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Sagot :
[tex]\text{cos A = }\frac{\sqrt[]{5}}{3}[/tex]Explanation:
We need an illustration:
To find cos A, we would apply cosine ratio (CAH)
[tex]\cos \text{ A = }\frac{adjacent}{\text{hypotenuse}}[/tex][tex]\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}[/tex][tex]\begin{gathered} \cos \text{ A = }\frac{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}[/tex][tex]\begin{gathered} 20=b^2 \\ b\text{ = }\sqrt[]{20}\text{ = }\sqrt[]{4\times5} \\ b\text{ = 2}\sqrt[]{5} \end{gathered}[/tex][tex]\begin{gathered} \cos \text{ A = }\frac{2\sqrt[]{5}}{6} \\ \cos \text{ A = }\frac{\sqrt[]{5}}{3} \end{gathered}[/tex]
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