If Sin A = \(3\over 4\), Calculate Cos A and Tan A.

Solution :

Consider a triangle ABC in which \(\angle\) B = 90triangle

For \(\angle\) A, we have :

Base = AB, Perpendicular = BC and Hypotenuse = AC,

\(\therefore\)  Sin A = \(perpendicular\over hypotenuse\) = \(BC\over AC\) = \(3\over 4\)

Let BC = 3k and AC = 4k,

Then,  \({AB}^2\) = \({AC}^2 – {BC}^2\) = \(\sqrt{7}\)

\(\therefore\)  cos A = \(Base\over Hypotenuse\) = \(AB\over AC\) = \(\sqrt{7}\over 4\)

and  tan A = \(perpendicular\over hypotenuse\) = \(BC\over AB\) = \(3\over \sqrt{7}\)

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