Consider the absolute value, because we only worry about the quadrant later.
[tex]\text{Consider: } sin(A) = |-\frac{\sqrt{3}}{4}|[/tex]
[tex]sin(A) = \frac{\sqrt{3}}{4}[/tex]
Thus, we know that the hypotenuse has a length of 4 units, and the side opposite the angle, A is √3, because this is the nature of the sine function in relation to its triangular component.
The missing side can be found using Pythagoras' Theorem:
4² - (√3)² = x²
16 - 3 = x²
13 = x²
x = √13
[tex]\therefore tan(A) = \pm \sqrt{\frac{3}{13}}[/tex]
Since angle A is in the third quadrant, the tangent function will produce a positive angle.
[tex]tan(A) = \sqrt{\frac{3}{13}}[/tex]
[tex]\text{Rationalise the denominator: } \frac{\sqrt{3}}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}}[/tex]
[tex]tan(A) = \frac{\sqrt{39}}{13}[/tex]
