Solve for θ in Degrees tan(theta)=-5

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Trigonometry

$\tan (θ) = - 5$

Take the inverse tangent of both sides of the equation to extract $θ$ from inside the tangent.

$θ = \arctan (- 5)$

Simplify the right side.

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Evaluate $\arctan (- 5)$ .

$θ = - 78.69006752$

The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the third quadrant.

$θ = - 78.69006752 - 180$

Simplify the expression to find the second solution.

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Add $360 °$ to $- 78.69006752 - 180 °$ .

$θ = - 78.69006752 - 180 ° + 360 °$

The resulting angle of $101.30993247 °$ is positive and coterminal with $- 78.69006752 - 180$ .

$θ = 101.30993247 °$

Find the period of $\tan (θ)$ .

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The period of the function can be calculated using $\frac{180}{| b |}$ .

$\frac{180}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{180}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{180}{1}$

Divide $180$ by $1$ .

$180$

Add $180$ to every negative angle to get positive angles.

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Add $180$ to $- 78.69006752$ to find the positive angle.

$- 78.69006752 + 180$

Subtract $78.69006752$ from $180$ .

$101.30993247$

List the new angles.

$θ = 101.30993247$

The period of the $\tan (θ)$ function is $180$ so values will repeat every $180$ degrees in both directions.

$θ = 101.30993247 + 180 n, 101.30993247 + 180 n$ , for any integer $n$

Consolidate $101.30993247 + 180 n$ and $101.30993247 + 180 n$ to $101.30993247 + 180 n$ .

$θ = 101.30993247 + 180 n$ , for any integer $n$

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