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Quiz 9
1. Convert to degrees: $3^{\circ} 12^{\prime} 15^{\prime \prime}$.
Solution: Recall that $1^{\prime}=\left(\dfrac{1}{60} \right)^{\circ}$ and $1^{\prime \prime} = \left( \dfrac{1}{3600} \right)^{\prime \prime}$. This means we can convert from minutes (i.e. ${}^{\prime}$) to degrees using the relationship $1=\dfrac{\frac{1}{60}^{\circ}}{1^{\prime}}$. Similarly we can convert seconds into degrees using the relationship $1= \dfrac{\frac{1}{3600}^{\circ}}{1^{\prime \prime}}$. So compute
$$12^{\prime} = 12^{\prime}\left( \dfrac{\frac{1}{60}^{\circ}}{1^{\prime}} \right)=\left( \dfrac{12}{60} \right)^{\circ} = 0.2^{\circ}$$
and
$$15^{\prime \prime} = 15^{\prime \prime} \left(\dfrac{\frac{1}{3600}^{\circ}}{1^{\prime \prime}} \right)=\left(\dfrac{15}{3600}\right)^{\circ}=0.004166\ldots$$
Therefore we see that
$$3^{\circ} 12^{\prime} 15^{\prime \prime} = 3^{\circ} + 0.2^{\circ} + 0.004166^{\circ} = 3.204166^{\circ}$$
2. Convert to radians: $60^{\circ}$. What quadrant does this angle lie in?
Solution: First, the angle is in quadrant I. Recall that $360^{\circ}=2\pi \mathrm{\hspace{2pt} radians}$. This means to convert from degrees to radians, we will use the conversion factor $1=\dfrac{2\pi \mathrm{\hspace{2pt} radians}}{360^{\circ}}$. So to convert, we write
$$60^{\circ} = 60^{\circ} \dfrac{2\pi \mathrm{\hspace{2pt} radians}}{360^{\circ}}=\dfrac{\pi}{3} \mathrm{radians}$$
3. Convert to degrees: $\dfrac{3\pi}{2} \mathrm{radians}$.
Solution: We will use the conversion factor $1= \dfrac{360^{\circ}}{2\pi \mathrm{\hspace{2pt}radians}}$. So compute
$$\dfrac{3\pi}{2} \mathrm{radians} = \dfrac{3\pi}{2} \mathrm{radians} \left( \dfrac{360^{\circ}}{2\pi \mathrm{\hspace{2pt} radians}} \right)=270^{\circ}.$$
4. Find the value of the six trig functions of an angle $\theta$ that goes through the point $(2,2)$.
Solution: Here $x=2$ and $y=2$. We must find $r$:
$$r=\sqrt{x^2+y^2}=\sqrt{4+4}=\sqrt{8}.$$
Therefore the six trig functions are:
$$\sin(\theta)=\dfrac{y}{r}=\dfrac{2}{\sqrt{8}},$$
$$\cos(\theta)=\dfrac{x}{r}=\dfrac{2}{\sqrt{8}},$$
$$\tan(\theta)=\dfrac{y}{x}=\dfrac{2}{2}=1,$$
$$\csc(\theta)=\dfrac{r}{y}=\dfrac{\sqrt{8}}{2},$$
$$\sec(\theta)=\dfrac{r}{x}=\dfrac{\sqrt{8}}{2},$$
and
$$\cot(\theta)=\dfrac{x}{y}=\dfrac{2}{2}=1.$$
5. Given $\cos(\theta)=\dfrac{3}{7}$ find $\sin(\theta)$ and $\cot(\theta)$.
Solution: Since $\cos(\theta)=\dfrac{x}{r}=\dfrac{3}{7}$ we may take $x=3$ and $r=7$. To find $y$ we use the formula $x^2+y^2=r^2$ and isolate $y$: $y^2=r^2-x^2$, or $y=\sqrt{r^2-x^2}$. Plug in the values of $r$ and $x$ to get
$$y=\sqrt{7^2-3^2}=\sqrt{49-9}=\sqrt{40}.$$
Now we see that
$$\sin(\theta) = \dfrac{y}{r}=\dfrac{\sqrt{40}}{7}$$
and
$$\cot(\theta)=\dfrac{x}{y}=\dfrac{3}{\sqrt{40}}.$$
