Question

I am having trouble solving this practiceI have attempted this but.. I am not sure if what I did was correct, need more clarificationI believe this is trigonometry

Answer 1

Solution

Recall: SOHCAHTOA

Part A

$Cot\text{ G=}\frac{1}{tan\text{ G}}$$\begin{gathered} tanG=\frac{2\sqrt{7}}{6}=\frac{\sqrt{7}}{3} \\ \\ Now \\ cot\text{ G=}\frac{1}{\frac{\sqrt{7}}{3}}=\frac{3}{\sqrt{7}} \end{gathered}$

The final answer

$\begin{gathered} Cot\text{ G=}\frac{3}{\sqrt{7}} \\ \\ rationlise \\ cot\text{ G=}\frac{3\times\sqrt{7}}{\sqrt{7}\times\sqrt{7}}=\frac{3\sqrt{7}}{7} \end{gathered}$

Part B

$sin\text{ E=}\frac{6}{8}=\frac{3}{4}$

The final answer

$sinE=\frac{3}{4}$

Part C

$\begin{gathered} sec\text{ G=}\frac{1}{cos\text{ G}} \\ \\ cos\text{ G=}\frac{6}{8}=\frac{3}{4} \\ now \\ sec\text{ G=}\frac{1}{\frac{3}{4}}=\frac{4}{3} \end{gathered}$

Answer summary