Question

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Answer 1

Answer:

The values of $r_{2}$ and $\alpha_{2}$ are 2 and 150º.

Step-by-step explanation:

The complete statement is:

Find $\alpha_{2}$ and $r_{2}$ such that  $\sin \theta - \sqrt{3}\cdot \cos \theta = r_{2}\cdot \cos (\theta - \alpha_{2})$.

We proceed to use the following trigonometric identity:

$\cos (\theta - \alpha_{2}) = \cos \theta \cdot \cos \alpha_{2} +\sin \theta \cdot \sin \alpha_{2}$ (1)

$\sin \theta -\sqrt{3}\cdot \cos \theta = r_{2}\cdot \cos \theta \cdot \cos \alpha_{2}+r_{2}\cdot \sin \theta \cdot \sin \alpha_{2}$

By direct comparison we derive these expressions:

$r_{2}\cdot \sin \alpha_{2} = 1$ (2)

$r_{2}\cdot \cos \alpha_{2} = -\sqrt{3}$ (3)

By dividing (2) by (3), we have the following formula:

$\tan \alpha_{2} = -\frac{1}{\sqrt{3}}$

$\tan \alpha_{2} = -\frac{\sqrt{3}}{3}$

The tangent function is negative at second and fourth quadrants. That is:

$\alpha_{2} = \tan^{-1} \left(-\frac{\sqrt{3}}{3} \right)$

There are at least two solutions:

$\alpha_{2,1} = 150^{\circ}$, $\alpha_{2,2} = 330^{\circ}$

And the value of $r_{2}$:

$r_{2}^{2}\cdot \sin^{2}\alpha_{2} + r_{2}^{2}\cdot \cos^{2}\alpha_{2} = 4$

$r_{2}^{2} = 4$

$r_{2} = 2$

The values of $r_{2}$ and $\alpha_{2}$ are 2 and 150º.