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

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Find limits using power and root laws

Full solution

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 $\cos A = \frac{3\sqrt{13}}{\sqrt{13} \times \sqrt{13}} = \frac{3\sqrt{13}}{13}$ .

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