Question

find the exact value of cos a in simplest radical form.

Explanation:

Step1: Recall cosine definition

In a right triangle, $\cos A = \frac{\text{adjacent to } A}{\text{hypotenuse}}$.

Step2: Identify sides

Adjacent to $A$ is $AC = 3$, hypotenuse $AB = 9$.

Step3: Calculate $\cos A$

$\cos A = \frac{3}{9} = \frac{1}{3}$? Wait, no, wait. Wait, check the triangle. Wait, $AC = 3$, $BC = \sqrt{72}$, $AB = 9$. Wait, let's verify the triangle with Pythagoras. $AC^2 + BC^2 = 3^2 + (\sqrt{72})^2 = 9 + 72 = 81 = 9^2$, so it's a right triangle. So adjacent to $A$ is $AC = 3$, hypotenuse $AB = 9$. Wait, but $\frac{3}{9} = \frac{1}{3}$? Wait, that seems too simple. Wait, no, maybe I mixed up. Wait, in angle $A$, the adjacent side is $AC$, opposite is $BC$, hypotenuse $AB$. So cosine is adjacent over hypotenuse, so $\cos A = \frac{AC}{AB} = \frac{3}{9} = \frac{1}{3}$. Wait, but let me check again. $AC = 3$, $AB = 9$, so yes, $\frac{3}{9} = \frac{1}{3}$. Wait, but maybe I made a mistake. Wait, $\sqrt{72}$ is $6\sqrt{2}$, so $3^2 + (6\sqrt{2})^2 = 9 + 72 = 81 = 9^2$, so the triangle is valid. So cosine of $A$ is adjacent over hypotenuse, which is $3/9 = 1/3$. Wait, but that seems too easy. Wait, maybe the problem is different? Wait, no, the diagram shows right triangle at $C$, so angle $A$ is at vertex $A$, so sides: $AC = 3$, $BC = \sqrt{72}$, $AB = 9$. So cosine of $A$ is adjacent (AC) over hypotenuse (AB), so $3/9 = 1/3$. Wait, but maybe I misread the diagram. Wait, the diagram: $C$ is right angle, so $AC$ and $BC$ are legs, $AB$ hypotenuse. So $AC = 3$, $AB = 9$, so $\cos A = AC/AB = 3/9 = 1/3$. Wait, but let me confirm with trigonometry. Yes, in right triangle, cosine of an angle is adjacent side over hypotenuse. So that's correct.

Answer:

$\frac{1}{3}$