Tan (theta - 30 degrees) = -5 ...say what?!

Question

Find all solutions of each equation in the interval [0º, 360º).

Round to the nearest tenth of a degree.

tan (Θ - 30º) = -5

Please show all steps and how this relates to the unit circle.

Answer 1

In order to solve tan(θ - 30°) = -5:

Step 1: Use the inverse function of tan to simplify the equation

Firstly, we can take arctan (inverse of tangent) on both sides:

tan^-1(tan(θ - 30°)) = tan^-1(-5)

What arctan does it it "undoes" the tan(x) for angles within the range of tan^-1x which is between -90° and 90°. So, it allows us to simplify this to:

θ - 30° = tan^-1(-5)

Using a calculator, we can find out what tan^-1(-5) evaluates to and so it becomes:

θ - 30° = -78.69°

Step 2: Solve for θ

Now, we can solve for θ by adding 30° to both sides which gives us:

θ = -78.69° + 30° = -48.69°

Step 3: Find the solutions in the interval [0º, 360º)

-48.69° would be our answer but the question states to find the solutions in the interval [0º, 360º) and in order to do that, we can find the reference angles by adding 180º to give us the angle in the 2^nd quadrant and add 180º again to give us the angle in the 4^th quadrant since this angle is negative and we know that tangent is negative in the 2^nd and the 4^th quadrants.

θ₁ = -48.69° + 180° = 131.31°

θ₂ = 131.31° + 180° = 311.31°

Since, we have to round off to the nearest tenth of a degree, it becomes:

θ_{1 =} 131.3°

θ₂ = 311.3°

Answer 2

Tan^-1(5) = 78.7° = reference angle

Tangent is negative in Quadrants 2 and 4.

So, θ - 30° = 180° - 78.7° or 360° - 78.7°

θ = 131.3° or 311.3°