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$PS$ is a vertical Modacell cell phone tower.

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Two people are standing at points $Q$ and $R$ (in the same horizontal plane as $S$) from where they are testing their cell phone reception.

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From $R$ the angle of elevation to $P$ is  $\\alpha =\\frac{\\pi}{6}$ and $QS$ is $t$ meter. $P$ is $h$  meters above ground level.

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3.1   In $\\Delta PSQ$

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        $PQ^2 \\ =\\ PS^2+SQ^2 = h^2+t^2$

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        $PQ\\ =\\ \\sqrt{h^2+t^2}$

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3.2   $PQ\\ =\\ \\sqrt{h^2+t^2}$

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                $=\\ \\sqrt{20^2+60^2}$

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                $=\\ \\sqrt{4000}=63.2455...\\approx 63.246\\ m$

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3.3  In $\\Delta PSR$

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                $\\sin(\\frac{\\pi}{6}) \\ =\\ \\frac{PS}{PR}$

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                $PR =\\ \\frac{h}{\\frac{1}{2}} \\ =\\ 2h \\ =\\ 40\\ m$

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3.4  In $\\Delta QPR$

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                $\\cos(Q\\hat{P}R) = \\frac{PQ^2+PR^2-QR^2}{2.PQ.PR}$

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                $\\cos(Q\\hat{P}R) = \\frac{63.25^2+40^2-50^2}{2(63.25)(40)} = 0.6127 ...$

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                $Q\\hat{P}R = 0.911...\\approx 0.91$

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3.1     Express $PQ$ in terms of $h$ and $t$.

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          [[0]]

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3.2     IF  $h=20\\ m$ and  $t = 60\\ m$, determine $PQ.$

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           [[1]]m

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3.3     Determine the length of $PR.$

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           [[2]]m

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3.4     If  $QR = 50\\ m$, determine $Q\\hat{P}R.$

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           [[3]]

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