Solve for B in Degrees 12tan(B)+5=5tan(B)+5

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Trigonometry

$12 \tan (B) + 5 = 5 \tan (B) + 5$

Move all terms containing $\tan (B)$ to the left side of the equation.

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Subtract $5 \tan (B)$ from both sides of the equation.

$12 \tan (B) + 5 - 5 \tan (B) = 5$

Subtract $5 \tan (B)$ from $12 \tan (B)$ .

$7 \tan (B) + 5 = 5$

Move all terms not containing $\tan (B)$ to the right side of the equation.

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Subtract $5$ from both sides of the equation.

$7 \tan (B) = 5 - 5$

Subtract $5$ from $5$ .

$7 \tan (B) = 0$

Divide each term in $7 \tan (B) = 0$ by $7$ and simplify.

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Divide each term in $7 \tan (B) = 0$ by $7$ .

$\frac{7 \tan (B)}{7} = \frac{0}{7}$

Simplify the left side.

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Cancel the common factor of $7$ .

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Cancel the common factor.

$\frac{7 \tan (B)}{7} = \frac{0}{7}$

Divide $\tan (B)$ by $1$ .

$\tan (B) = \frac{0}{7}$

Simplify the right side.

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Divide $0$ by $7$ .

$\tan (B) = 0$

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

$B = \arctan (0)$

Simplify the right side.

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The exact value of $\arctan (0)$ is $0$ .

$B = 0$

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

$B = 180 + 0$

Add $180$ and $0$ .

$B = 180$

Find the period of $\tan (B)$ .

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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 $1$ .

$180$

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

$B = 180 n, 180 + 180 n$ , for any integer $n$

Consolidate the answers.

$B = 180 n$ , for any integer $n$

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