Question

In triangle ABC, with right angle at C, if c=6 and a =4, the CosA =

Answer 1

`\text{cos A = }\frac{\sqrt[]{5}}{3}`

Step-by-step explanation:

We need an illustration:

To find cos A, we would apply cosine ratio (CAH)

`\cos \text{ A = }\frac{adjacent}{\text{hypotenuse}}`
`\begin{gathered} angle\text{ = A} \\ \text{opposite = side opposite the angle = a} \\ a\text{ = 4} \\ \text{adjacent = CA = b} \\ \text{hypotenuse = c = 6} \end{gathered}`
`\begin{gathered} \cos \text{ A = }(b)/(6) \\ u\sin g\text{ pythagoras theorem:} \\ Hypotenuse^(2)=opposite^(2)+adjacent^(2) \\ 6^2=4^2+b^2 \\ 36=16+b^2 \\ 36-16=b^2 \end{gathered}`
`\begin{gathered} 20=b^2 \\ b\text{ = }\sqrt[]{20}\text{ = }\sqrt[]{4*5} \\ b\text{ = 2}\sqrt[]{5} \end{gathered}`
`\begin{gathered} \cos \text{ A = }\frac{2\sqrt[]{5}}{6} \\ \cos \text{ A = }\frac{\sqrt[]{5}}{3} \end{gathered}`