Question

Help appreciated please and thank you​

Answer 1

Answer:

D. $\frac{4 \sqrt{33} }{33}$

Step-by-step explanation:

$\tan \alpha = \frac{4}{ \sqrt{33} } \\ = \frac{4}{ \sqrt{33} } \times \frac{ \sqrt{33} }{ \sqrt{33} } \\ \\ = \frac{4 \sqrt{33} }{ {( \sqrt{33)} }^{2} } \\ \\ \tan \alpha = \frac{4 \sqrt{33} }{33}$

Answer 2

The answer I think is d