SOLUTION
Given the question in the image, the following are the solution steps to answer the question
STEP 1: Draw the described triangle
STEP 2: Find the third side
[tex]\begin{gathered} \text{third side is the hypotenuse} \\ \text{opposite}=8,\text{adjacent}=4,\text{hypotenuse}=\text{?} \\ By\text{ pythagorean theorem,} \\ \text{hypotenuse}^2=opposite^2+adjacent^2 \\ \text{hypotenuse}^2=8^2+4^2 \\ \text{hypotenuse}^2=64+16=80 \\ \text{hypotenuse}=\sqrt[]{80}=\sqrt[]{16\times5}=\sqrt[]{16}\times\sqrt[]{5}=4\times\sqrt[]{5}=4\sqrt[]{5} \end{gathered}[/tex]STEP 3: Redraw the complete triangles
STEP 4: Find the trigonometric ratio for csc B
The cosecant of angle B in a right triangle is equal to the length of the hypotenuse divided by the opposite side. To solve csc, simply enter the length of the hypotenuse and opposite side, then solve.
[tex]\begin{gathered} \csc B\text{ stands for Cosec B} \\ \csc B=\frac{1}{\sin\theta}=\frac{hypotenuse}{opposite} \\ \text{For the angle B,} \\ \text{hypoteunse}=4\sqrt[]{5},opposite=8,adjacent=4 \\ By\text{ substituting into the formula, we have} \\ \csc B=\frac{4\sqrt[]{5}}{8}=\frac{1\sqrt[]{5}}{2}=\frac{\sqrt[]{5}}{2} \end{gathered}[/tex]STEP 5: Find the trigonometric ratio for cot B
The cotangent formula for an angle θ is: cot θ = (Adjacent side) / (Opposite side)
[tex]\begin{gathered} \text{cot B stands for cotangent B} \\ \cot B=\frac{\text{adjacent}}{\text{opposite}} \\ \text{opposite}=8,\text{adjacent}=4 \\ By\text{ substitution.} \\ \cot B=\frac{4}{8}=\frac{1}{2} \end{gathered}[/tex]STEP 5: Find the trigonometric ratio for cos A
[tex]\begin{gathered} \cos A\text{ stands for Cosine A} \\ \cos A=\frac{\text{adjacent}}{\text{hypotenuse}} \\ \text{adjacent}=8,\text{hypotenuse}=4\sqrt[]{5} \\ By\text{ substitution,} \\ \cos A=\frac{8}{4\sqrt[]{5}}=\frac{2}{\sqrt[]{5}} \\ By\text{ rationalization of surds, we have;} \\ \frac{2}{\sqrt[]{5}}\times\frac{\sqrt[]{5}}{\sqrt[]{5}}=\frac{2\times\sqrt[]{5}}{\sqrt[]{5}\times\sqrt[]{5}}=\frac{2\sqrt[]{5}}{\sqrt[]{25}} \\ \cos A=\frac{2\sqrt[]{5}}{5} \end{gathered}[/tex]
