computed as well or plotted manually with pencil and graph?

    – subham soni
    Feb 23 at 8:32











  • @subhamsoni I notice that the green angle is greeter than 30° I suppose you needed to be 30°

    – Hafid Boukhoulda
    Feb 23 at 8:38













  • I have mentioned end angle as 30. Is there anything I am missing

    – subham soni
    Feb 23 at 8:39











  • @subhamsoni Yes you need the specify the correct radius 0.75 cm and not 1cm since the start point is (0.75,0). Also in Tikz there is another manner to draw angles.

    – Hafid Boukhoulda
    Feb 23 at 8:43













  • it would be great if you can share the other way because precision is always better.

    – subham soni
    Feb 23 at 8:46



















  • Will this ensure that the length is actual value when computed as well or plotted manually with pencil and graph?

    – subham soni
    Feb 23 at 8:32











  • @subhamsoni I notice that the green angle is greeter than 30° I suppose you needed to be 30°

    – Hafid Boukhoulda
    Feb 23 at 8:38













  • I have mentioned end angle as 30. Is there anything I am missing

    – subham soni
    Feb 23 at 8:39











  • @subhamsoni Yes you need the specify the correct radius 0.75 cm and not 1cm since the start point is (0.75,0). Also in Tikz there is another manner to draw angles.

    – Hafid Boukhoulda
    Feb 23 at 8:43













  • it would be great if you can share the other way because precision is always better.

    – subham soni
    Feb 23 at 8:46

















Will this ensure that the length is actual value when computed as well or plotted manually with pencil and graph?

– subham soni
Feb 23 at 8:32





Will this ensure that the length is actual value when computed as well or plotted manually with pencil and graph?

– subham soni
Feb 23 at 8:32













@subhamsoni I notice that the green angle is greeter than 30° I suppose you needed to be 30°

– Hafid Boukhoulda
Feb 23 at 8:38







@subhamsoni I notice that the green angle is greeter than 30° I suppose you needed to be 30°

– Hafid Boukhoulda
Feb 23 at 8:38















I have mentioned end angle as 30. Is there anything I am missing

– subham soni
Feb 23 at 8:39





I have mentioned end angle as 30. Is there anything I am missing

– subham soni
Feb 23 at 8:39













@subhamsoni Yes you need the specify the correct radius 0.75 cm and not 1cm since the start point is (0.75,0). Also in Tikz there is another manner to draw angles.

– Hafid Boukhoulda
Feb 23 at 8:43







@subhamsoni Yes you need the specify the correct radius 0.75 cm and not 1cm since the start point is (0.75,0). Also in Tikz there is another manner to draw angles.

– Hafid Boukhoulda
Feb 23 at 8:43















it would be great if you can share the other way because precision is always better.

– subham soni
Feb 23 at 8:46





it would be great if you can share the other way because precision is always better.

– subham soni
Feb 23 at 8:46











3














Here is an alternative to Hafid's nice answer, going further into the direction you started. You do not need to use any trigonometric function. Just using polar coordinates and projections is sufficient.



documentclass{article}
usepackage{tikz}
usetikzlibrary{calc,angles,quotes}
tikzset{help lines/.style=very thin}
tikzset{My Grid/.style={help lines,color=blue!50}}

begin{document}
begin{tikzpicture}
draw[My Grid] (-4,-4) grid (4,4);
draw (-5,0) node[left] {$(-5,0)$} -- (5,0) node[right] {$(5,0)$};
draw (0,-5) node[below] {$(0,-5)$} -- (0,5) node[above] {$(0,5)$};
draw (0,0) coordinate (O) circle [radius=3cm];
draw[red, very thick] (30:3cm) coordinate (A)
% (30:3cm) is a polar coordinate with angle 30 (degrees) and radius 3cm
-- (0,0-|A) coordinate(Ax)
% (0,0-|30:3cm) is a point that has the x coordinate of A and y=0
% see https://tex.stackexchange.com/a/401429/121799 for more details
node[midway,left]{$sinalpha$};
draw [very thick,orange] (3,0) -- (intersection cs:
first line={(O)--(A)},second line={(3,0)--(3,3)}) coordinate(A')
% (A') is at the intersections of the lines OA and the vertical line through (3,0)
node[midway,right]{$tanalpha$};
pic[fill=green!50,angle radius=1cm,
angle eccentricity=0.6, "$alpha$"] {angle=Ax--O--A};
% that's almost a 1-1 copy of what you can find on p. 560 of the manual
draw (O) -- (A');
draw[very thick,blue] (O) -- (Ax) node[midway,below]{$cosalpha$};
end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • Thanks marmot . Can you please add comments to each line of the tikz code which will help me understand better. Especially the relationship between polar coordinates and projection

    – subham soni
    Feb 23 at 15:12






  • 1





    @subhamsoni I added explanations in the comments in the code.

    – marmot
    Feb 23 at 15:24
















3














Here is an alternative to Hafid's nice answer, going further into the direction you started. You do not need to use any trigonometric function. Just using polar coordinates and projections is sufficient.



documentclass{article}
usepackage{tikz}
usetikzlibrary{calc,angles,quotes}
tikzset{help lines/.style=very thin}
tikzset{My Grid/.style={help lines,color=blue!50}}

begin{document}
begin{tikzpicture}
draw[My Grid] (-4,-4) grid (4,4);
draw (-5,0) node[left] {$(-5,0)$} -- (5,0) node[right] {$(5,0)$};
draw (0,-5) node[below] {$(0,-5)$} -- (0,5) node[above] {$(0,5)$};
draw (0,0) coordinate (O) circle [radius=3cm];
draw[red, very thick] (30:3cm) coordinate (A)
% (30:3cm) is a polar coordinate with angle 30 (degrees) and radius 3cm
-- (0,0-|A) coordinate(Ax)
% (0,0-|30:3cm) is a point that has the x coordinate of A and y=0
% see https://tex.stackexchange.com/a/401429/121799 for more details
node[midway,left]{$sinalpha$};
draw [very thick,orange] (3,0) -- (intersection cs:
first line={(O)--(A)},second line={(3,0)--(3,3)}) coordinate(A')
% (A') is at the intersections of the lines OA and the vertical line through (3,0)
node[midway,right]{$tanalpha$};
pic[fill=green!50,angle radius=1cm,
angle eccentricity=0.6, "$alpha$"] {angle=Ax--O--A};
% that's almost a 1-1 copy of what you can find on p. 560 of the manual
draw (O) -- (A');
draw[very thick,blue] (O) -- (Ax) node[midway,below]{$cosalpha$};
end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • Thanks marmot . Can you please add comments to each line of the tikz code which will help me understand better. Especially the relationship between polar coordinates and projection

    – subham soni
    Feb 23 at 15:12






  • 1





    @subhamsoni I added explanations in the comments in the code.

    – marmot
    Feb 23 at 15:24














3












3








3







Here is an alternative to Hafid's nice answer, going further into the direction you started. You do not need to use any trigonometric function. Just using polar coordinates and projections is sufficient.



documentclass{article}
usepackage{tikz}
usetikzlibrary{calc,angles,quotes}
tikzset{help lines/.style=very thin}
tikzset{My Grid/.style={help lines,color=blue!50}}

begin{document}
begin{tikzpicture}
draw[My Grid] (-4,-4) grid (4,4);
draw (-5,0) node[left] {$(-5,0)$} -- (5,0) node[right] {$(5,0)$};
draw (0,-5) node[below] {$(0,-5)$} -- (0,5) node[above] {$(0,5)$};
draw (0,0) coordinate (O) circle [radius=3cm];
draw[red, very thick] (30:3cm) coordinate (A)
% (30:3cm) is a polar coordinate with angle 30 (degrees) and radius 3cm
-- (0,0-|A) coordinate(Ax)
% (0,0-|30:3cm) is a point that has the x coordinate of A and y=0
% see https://tex.stackexchange.com/a/401429/121799 for more details
node[midway,left]{$sinalpha$};
draw [very thick,orange] (3,0) -- (intersection cs:
first line={(O)--(A)},second line={(3,0)--(3,3)}) coordinate(A')
% (A') is at the intersections of the lines OA and the vertical line through (3,0)
node[midway,right]{$tanalpha$};
pic[fill=green!50,angle radius=1cm,
angle eccentricity=0.6, "$alpha$"] {angle=Ax--O--A};
% that's almost a 1-1 copy of what you can find on p. 560 of the manual
draw (O) -- (A');
draw[very thick,blue] (O) -- (Ax) node[midway,below]{$cosalpha$};
end{tikzpicture}
end{document}


enter image description here






share|improve this answer















Here is an alternative to Hafid's nice answer, going further into the direction you started. You do not need to use any trigonometric function. Just using polar coordinates and projections is sufficient.



documentclass{article}
usepackage{tikz}
usetikzlibrary{calc,angles,quotes}
tikzset{help lines/.style=very thin}
tikzset{My Grid/.style={help lines,color=blue!50}}

begin{document}
begin{tikzpicture}
draw[My Grid] (-4,-4) grid (4,4);
draw (-5,0) node[left] {$(-5,0)$} -- (5,0) node[right] {$(5,0)$};
draw (0,-5) node[below] {$(0,-5)$} -- (0,5) node[above] {$(0,5)$};
draw (0,0) coordinate (O) circle [radius=3cm];
draw[red, very thick] (30:3cm) coordinate (A)
% (30:3cm) is a polar coordinate with angle 30 (degrees) and radius 3cm
-- (0,0-|A) coordinate(Ax)
% (0,0-|30:3cm) is a point that has the x coordinate of A and y=0
% see https://tex.stackexchange.com/a/401429/121799 for more details
node[midway,left]{$sinalpha$};
draw [very thick,orange] (3,0) -- (intersection cs:
first line={(O)--(A)},second line={(3,0)--(3,3)}) coordinate(A')
% (A') is at the intersections of the lines OA and the vertical line through (3,0)
node[midway,right]{$tanalpha$};
pic[fill=green!50,angle radius=1cm,
angle eccentricity=0.6, "$alpha$"] {angle=Ax--O--A};
% that's almost a 1-1 copy of what you can find on p. 560 of the manual
draw (O) -- (A');
draw[very thick,blue] (O) -- (Ax) node[midway,below]{$cosalpha$};
end{tikzpicture}
end{document}


enter image description here







share|improve this answer














share|improve this answer



share|improve this answer








edited Feb 23 at 15:23

























answered Feb 23 at 15:06









marmotmarmot

112k5142268




112k5142268













  • Thanks marmot . Can you please add comments to each line of the tikz code which will help me understand better. Especially the relationship between polar coordinates and projection

    – subham soni
    Feb 23 at 15:12






  • 1





    @subhamsoni I added explanations in the comments in the code.

    – marmot
    Feb 23 at 15:24



















  • Thanks marmot . Can you please add comments to each line of the tikz code which will help me understand better. Especially the relationship between polar coordinates and projection

    – subham soni
    Feb 23 at 15:12






  • 1





    @subhamsoni I added explanations in the comments in the code.

    – marmot
    Feb 23 at 15:24

















Thanks marmot . Can you please add comments to each line of the tikz code which will help me understand better. Especially the relationship between polar coordinates and projection

– subham soni
Feb 23 at 15:12





Thanks marmot . Can you please add comments to each line of the tikz code which will help me understand better. Especially the relationship between polar coordinates and projection

– subham soni
Feb 23 at 15:12




1




1





@subhamsoni I added explanations in the comments in the code.

– marmot
Feb 23 at 15:24





@subhamsoni I added explanations in the comments in the code.

– marmot
Feb 23 at 15:24


















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