Full solution

Q. Example \#

2

: The angle

\theta

is in standard position and

\tan \theta=-\frac{4}{3}

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a) In which quadrants could the terminal arm of the angle lie?

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Quadrant

3

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b) draw a diagram for each case

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x^{2}+y^{2}=c^{2}

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c) determine all possible values for

\theta

Identify Quadrants for $\tan \theta$ : Identify the quadrants where $\tan \theta$ is negative. $\newline$ Since $\tan \theta$ is the ratio of $y$ over $x$ (opposite over adjacent in a right triangle), it's negative when $y$ and $x$ have opposite signs. $\newline$ This happens in Quadrant $\text{II}$ (where $x$ is negative and $y$ is positive) and Quadrant $\tan \theta$ $0$ (where $x$ is positive and $y$ is negative).
Draw Diagrams for Cases: Draw a diagram for each case. $\newline$ For Quadrant II, draw a right triangle with the terminal arm extending into the quadrant, the opposite side going up (positive $y$ ), and the adjacent side going left (negative $x$ ). $\newline$ For Quadrant IV, draw a right triangle with the terminal arm extending into the quadrant, the opposite side going down (negative $y$ ), and the adjacent side going right (positive $x$ ).
Use Pythagorean Identity for $c$ : Use the Pythagorean identity to find $c$ . Given $\tan \theta = -\frac{4}{3}$ , we can assume the opposite side is $4$ and the adjacent side is $-3$ (or $3$ and $-4$ for Quadrant IV). Using the Pythagorean theorem, $x^2 + y^2 = c^2$ , we get $(-3)^2 + 4^2 = c^2$ , which simplifies to $9 + 16 = c^2$ , so $c$ $0$ . Therefore, $c$ $1$ or $c$ $2$ , but since $c$ represents the hypotenuse of a right triangle, it must be positive, so $c$ $1$ .
Determine Values for theta: Determine all possible values for $\theta$ . Since $\tan \theta = -\frac{4}{3}$ , we can use the $\arctan$ function to find the reference angle. The reference angle is $\arctan(\frac{4}{3})$ , which is approximately $53.13$ degrees. In Quadrant II, $\theta$ would be $180 - 53.13 = 126.87$ degrees. In Quadrant IV, $\theta$ would be $360 - 53.13 = 306.87$ degrees.