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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.
