Question

given Sin A =1/3 find Cos A and Tan A​

Answer 1

$\bf\huge\blue{\underline{\underline{ Question : }}}$

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,

$\sf\:\implies \cos\:A = \cfrac{Adjacent}{Hypotenuse}$

$\sf\:\implies \cos\:A = \cfrac{2\sqrt{2}}{3}$

$\sf\:\implies \tan\:A = \cfrac{Opposite}{Adjacent}$

$\sf\:\implies \tan\:A = \cfrac{1}{2\sqrt{2}}$

$\underline{\boxed{\bf{\purple{\therefore \cos\:A = \cfrac{2\sqrt{2}}{3} \: ; \: \tan\:A = \cfrac{1}{2\sqrt{2}}}}}}\:\orange{\bigstar}$