Question

4.
12
If cos A=
13
then find sin A and tan A (A<90°).
​

Answer 1

$cos \: A = \frac{1}{13}$

$\huge\green{USING \:IDENTITY}$

$sin {}^{2} A + cos {}^{2} A = 1$

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$sin {}^{2} A + ( \frac{1}{13} ) {}^{2} = 1$

$sin {}^{2} A = 1 - \frac{1}{13}$

$sin {}^{2} A = \frac{13 - 1}{13}$

$sin {}^{2} A = \frac{12}{13}$

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Second answer

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$\large{tan \: A= \frac{sin \: A}{cos \: A}}$

$\large{tan \: A = \frac{ \frac{12}{13} }{ \frac{1}{13}}}$

$\large{tan \: A = 12}$

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$\large\red{\underline{\underline{\mathbb{MARk\:BRAINLIEST}}}}$