Cone in Latex with right angles and Label












1















I am new LaTex user and I have hard time plot the following shape in LaTex. I started the shape but I can't find way to finish it if you can help me please



Here is my code



documentclass{article}
usepackage{tikz}
usepackage{mathrsfs}
begin{document}
begin{tikzpicture}
draw (3,0) node {$mathcal{C}$};
draw (-2,2)--(0,0);
draw (3,1.5)--(0,0);
draw (-2,-2)--(0,0);
draw (3,-1.5)--(0,0);
end{tikzpicture}

end{document}


Here is the shape I am aiming to get if you can help me finish it please.



enter image description here










share|improve this question


















  • 2





    Welcome to TeX.SE! Please explain a bit more the context of the figure. Is this a 3d figure? If so, there is the possibility to draw this in 3d.

    – marmot
    Jan 14 at 2:21











  • @marmot Thank you. Yes it is 3d figure where the point X is projected to the Cone C

    – JesZ
    Jan 14 at 2:35


















1















I am new LaTex user and I have hard time plot the following shape in LaTex. I started the shape but I can't find way to finish it if you can help me please



Here is my code



documentclass{article}
usepackage{tikz}
usepackage{mathrsfs}
begin{document}
begin{tikzpicture}
draw (3,0) node {$mathcal{C}$};
draw (-2,2)--(0,0);
draw (3,1.5)--(0,0);
draw (-2,-2)--(0,0);
draw (3,-1.5)--(0,0);
end{tikzpicture}

end{document}


Here is the shape I am aiming to get if you can help me finish it please.



enter image description here










share|improve this question


















  • 2





    Welcome to TeX.SE! Please explain a bit more the context of the figure. Is this a 3d figure? If so, there is the possibility to draw this in 3d.

    – marmot
    Jan 14 at 2:21











  • @marmot Thank you. Yes it is 3d figure where the point X is projected to the Cone C

    – JesZ
    Jan 14 at 2:35
















1












1








1








I am new LaTex user and I have hard time plot the following shape in LaTex. I started the shape but I can't find way to finish it if you can help me please



Here is my code



documentclass{article}
usepackage{tikz}
usepackage{mathrsfs}
begin{document}
begin{tikzpicture}
draw (3,0) node {$mathcal{C}$};
draw (-2,2)--(0,0);
draw (3,1.5)--(0,0);
draw (-2,-2)--(0,0);
draw (3,-1.5)--(0,0);
end{tikzpicture}

end{document}


Here is the shape I am aiming to get if you can help me finish it please.



enter image description here










share|improve this question














I am new LaTex user and I have hard time plot the following shape in LaTex. I started the shape but I can't find way to finish it if you can help me please



Here is my code



documentclass{article}
usepackage{tikz}
usepackage{mathrsfs}
begin{document}
begin{tikzpicture}
draw (3,0) node {$mathcal{C}$};
draw (-2,2)--(0,0);
draw (3,1.5)--(0,0);
draw (-2,-2)--(0,0);
draw (3,-1.5)--(0,0);
end{tikzpicture}

end{document}


Here is the shape I am aiming to get if you can help me finish it please.



enter image description here







tikz-pgf






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Jan 14 at 1:37









JesZJesZ

82




82








  • 2





    Welcome to TeX.SE! Please explain a bit more the context of the figure. Is this a 3d figure? If so, there is the possibility to draw this in 3d.

    – marmot
    Jan 14 at 2:21











  • @marmot Thank you. Yes it is 3d figure where the point X is projected to the Cone C

    – JesZ
    Jan 14 at 2:35
















  • 2





    Welcome to TeX.SE! Please explain a bit more the context of the figure. Is this a 3d figure? If so, there is the possibility to draw this in 3d.

    – marmot
    Jan 14 at 2:21











  • @marmot Thank you. Yes it is 3d figure where the point X is projected to the Cone C

    – JesZ
    Jan 14 at 2:35










2




2





Welcome to TeX.SE! Please explain a bit more the context of the figure. Is this a 3d figure? If so, there is the possibility to draw this in 3d.

– marmot
Jan 14 at 2:21





Welcome to TeX.SE! Please explain a bit more the context of the figure. Is this a 3d figure? If so, there is the possibility to draw this in 3d.

– marmot
Jan 14 at 2:21













@marmot Thank you. Yes it is 3d figure where the point X is projected to the Cone C

– JesZ
Jan 14 at 2:35







@marmot Thank you. Yes it is 3d figure where the point X is projected to the Cone C

– JesZ
Jan 14 at 2:35












1 Answer
1






active

oldest

votes


















2














It is always a bit hard to undo the projection, i.e. to guess the 3d coordinates from their projections on the screen. So most likely I got some wrong, yet this may serve as a start. I recommend using tikz-3dplot because it allows you to do orthographic projections from 3d to the screen. That is, you can adjust the view angles to your needs.



documentclass[tikz,border=3.14mm]{standalone}
usepackage{tikz-3dplot}
begin{document}
tdplotsetmaincoords{70}{20}
begin{tikzpicture}[dot/.style={circle,fill,inner sep=1pt}]
tdplotsetrotatedcoords{30}{20}{0}
begin{scope}[tdplot_rotated_coords]
draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
-- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
draw[dotted,thick] (0,0,3) node[dot,label=right:{$widetilde{theta}$}]{}
-- (0,5,3) node[dot,label=above:{$X$}]{};
draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
draw[dotted,thick] (1.5,0,0) node[dot,label=above:{$widetilde{theta}$}]{}
-- ++ (0,-3,-3) node[dot,label=right:{$X$}]{};
draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
draw[dotted,thick] (O) -- (0,3,-3) node[dot,label=left:{$X$}]{};
end{scope}
begin{scope}[xshift=7cm,tdplot_rotated_coords]
draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
-- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
draw (0,0,3) node[dot,label=right:{$Q$}] (Q) {}
-- (0,5,3) node[dot,label=above:{$P$}]{};
draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
draw (1.5,0,0) node[dot,label=above right:{$R$}] (R) {}
-- ++ (0,-3,-3) node[dot,label=right:{$S$}]{};
draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
draw (R) to[out=80,in=-40] (Q);
end{scope}
end{tikzpicture}
end{document}


enter image description here






share|improve this answer























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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    It is always a bit hard to undo the projection, i.e. to guess the 3d coordinates from their projections on the screen. So most likely I got some wrong, yet this may serve as a start. I recommend using tikz-3dplot because it allows you to do orthographic projections from 3d to the screen. That is, you can adjust the view angles to your needs.



    documentclass[tikz,border=3.14mm]{standalone}
    usepackage{tikz-3dplot}
    begin{document}
    tdplotsetmaincoords{70}{20}
    begin{tikzpicture}[dot/.style={circle,fill,inner sep=1pt}]
    tdplotsetrotatedcoords{30}{20}{0}
    begin{scope}[tdplot_rotated_coords]
    draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
    -- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
    draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
    draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
    draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
    draw[dotted,thick] (0,0,3) node[dot,label=right:{$widetilde{theta}$}]{}
    -- (0,5,3) node[dot,label=above:{$X$}]{};
    draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
    draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
    draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
    draw[dotted,thick] (1.5,0,0) node[dot,label=above:{$widetilde{theta}$}]{}
    -- ++ (0,-3,-3) node[dot,label=right:{$X$}]{};
    draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
    path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
    draw[dotted,thick] (O) -- (0,3,-3) node[dot,label=left:{$X$}]{};
    end{scope}
    begin{scope}[xshift=7cm,tdplot_rotated_coords]
    draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
    -- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
    draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
    draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
    draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
    draw (0,0,3) node[dot,label=right:{$Q$}] (Q) {}
    -- (0,5,3) node[dot,label=above:{$P$}]{};
    draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
    draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
    draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
    draw (1.5,0,0) node[dot,label=above right:{$R$}] (R) {}
    -- ++ (0,-3,-3) node[dot,label=right:{$S$}]{};
    draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
    path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
    draw (R) to[out=80,in=-40] (Q);
    end{scope}
    end{tikzpicture}
    end{document}


    enter image description here






    share|improve this answer




























      2














      It is always a bit hard to undo the projection, i.e. to guess the 3d coordinates from their projections on the screen. So most likely I got some wrong, yet this may serve as a start. I recommend using tikz-3dplot because it allows you to do orthographic projections from 3d to the screen. That is, you can adjust the view angles to your needs.



      documentclass[tikz,border=3.14mm]{standalone}
      usepackage{tikz-3dplot}
      begin{document}
      tdplotsetmaincoords{70}{20}
      begin{tikzpicture}[dot/.style={circle,fill,inner sep=1pt}]
      tdplotsetrotatedcoords{30}{20}{0}
      begin{scope}[tdplot_rotated_coords]
      draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
      -- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
      draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
      draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
      draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
      draw[dotted,thick] (0,0,3) node[dot,label=right:{$widetilde{theta}$}]{}
      -- (0,5,3) node[dot,label=above:{$X$}]{};
      draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
      draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
      draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
      draw[dotted,thick] (1.5,0,0) node[dot,label=above:{$widetilde{theta}$}]{}
      -- ++ (0,-3,-3) node[dot,label=right:{$X$}]{};
      draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
      path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
      draw[dotted,thick] (O) -- (0,3,-3) node[dot,label=left:{$X$}]{};
      end{scope}
      begin{scope}[xshift=7cm,tdplot_rotated_coords]
      draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
      -- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
      draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
      draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
      draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
      draw (0,0,3) node[dot,label=right:{$Q$}] (Q) {}
      -- (0,5,3) node[dot,label=above:{$P$}]{};
      draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
      draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
      draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
      draw (1.5,0,0) node[dot,label=above right:{$R$}] (R) {}
      -- ++ (0,-3,-3) node[dot,label=right:{$S$}]{};
      draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
      path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
      draw (R) to[out=80,in=-40] (Q);
      end{scope}
      end{tikzpicture}
      end{document}


      enter image description here






      share|improve this answer


























        2












        2








        2







        It is always a bit hard to undo the projection, i.e. to guess the 3d coordinates from their projections on the screen. So most likely I got some wrong, yet this may serve as a start. I recommend using tikz-3dplot because it allows you to do orthographic projections from 3d to the screen. That is, you can adjust the view angles to your needs.



        documentclass[tikz,border=3.14mm]{standalone}
        usepackage{tikz-3dplot}
        begin{document}
        tdplotsetmaincoords{70}{20}
        begin{tikzpicture}[dot/.style={circle,fill,inner sep=1pt}]
        tdplotsetrotatedcoords{30}{20}{0}
        begin{scope}[tdplot_rotated_coords]
        draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
        -- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
        draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
        draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
        draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
        draw[dotted,thick] (0,0,3) node[dot,label=right:{$widetilde{theta}$}]{}
        -- (0,5,3) node[dot,label=above:{$X$}]{};
        draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
        draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
        draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
        draw[dotted,thick] (1.5,0,0) node[dot,label=above:{$widetilde{theta}$}]{}
        -- ++ (0,-3,-3) node[dot,label=right:{$X$}]{};
        draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
        path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
        draw[dotted,thick] (O) -- (0,3,-3) node[dot,label=left:{$X$}]{};
        end{scope}
        begin{scope}[xshift=7cm,tdplot_rotated_coords]
        draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
        -- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
        draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
        draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
        draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
        draw (0,0,3) node[dot,label=right:{$Q$}] (Q) {}
        -- (0,5,3) node[dot,label=above:{$P$}]{};
        draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
        draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
        draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
        draw (1.5,0,0) node[dot,label=above right:{$R$}] (R) {}
        -- ++ (0,-3,-3) node[dot,label=right:{$S$}]{};
        draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
        path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
        draw (R) to[out=80,in=-40] (Q);
        end{scope}
        end{tikzpicture}
        end{document}


        enter image description here






        share|improve this answer













        It is always a bit hard to undo the projection, i.e. to guess the 3d coordinates from their projections on the screen. So most likely I got some wrong, yet this may serve as a start. I recommend using tikz-3dplot because it allows you to do orthographic projections from 3d to the screen. That is, you can adjust the view angles to your needs.



        documentclass[tikz,border=3.14mm]{standalone}
        usepackage{tikz-3dplot}
        begin{document}
        tdplotsetmaincoords{70}{20}
        begin{tikzpicture}[dot/.style={circle,fill,inner sep=1pt}]
        tdplotsetrotatedcoords{30}{20}{0}
        begin{scope}[tdplot_rotated_coords]
        draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
        -- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
        draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
        draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
        draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
        draw[dotted,thick] (0,0,3) node[dot,label=right:{$widetilde{theta}$}]{}
        -- (0,5,3) node[dot,label=above:{$X$}]{};
        draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
        draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
        draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
        draw[dotted,thick] (1.5,0,0) node[dot,label=above:{$widetilde{theta}$}]{}
        -- ++ (0,-3,-3) node[dot,label=right:{$X$}]{};
        draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
        path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
        draw[dotted,thick] (O) -- (0,3,-3) node[dot,label=left:{$X$}]{};
        end{scope}
        begin{scope}[xshift=7cm,tdplot_rotated_coords]
        draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
        -- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
        draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
        draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
        draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
        draw (0,0,3) node[dot,label=right:{$Q$}] (Q) {}
        -- (0,5,3) node[dot,label=above:{$P$}]{};
        draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
        draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
        draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
        draw (1.5,0,0) node[dot,label=above right:{$R$}] (R) {}
        -- ++ (0,-3,-3) node[dot,label=right:{$S$}]{};
        draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
        path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
        draw (R) to[out=80,in=-40] (Q);
        end{scope}
        end{tikzpicture}
        end{document}


        enter image description here







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jan 14 at 2:38









        marmotmarmot

        93.3k4109204




        93.3k4109204






























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