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%\header{MATH 11}{Exam 1}{Fall 2016}
\begin{document}
\begin{coverpages}
\begin{center}\huge MATH 1540 - EXAM 1 - FALL 2017\\
SOLUTION\\
\vspace*{2in}
\large Name: \makebox[3in]{\hrulefill}\\
\end{center}
\begin{flushleft}\vspace{0.5in}
Thursday 7/Friday 8 September 2017\\
Instructor: Tom Cuchta
\end{flushleft}
\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}
\vspace*{10pt}
\end{coverpages}

\begin{questions}
\question[5] Sketch a graph of the equation $y=2+\sqrt{x-1}$. Include at last two labeled points in your plot to guarantee full credit.\\
\textit{Solution:} \\
		\begin{tabular}{lll}
			\begin{minipage}{0.45\textwidth}
				\begin{tikzpicture}
					\begin{axis}[axis lines=middle,xmin=-1,xmax=6,ymax=5,ymin=-1]
					\addplot[very thick,->,samples=100][domain=0:5] {sqrt(x)};
					\node[label={270:{$(0,0)$}},circle,fill,inner sep=1.5pt] at (axis cs:0,0) {};
					\node[label={270:{$(1,1)$}},circle,fill,inner sep=1.5pt] at (axis cs:1,1) {};
					\node[label={270:{$(4,2)$}},circle,fill,inner sep=1.5pt] at (axis cs:4,2) {};
					\node[label={180:{$y=\sqrt{x}$}}] at (axis cs:3.5,4.5) {};
					\end{axis}
				\end{tikzpicture} 
			\end{minipage}
			&	\begin{tikzpicture}
					\draw node at (2,6) {\scalebox{2}{$\stackrel{\mathrm{h.shift}}{\longrightarrow}$}};
				\end{tikzpicture}
			&\begin{minipage}{0.45\textwidth}
				\begin{tikzpicture}
					\begin{axis}[axis lines=middle,xmin=-1,xmax=6,ymax=5,ymin=-1]
					\addplot[very thick,->,samples=100][domain=1:5.5] {sqrt(x-1)};
					\node[label={135:{$(1,0)$}},circle,fill,inner sep=1.5pt] at (axis cs:1,0) {};
					\node[label={270:{$(2,1)$}},circle,fill,inner sep=1.5pt] at (axis cs:2,1) {};
					\node[label={270:{$(5,2)$}},circle,fill,inner sep=1.5pt] at (axis cs:5,2) {};
					\node[label={180:{$y=\sqrt{x}$}}] at (axis cs:3.5,4.5) {};
					\end{axis}					
				\end{tikzpicture}
			\end{minipage}\\
			& \begin{tikzpicture}
					\draw node at (2,6) {\scalebox{2}{$\stackrel{\mathrm{v.shift}}{\longrightarrow}$}};
				\end{tikzpicture}			
			&\begin{minipage}{0.45\textwidth}
				\begin{tikzpicture}
					\begin{axis}[axis lines=middle,xmin=-1,xmax=6,ymax=5,ymin=-1]
					\addplot[very thick,->,samples=100][domain=1:5.5] {2+sqrt(x-1)};
					\node[label={270:{$(1,2)$}},circle,fill,inner sep=1.5pt] at (axis cs:1,2) {};
					\node[label={270:{$(2,3)$}},circle,fill,inner sep=1.5pt] at (axis cs:2,3) {};
					\node[label={270:{$(5,4)$}},circle,fill,inner sep=1.5pt] at (axis cs:5,4) {};
					\node[label={180:{$y=\sqrt{x}$}}] at (axis cs:3.5,4.5) {};
					\end{axis}					
				\end{tikzpicture}
			\end{minipage}
	\end{tabular}
\question[6] Draw the specified angle.
\begin{parts}
\part[2] $\dfrac{3\pi}{4}$ radians\\
\textit{Solution:}
	\begin{tikzpicture}
		\draw[<->] (-1,0) -- (1,0);
		\draw[<->] (0,-1) -- (0,1);
		\draw[thick,->] (0,0) -- (-0.71,0.71);
		\draw[thin,->] (0.3,0) arc (0:130:1em);
	\end{tikzpicture}
\part[2] $237^{\circ}$\\
\textit{Solution:}
	\begin{tikzpicture}
		\draw[<->] (-1,0) -- (1,0);
		\draw[<->] (0,-1) -- (0,1);
		\draw[thick,->] (0,0) -- (-0.83,-0.83);
		\draw[thin,->] (0.3,0) arc (0:233:1em);
	\end{tikzpicture}
\part[2] $-\dfrac{13\pi}{6}$ radians\\
\textit{Solution:}
	\begin{tikzpicture}
		\draw[<->] (-1,0) -- (1,0);
		\draw[<->] (0,-1) -- (0,1);
		\draw[thick,->] (0,0) -- (0.87,-0.5);
		\bigangle{-390};
	\end{tikzpicture}
\end{parts}

\question[6] Here is a portion of the unit circle, partially labeled: \\
\begin{center}
    \begin{tikzpicture}[scale=2,cap=round,>=latex]
        \draw[->] (-1.5cm,0cm) -- (1.5cm,0cm) node[right,fill=white] {$x$};
        \draw[->] (0cm,-1.5cm) -- (0cm,1.5cm) node[above,fill=white] {$y$};
        \draw[thick] (0cm,0cm) circle(1cm);

		\coordinate (a) at (0,0);
		\coordinate [label=$(\frac{1}{\sqrt{5}}{,}\frac{2}{\sqrt{5}})$](b) at (0.447213,0.894427);
		\coordinate (c) at (0,0);
		\draw (a) -- (b) -- (c) --cycle;
		\draw [black,thick](a)+(0.2,0) arc (15:29:0.95cm);

		\draw (0.45,0.2) node {$\theta$};
		\draw (1.2,-0.2) node {$(1,0)$};
    \end{tikzpicture}
\end{center}

\begin{parts}
\part[2] What is $\cos(\theta)$?\\
\textit{Solution:} $\dfrac{1}{\sqrt{5}}$
\part[2] What is $\cot(\theta)$?\\
\textit{Solution:} $\dfrac{1/\sqrt{5}}{2/\sqrt{5}}=\dfrac{1}{\sqrt{5}}\cdot \dfrac{\sqrt{5}}{2}=\dfrac{1}{2}$
\part[2] What is $\csc(\theta)$?\\
\textit{Solution:} $\dfrac{1}{2/\sqrt{5}}=\dfrac{\sqrt{5}}{2}$
\end{parts}

\newpage

\question[12] 
\begin{parts}
\part[4] Convert $119^{\circ}$ into radians.\\

\textit{Solution:} $119^{\circ}=119^{\circ}\cdot\dfrac{\pi}{180^{\circ}}=\dfrac{119\pi}{180}$\\

\part[4] Convert $\dfrac{7\pi}{13}$ radians into degrees.\\

\textit{Solution:} $\dfrac{7\pi}{13}=\dfrac{7\pi}{13}\cdot \dfrac{180^{\circ}}{\pi}=\dfrac{7\cdot 180}{13}=\dfrac{1260}{13}$\\

\part[4] Convert $5$ radians to degrees.\\

\textit{Solution:} $5=5\cdot\dfrac{180}{\pi}=\dfrac{900}{\pi}$

\end{parts}

\question[4] What quadrant is $\theta$ in if...
\begin{parts}
\part[2] $\cos(\theta)>0$ and $\sin(\theta)<0$.\\

\textit{Solution:} IV.
\part[2] $\tan(\theta)<0$ and $\csc(\theta)>0$\\

\textit{Solution:} II
\end{parts}

\question[12] Find an exact value for...
\begin{parts}
\part[4] $\sin \left( \dfrac{\pi}{2} \right)$\\

\textit{Solution:} $\sin\left(\dfrac{\pi}{2}\right)=1$

\part[4] $\cot \left( \dfrac{\pi}{3} \right)$\\

\textit{Solution:} $\cot\left(\dfrac{\pi}{3}\right)=\dfrac{1/2}{\sqrt{3}/2}=\dfrac{1}{\sqrt{3}}$

\part[4] $\cos \left( \dfrac{11\pi}{6} \right)$\\

\textit{Solution:} $\cos\left(\dfrac{11\pi}{6}\right)=\dfrac{\sqrt{3}}{2}$

\end{parts}

\newpage 

\question[12] 
\begin{parts}
\part[6] Solve the triangle. \\
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (4,4) -- cycle;
\draw (4.5,1.7) node {$b$};
\draw (1.8,2.6) node {$7$};
\draw (2,-0.5) node {$5$};
\draw (3.7,3.3) node {$A$};
\draw (0.6,0.2) node {$21^{\circ}$};
\draw (3.8,0) -- (3.8,0.2) -- (4,0.2);
\end{tikzpicture}
\end{center}

\textit{Solution:} \\
$A=180^{\circ}-90^{\circ}-21^{\circ}=\boxed{69^{\circ}}$\\
\begin{flalign*}
	25+b^2&=49&\\
	b^2&=24&\\
	\Aboxed{b&=\sqrt{24}}&
\end{flalign*}

\part[6] Solve the triangle. \\
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (4,4) -- cycle;
\draw (4.5,1.7) node {$11$};
\draw (1.8,2.6) node {$c$};
\draw (2,-0.5) node {$3$};
\draw (3.7,3.25) node {$13^{\circ}$};
\draw (0.6,0.2) node {$B$};
\draw (3.8,0) -- (3.8,0.2) -- (4,0.2);
\end{tikzpicture}
\end{center}
\textit{Solution:}\\
$B=180^{\circ}-90^{\circ}-13^{\circ}=\boxed{77^{\circ}}$\\
\begin{flalign*}
	3^2+11^2&=c^2&\\
	130&=c^2&\\
	\Aboxed{c&=\sqrt{130}}&
\end{flalign*}
\end{parts}

\newpage

\question[15] If $\cos(t)=\dfrac{5}{9}$ and $t$ is in quadrant IV, find the other five trigonometric functions.\\

\textit{Solution:} \\
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (4,4) -- cycle;
\draw (4.5,1.7) node {$\sqrt{56}$};
\draw (1.8,2.6) node {$9$};
\draw (2,-0.5) node {$5$};
\draw (0.6,0.2) node {$t$};
\draw (3.8,0) -- (3.8,0.2) -- (4,0.2);
\end{tikzpicture}

$a^2=9^2-5^2=81-25=56\rightarrow a=\sqrt{56}$	
\end{minipage}
\begin{minipage}{0.45\textwidth}
$\begin{array}{ll}
	\sin(t)=-\dfrac{\sqrt{56}}{9} & \csc(t)=-\dfrac{9}{\sqrt{56}}\\
	\cos(t)=\dfrac{5}{9} & \sec(t)=\dfrac{9}{5}\\
	\tan(t)=-\dfrac{\sqrt{56}}{5} & \cot(t)=-\dfrac{5}{\sqrt{56}}
\end{array}$
\end{minipage}

\question[8] Find the arc length subtended by an angle of $\theta=37^{\circ}$ of a circle of radius 3. Express your final answer accurate to at least two decimal places.\\

\textit{Solution:}\\
First, convert $\theta$ to radians:\\
\begin{align*}
	\theta=37^{\circ}=37^{\circ}\cdot\dfrac{\pi}{180^{\circ}}=\dfrac{37\pi}{180}.
\end{align*}
\begin{align*}
	s&=r\theta\\
	s&=3\cdot\dfrac{37\pi}{180}\\
	s&=\dfrac{111\pi}{180}\\
	\Aboxed{s&=1.9373}
\end{align*}
\question[16] There is an antenna on the top of a building. From a location 500 feet from the base of the building, the angle of elevation to the top of the building is measured to be $42^{\circ}$. From the same location, the angle of elevation to the top of the antenna is measured to be $47^{\circ}$. Find the height of the antenna. Express your final answer accurate to at least two decimal places.\\

\textit{Solution:} We start with a figure to describe the problem:\\


\begin{tikzpicture}[scale=1.5]
	\draw (1,0) rectangle (1.5,1);
	\draw (1.25,1) -- (1.25,2);
	\draw[fill=black] (1.25,2) circle (3pt);
	\draw (0.5,0) -- (5,0);
	\draw[dashed] (5,0) -- (0.5,2.5);
	\draw[dashed] (5,0) -- (0.5,1.3);
	\draw (4,0) arc (180:165:1cm) node[midway, left] {$42^{\circ}$};
	\draw (3.25,0) arc (180:155:2cm) node[midway,left] {$47^{\circ}$};
	\draw[|-|] (1.25,-0.15) -- (5,-0.15) node[midway,below] {500};
	\draw[|-|] (0.9,0.05) -- (0.9,1) node[midway, left] {$b$};
	\draw[|-|] (0.9,1.05) -- (0.9,2.1) node[midway, left] {$a$};
\end{tikzpicture}

Thus, we seek $a$.

Note that we have \textit{two} right triangles in this picture, reproduced (in a simplified form) below:\\

\begin{tikzpicture}
	\draw (0,0) -- (0,2) node[midway, left] {$b$} -- (4,0) -- cycle node[midway, below] {500};
	\draw (0,0) rectangle (0.2,0.2);
	\node at (3,0.25) {$42^{\circ}$};
	\draw (6,0) -- (6,2) node[midway, left] {$b+a$}-- (10,0) -- cycle node[midway, below] {500}; 	
	\draw (6,0) rectangle (6.2,0.2);
	\node at (9,0.25) {$47^{\circ}$};
\end{tikzpicture}

Thus,
\begin{flalign*}
	\tan\left(42^{\circ}\right)&=\dfrac{b}{500}&\\
	b&=500\tan\left(42^{\circ}\right)&
\end{flalign*}
and
\begin{flalign*}
	\tan\left(47^\circ\right)&=\dfrac{b+a}{500}&\\
	b+a&=500\tan\left(47^{\circ}\right)&
\end{flalign*}
So that
\begin{flalign*}
	a&=\left(b+a\right)-b&\\
	&=500\tan\left(47^{\circ}\right)-500\tan\left(42^{\circ}\right)&\\
	\Aboxed{a&\approx 85.98\text{ feet}.}&
\end{flalign*}
\question[4] Use your calculator to compute the following trig functions accurate to at least two decimal places.
\begin{parts}
\part[1] $\tan(21^{\circ})$\\

\textit{Solution:}\\
$\tan(21^{\circ})=0.3838$\\

\part[1] $\cos\left( \dfrac{8\pi}{13} \right)$\\

\textit{Solution:}\\
$\cos\left( \dfrac{8\pi}{13} \right)=-0.3546$\\

\part[1] $\csc(57^{\circ})$\\

\textit{Solution:}\\
$\csc(57^{\circ})=1.192$\\

\part[1] $\tan(21)$\\

\textit{Solution:}\\
$\tan(21)=-1.527$

\end{parts}

\end{questions}
\end{document}