Right ABC has its right angle at C, BC=4 , and AC=8 .

Start with drawing that right angle triangle 

 

The sides are 

BC=4

AC=8

AB=\(\sqrt{80}\) (applying Pythagoras theorem,

 \(\sqrt{(AC)^2+(BC)^2}=AB\) "Addition since AB is the hypotenuse" 

So we know the sides and they ask for trigonometric ratios 

sin A =??

I assume A is angle "A"

\(sin(A)=\frac{opposite}{hypotenuse}\)

The opposite side to angle A is BC and the hypotenuse is AB

We know BC=4 and AB= \(\sqrt{80}\)

\(sin(A)=\frac{4}{\sqrt{80}}\)

 

Here is hints for the next 2 questions:

\(tan(B)=\frac{opposite}{adjacent}\)

\(sec(A)=\frac{hypotenuse}{adjacent}\)Proof: (Notice that \(sec(A)=\frac{1}{cos(A)}\), we know that \(cos(A)=\frac{Adjacent}{Hypotenuse}\) , substitute: \(sec(A)=\frac{1}{\frac{adjacent}{hypotenuse}}\) Multiply the numerator and denominator by the hypotenuse, \(sec(A)=\frac{hypotenuse}{adjacent}\)