Answer:
The values of [tex]r_{2}[/tex] and [tex]\alpha_{2}[/tex] are 2 and 150º.
Step-by-step explanation:
The complete statement is:
Find [tex]\alpha_{2}[/tex] and [tex]r_{2}[/tex] such that [tex]\sin \theta - \sqrt{3}\cdot \cos \theta = r_{2}\cdot \cos (\theta - \alpha_{2})[/tex].
We proceed to use the following trigonometric identity:
[tex]\cos (\theta - \alpha_{2}) = \cos \theta \cdot \cos \alpha_{2} +\sin \theta \cdot \sin \alpha_{2}[/tex] (1)
[tex]\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}[/tex]
By direct comparison we derive these expressions:
[tex]r_{2}\cdot \sin \alpha_{2} = 1[/tex] (2)
[tex]r_{2}\cdot \cos \alpha_{2} = -\sqrt{3}[/tex] (3)
By dividing (2) by (3), we have the following formula:
[tex]\tan \alpha_{2} = -\frac{1}{\sqrt{3}}[/tex]
[tex]\tan \alpha_{2} = -\frac{\sqrt{3}}{3}[/tex]
The tangent function is negative at second and fourth quadrants. That is:
[tex]\alpha_{2} = \tan^{-1} \left(-\frac{\sqrt{3}}{3} \right)[/tex]
There are at least two solutions:
[tex]\alpha_{2,1} = 150^{\circ}[/tex], [tex]\alpha_{2,2} = 330^{\circ}[/tex]
And the value of [tex]r_{2}[/tex]:
[tex]r_{2}^{2}\cdot \sin^{2}\alpha_{2} + r_{2}^{2}\cdot \cos^{2}\alpha_{2} = 4[/tex]
[tex]r_{2}^{2} = 4[/tex]
[tex]r_{2} = 2[/tex]
The values of [tex]r_{2}[/tex] and [tex]\alpha_{2}[/tex] are 2 and 150º.
