Problems from Right triangle resolution using trigonometric identities

Development:

In this case we know one of the cathetus and one of the angles, therefore we can follow fourth case. As such:

The angle $C$ is, $$C={90}^{\circ }-{30}^{\circ }={60}^{\circ }$$

The side $a$ is, $$a={\displaystyle \frac{6}{\sin ({30}^{\circ })}}={\displaystyle \frac{6}{{\displaystyle \frac{1}{2}}}}=12$$

The side $c$ is, $$c={\displaystyle \frac{6}{\tan ({30}^{\circ })}}={\displaystyle \frac{6}{{\displaystyle \frac{\sqrt{3}}{3}}}}={\displaystyle \frac{18}{\sqrt{3}}}$$

Solution:

Thus, the solution is:

$$C={60}^{\circ }a=12c={\displaystyle \frac{18}{\sqrt{3}}}$$

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