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Quiz 10
1. Suppose that $\theta$ is an angle in standard position that passes through $(-2,5)$. Find the value of the six trigonometric functions of $\theta$.
Solution: The given point $(-2,5)$ tells us that $x=-2$ and $y=5$. By definition, $r=\sqrt{(-2)^2+5^2}=\sqrt{4+25}=\sqrt{29}$. By definition, $\sin(\theta) = \dfrac{y}{r}$, $\cos(\theta)=\dfrac{x}{r}$, $\tan(\theta)=\dfrac{y}{x}$, $\sec(\theta)=\dfrac{r}{x}$, $\csc(\theta)=\dfrac{x}{y}$, and $\sec(\theta)=\dfrac{r}{x}$. Therefore for this problem, $$\sin(\theta) = \dfrac{-2}{\sqrt{29}},$$ $$\cos(\theta) =\dfrac{5}{\sqrt{29}},$$ $$\tan(\theta) = \dfrac{5}{-2},$$ $$\csc(\theta) = \dfrac{\sqrt{29}}{5},$$ $$\sec(\theta) = \dfrac{\sqrt{29}}{-2},$$ and $$\cot(\theta) = \dfrac{-2}{5}.$$

2. Use a calculator to compute $\tan(0.7311^{\circ})$.
Solution: $$\tan(0.7311^{\circ})=0.0127608\ldots$$

3. Solve the right triangle:

Solution:
Find $A$
Use the fact that the sum of angles in a triangle is $180^{\circ}$ to write $$90^{\circ} + 32^{\circ} + A = 180^{\circ}.$$ Rearrange the equation to get $$A =180^{\circ} - 90^{\circ} - 32^{\circ}=58^{\circ}.$$ Find $c$
Use the sine function to write $$\sin(32^{\circ})=\dfrac{5}{c},$$ so algebra gives $$c = \dfrac{5}{\sin(32^{\circ})} = 9.435.$$

Find $a$
Use the tangent function to write $$\tan(32^{\circ})=\dfrac{5}{a},$$ so algebra gives $$a =\dfrac{5}{\tan(32^{\circ})}=8.002.$$ 4. Solve the right triangle if possible. If not possible, say so:

Solution:
Find $B$
We know that $$\tan(B)=\dfrac{2}{3},$$ so $$B=\tan^{-1}\left( \dfrac{2}{3} \right)=33.69^{\circ}.$$
Find $A$
Use the fact that the sum of angles in a triangle is $180^{\circ}$ to write $$33.69^{\circ} + A + 90^{\circ}=180^{\circ},$$ and solve for $A$ to get $$A = 180^{\circ} - 33.69^{\circ} - 90^{\circ}=56.31^{\circ}.$$
Find $c$
Can use Pythagorean theorem to write $$3^2+2^2=c^2,$$ therefore $$c=\sqrt{9+4}=\sqrt{13}.$$ Could also use trigonometric functions to do it.


5. Solve the right triangle if possible. If not possible, say so:

Solution: It is not possible. Imagine blowing the triangle up to be twice as big (with the same angles) -- this shows us there's no single answer to this problem, so it cannot be solved.