Question

I need help with my pre-calculus homework, please show me how to solve them step by step if possible. The image of the problem is attached.

Answer 1

`\frac{5\sqrt[]{39}}{39}`

1) We need to make use of one Pythagorean Trig Identity so that we can find the cotangent of A.

`\begin{gathered} \sin ^2(A)+\cos ^2(A)=1 \\ \sin ^2(A)+((5)/(8))^2=1 \\ \sin ^2(A)=1-(25)/(64) \\ \sin (A)=\sqrt[]{(64)/(64)-(25)/(64)} \\ \sin (A)=\sqrt[]{(39)/(64)}=\frac{\sqrt[]{39}}{8} \end{gathered}`

2) Now, we can have the cotangent since we know the sine, the cosine:

`\cot (A)=(\cos(A))/(\sin(A))=\frac{(5)/(8)}{\frac{\sqrt[]{39}}{8}}=(5)/(8)*\frac{8}{\sqrt[]{39}}=\frac{5}{\sqrt[]{39}}\cdot\frac{\sqrt[]{39}}{\sqrt[]{39}}=\frac{5\sqrt[]{39}}{39}`