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Basic trigonometry question, students are asked to find the width of a roadway in a circular tunnel given distance from edge of road to centre top of tunnel (randomised) and angle from edge of road to centre top of tunnel (randomised).

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Let the point of intersection of \\(AC\\) and \\(BZ\\) be \\(D\\). Since \\(BZ\\) is a diameter of the circle, the distance \\(AD\\) will be half of \\(AC\\). Further triangle \\(ADB\\) is a right angled triangle and we have

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\\begin{align}\\cos A &=\\frac{AD}{AB}\\\\\\cos\\var{A}^\\circ&=\\frac{AD}{\\var{d}}\\end{align}

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Hence \\begin{align}AD&=\\var{d}\\cos\\var{A}^\\circ\\\\&\\approx \\var{precround(d*cos(pi*A/180),3)}\\text{ m}\\end{align}

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Thus \\(AC=2\\times AD\\approx\\var{precround(2*d*cos(pi*A/180),2)}\\text{ m}\\).

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The length of the platform is approximately \\( \\var{precround(2*d*cos(pi*A/180),2)}\\text{ m.}\\)

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Distance from A to B in m in diagram, random integer from 4 to 9

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Angle BAC in diagram in degrees, random 45 to 80 in steps of 5.

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The diagram below shows a circular tunnel with \\(AB=\\var{d}\\) metres and angle \\(BAC=\\var{A}^\\circ\\). A platform is built in the tunnel from \\(A\\) to \\(C\\). Assume that angle \\(BAZ=\\) angle \\(BCZ=90^\\circ\\). Find the width, \\(AC\\), of the platform in the tunnel. You may also assume that \\(BZ\\) is a diameter and is perpendicular to \\(AC\\). Write the width in the box below and show your full working on your working paper.

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Platform width = [[0]] m (give your answer correct to 2 decimal places)

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