// Numbas version: exam_results_page_options {"name": "13: Vectors and Trigonometry: Subduction Zone", "extensions": [], "custom_part_types": [], "resources": [["question-resources/93e3bc2ef7cb323f434ca53208c4a9ad.jpg", "/srv/numbas/media/question-resources/93e3bc2ef7cb323f434ca53208c4a9ad.jpg"], ["question-resources/2016-08-18_3.png", "/srv/numbas/media/question-resources/2016-08-18_3.png"], ["question-resources/2016-08-18_6.png", "/srv/numbas/media/question-resources/2016-08-18_6.png"], ["question-resources/2016-08-19_3.png", "/srv/numbas/media/question-resources/2016-08-19_3.png"], ["question-resources/2016-08-19_3_QIXNOoD.png", "/srv/numbas/media/question-resources/2016-08-19_3_QIXNOoD.png"], ["question-resources/2016-08-19_4.png", "/srv/numbas/media/question-resources/2016-08-19_4.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"preamble": {"css": "", "js": ""}, "statement": "

Below is an image illustrating Plate A being subducted beneath Plate B.

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S$^\\circ%$ = The subduction angle

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$\\overrightarrow{XY}$ = The vector from the GPS tracker at location (X) to the location of the end point (Y).

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Plate A subducts beneath Plate B. From the starting point X, the plate moves laterally $\\var{a} km$ east and then subducts $\\var{b} km$ downwards until point Y. The angle of subduction, $S^\\circ$, is $\\var{x}^\\circ$.

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Deduce the magnitude of $\\overrightarrow{XY}$

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[[0]] km

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Give your answer to two decimal places.

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To find the magnitute of the vector $\\overrightarrow{XY}$, we use the cosine rule:

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$c^2=a^2+b^2-2\\space{a}\\space{b}\\space{cos(C)}$

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If we consider $\\overrightarrow{XY}$ to be side $c$, and $a$ is $\\var{a}$ and $b$ is $\\var{b}$, you can substitute in the known values to obtain a value for the magnitude. 

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From location X, plate A moves $\\var{h} km$ east and then subducts $\\var{b} km$ downwards beneath plate B. If the overall magnitude of $\\overrightarrow{XY}$ is $\\var{d} km$, what is the angle of subduction $S^\\circ$? 

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Give your answer to the nearest whole number.

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[[0]]$^\\circ$

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To calculate the subduction angle, $S$, we use the cosine rule.

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If we take side $c$ to be the vector $\\overrightarrow{XY}$, then starting from the original equation with side $c$ being the subject; $c^2=a^2+b^2-2\\space{a}\\space{b}\\space{cos(C)}$, we need to make $cos(C)$ the subject.

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$cos(C^\\circ)= \\frac{a^2+b^2-c^2}{2ab}$

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$C^\\circ = cos^{-1}[\\frac{a^2+b^2-c^2}{2ab}]$

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where $a$ is $\\var{h} km$, and $b$ is $\\var{b} km$ and $c$ is $\\var{d} km$.

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where $C^\\circ$ = subduction angle $S$.

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You can alternatively view the subduction zone in graphical format as illustrated below.

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You are given the coordinates for X, Y and Z.
X = ($\\var{coordx1},\\var{coordx2}$)
Y = ($\\var{coordy1},\\var{coordy2}$)
Z = ($\\var{coordz1},\\var{coordz2}$)

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Find the magnitude of $\\overrightarrow{ZY}$ =[[0]]
Find the overal magnitude of $\\overrightarrow{XY}$ =[[1]]

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Give your answers to 1 decimal place.

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Part one

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To find the magnitude of vector $\\overrightarrow{ZY}$, it will help to construct a right angled triangle as shown in the illustration. You know the depth of Y as it is the $y$ variable in the coordinates ($x,y$), and you can work out the distance horizontally between Z and Y as this is the $x$ coordinate of Y minus the $x$ coordinate of Z, as the $x$ coordinate of each point is the distance from X, hence giving the distance between Z and Y.
You can now use Pythagoras' theorem $a^2 + b^2 = c^2$ to determine the magnitude of $\\overrightarrow{ZY}$

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Part two

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Now construct a large right angled triangle to deduce the magnitude of vector $\\overrightarrow{XY}$ like the one illustrated.The overall length is distance Y (from the x-coordinate of the point Y), and the depth is the depth of Y (from the y-coordinate of the point Y) Use Pythagoras' theorem to find the magnitude of $\\overrightarrow{XY}$.

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This exercise will help you solve questions based on subduction zones using  the Cosine rule.

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Part a)

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To find the magnitute of the vector $\\overrightarrow{XY}$, we use the cosine rule:

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$c^2=a^2+b^2-2\\space{a}\\space{b}\\space{cos(C)}$

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If we consider $\\overrightarrow{XY}$ to be side $c$, you can substitute in the known values to obtain a value for the magnitude.

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$c^2=\\sqrt{\\var{a}^2+\\var{b}^2-2\\times\\var{a}\\times\\var{b}\\times{cos\\var{x}}}$

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$c = \\var{ans1}$km.

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Part b)

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To calculate the subduction angle, $S$, we use the cosine rule.

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If we take side $c$ to be the vector $\\overrightarrow{XY}$, then starting from the original equation with side $c$ being the subject; $c^2=a^2+b^2-2\\space{a}\\space{b}\\space{cos(C)}$, we need to make $cos(C)$ the subject.

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$cos(C^\\circ)= \\frac{a^2+b^2-c^2}{2ab}$

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$C^\\circ = cos^{-1}\\left[\\frac{a^2+b^2-c^2}{2ab}\\right]$, where $C^\\circ = S^\\circ$

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$S^\\circ = cos^{-1}\\left[\\frac{\\var{h}^2+\\var{b}^2-\\var{d}^2}{2\\times\\var{h}\\times\\var{b}}\\right]$

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= $\\var{g}^\\circ$ to the nearest whole number.

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Part c)

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Using the diagrams and the explanations illustrated in the steps to part c we obtain the following solutions:

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