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Calculating the angle between a vector and the positive $x$-axis.

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Calculate the angle between the vector $\\mathbf a = \\var{v}$ and the right horizontal:

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{geogq[which]}

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{advice1[which]}

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{geogans[which]}

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{advice2[which]}

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To calculate the angle between the vector $\\\\mathbf a$ and the right horizontal, it can be helpful to draw the vector with the $x$ and $y$ components also included. This allows us to view the vector and its components as a right-angled triangle, and calculate the angle using trigonometry.

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Since this vector is in the second quadrant, to calculate the angle it forms with the right horizontal it can be easier to first find the angle it forms with the left horizontal, and then subtract this answer from $180^\\\\circ$.

\\n

To calculate the angle it forms with the left horizontal, it can be helpful to draw the vector with the magnitudes of the $x$ and $y$ components also included. This allows us to view the vector and its components as a right-angled triangle, and calculate the angle using trigonometry.

\\n

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Since this vector is in the third quadrant, to calculate the angle it forms with the right horizontal it can be easier to first find the angle it forms with the left horizontal, and then add this answer to $180^\\\\circ$.

\\n

To calculate the angle it forms with the left horizontal, it can be helpful to draw the vector with the magnitudes of the $x$ and $y$ components also included. This allows us to view the vector and its components as a right-angled triangle, and calculate the angle using trigonometry.

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Since this vector is in the fourth quadrant, to calculate the angle it forms with the right horizontal from above, it can be easier to first find the angle it forms with the right horizontal from below, and then subtract this answer from $360^\\\\circ$.

\\n

To calculate the angle it forms with the right horizontal from below, it can be helpful to draw the vector with the magnitudes of the $x$ and $y$ components also included. This allows us to view the vector and its components as a right-angled triangle, and calculate the angle using trigonometry.

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We can see that we have a triangle which has an adjacent edge of length $\\\\var{a}$ and an opposite edge of length $\\\\var{b}$. Therefore,

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\\\\[ \\\\begin{split} \\\\tan(\\\\theta) &\\\\,= \\\\frac{\\\\var{b}}{\\\\var{a}} \\\\\\\\ \\\\implies \\\\theta &\\\\,= \\\\tan^{-1}\\\\left(\\\\simplify[fractionNumbers]{{b/a}}\\\\right) \\\\\\\\ &\\\\,=\\\\var{angle[which]}^\\\\circ \\\\, \\\\text{ (2 d.p.)} \\\\end{split} \\\\]

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We can see that we have a triangle which has an adjacent edge of length $\\\\var{abs(a)}$ and an opposite edge of $\\\\var{abs(b)}$. Therefore,

\\n

\\\\[ \\\\begin{split} \\\\tan(\\\\phi) &\\\\,= \\\\frac{\\\\var{abs(b)}}{\\\\var{abs(a)}} \\\\\\\\ \\\\implies \\\\phi &\\\\,= \\\\tan^{-1} \\\\left(\\\\simplify[fractionNumbers]{{abs(b)/abs(a)}}\\\\right) \\\\\\\\ &\\\\,=\\\\var{180-angle[which]}^\\\\circ \\\\, \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

\\n

Since \\\\[ \\\\begin{split} \\\\theta+\\\\phi &\\\\,= 180^\\\\circ \\\\\\\\ \\\\implies \\\\theta &\\\\,= 180^\\\\circ - \\\\phi \\\\\\\\ &\\\\,=\\\\var{angle[which]}^\\\\circ \\\\, \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

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We can see that we have a triangle which has an adjacent edge of length $\\\\var{abs(a)}$ and an opposite edge of $\\\\var{abs(b)}$. Therefore,

\\n

\\\\[ \\\\begin{split} \\\\tan(\\\\phi) &\\\\,= \\\\frac{\\\\var{abs(b)}}{\\\\var{abs(a)}} \\\\\\\\ \\\\implies \\\\phi &\\\\,= \\\\tan^{-1} \\\\left(\\\\simplify[fractionNumbers]{{abs(b)/abs(a)}}\\\\right) \\\\\\\\ &\\\\,=\\\\var{angle[which]-180}^\\\\circ \\\\, \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

\\n

Hence,

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\\\\[ \\\\begin{split} \\\\theta &\\\\,= 180^\\\\circ + \\\\phi \\\\\\\\ &\\\\,=\\\\var{angle[which]}^\\\\circ \\\\, \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

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We can see that we have a triangle which has an adjacent edge of length $\\\\var{abs(a)}$ and an opposite edge of $\\\\var{abs(b)}$. Therefore,

\\n

\\\\[ \\\\begin{split} \\\\tan(\\\\phi) &\\\\,= \\\\frac{\\\\var{abs(b)}}{\\\\var{abs(a)}} \\\\\\\\ \\\\implies \\\\phi &\\\\,= \\\\tan^{-1} \\\\left(\\\\simplify[fractionNumbers]{{abs(b)/abs(a)}}\\\\right) \\\\\\\\ &\\\\,=\\\\var{360-angle[which]}^\\\\circ \\\\, \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

\\n

Since \\\\[ \\\\begin{split} \\\\theta+\\\\phi &\\\\,= 360^\\\\circ \\\\\\\\ \\\\implies \\\\theta &\\\\,= 360^\\\\circ - \\\\phi \\\\\\\\ &\\\\,=\\\\var{angle[which]}^\\\\circ \\\\, \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

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$\\theta= $[[0]]$^\\circ$

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(Give your answer in degrees, to 2 decimal places where necessary)

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