If tan A = 2/3 , find cos A

Identify tan A: We know that tan A = opposite/adjacent, and we are given tan A = 2/3. To find cos A, which is adjacent/hypotenuse, we need to find the length of the hypotenuse using the Pythagorean theorem.Calculate Hypotenuse: Let's assume the opposite side (O) is 2 units and the adjacent side (A) is 3 units. According to the Pythagorean theorem, the hypotenuse (H) can be calculated as H = √(O^2 + A^2).Find cos A: Plugging in the values, we get H = √(2^2 + 3^2) = √(4 + 9) = √13.Rationalize Denominator: Now we can find cos A, which is adjacent/hypotenuse, so cos A = A/H = 3/√13. To rationalize the denominator, we multiply the numerator and denominator by √13.After rationalizing, we get cos A = (3√13)/(√13 * √13) = 3√13/13.

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If $\tan A = \frac{2}{3}$ , find $\cos A$

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Q. If

\tan A = \frac{2}{3}

, find

\cos A

Identify $\tan A$ : We know that $\tan A = \frac{\text{opposite}}{\text{adjacent}}$ , and we are given $\tan A = \frac{2}{3}$ . To find $\cos A$ , which is $\frac{\text{adjacent}}{\text{hypotenuse}}$ , we need to find the length of the hypotenuse using the Pythagorean theorem.
Calculate Hypotenuse: Let's assume the opposite side $O$ is $2$ units and the adjacent side $A$ is $3$ units. According to the Pythagorean theorem, the hypotenuse $H$ can be calculated as $H = \sqrt{O^2 + A^2}$ .
Find $\cos A$ : Plugging in the values, we get $H = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$ .
Rationalize Denominator: Now we can find $\cos A$ , which is adjacent/hypotenuse, so $\cos A = \frac{A}{H} = \frac{3}{\sqrt{13}}$ . To rationalize the denominator, we multiply the numerator and denominator by $\sqrt{13}$ .
Rationalize Denominator: Now we can find $\cos A$ , which is adjacent/hypotenuse, so $\cos A = \frac{A}{H} = \frac{3}{\sqrt{13}}$ . To rationalize the denominator, we multiply the numerator and denominator by $\sqrt{13}$ .After rationalizing, we get $\cos A = \frac{3\sqrt{13}}{\sqrt{13} \times \sqrt{13}} = \frac{3\sqrt{13}}{13}$ .