Given Sin A =1/3 find Cos A and Tan A

1.	Given Sin A =1/3 find Cos A and Tan A
Answer» $\bf\huge\blue{\underline{\underline{ Question : }}}$ Sin A =1/3 find Cos A and TAN A. $\bf\huge\blue{\underline{\underline{ Solution : }}}$ ★ GIVEN that, sin A = 1/3 ★ To find, cos A tan A ★ LET, We know that, $\tt\:\rightarrow \sin\:A = \cfrac{Opposite}{Hypotenuse} = \cfrac{1}{3}$ ★ Now, Opposite = 1 Hypotenuse = 3 Adjacent = ? By using Pythagoras theorem, $\tt\:(Hypotenuse)^{2} = (Opposite)^{2} + (Adjacent)^{2}$ $\sf\:\implies (AC)^{2} = (AB)^{2} + (BC)^{2}$ $\sf\:\implies (3)^{2} = (AB)^{2} + (1)^{2}$ $\sf\:\implies 9 = (AB)^{2} + 1$ $\sf\:\implies 9 - 1= (AB)^{2}$ $\sf\:\implies (AB)^{2} = 8$ $\sf\:\implies (AB) = \sqrt{8}$ $\sf\:\implies (AB) =2 \sqrt{2}$ ★ Now, $\sf\:\implies \cos\:A = \cfrac{Adjacent}{Hypotenuse}$ $\sf\:\implies \cos\:A = \cfrac{2\sqrt{2}}{3}$ $\sf\:\implies \tan\:A = \cfrac{Opposite}{Adjacent}$ $\sf\:\implies \tan\:A = \cfrac{1}{2\sqrt{2}}$ $\underline{\boxed{\bf{\purple{\therefore \cos\:A = \cfrac{2\sqrt{2}}{3} \: ; \: \tan\:A = \cfrac{1}{2\sqrt{2}}}}}}\:\orange{\bigstar}$

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