Question

if sin 3/5 then find cos and tan

Answer 1

$\sin( \alpha ) = \frac{perpendicular}{hypotenuse} \\ = \frac{AC}{AB} = \frac{3}{5} \\ hence \: bc = 4 \: by \: pythagoras \\ hence \: \cos( \alpha ) = \frac{base}{hypotenuse} \\ = \frac{4}{5} \\ tan \alpha = \frac{perpendicular}{base} = \frac{3}{4}$

Answer 2

    According to the question

sinA = $\dfrac{3}{5}$

          Method : 1( By Theorem )

sinA = $\dfrac{3}{5}$

$\dfrac{height}{hypotenuse} = \dfrac{3}{5}$

Let,

      Height from angle A = 3 x

      Hypotenuse of triangle = 5 x

         By Pythagoras Theorem

Hypotenuse^2 = height^2 + base^2

( 5 x )^2 = ( 3 x )^2 + ( base )^2

25x^2 - 3x^2 = ( base )^2

16x^2 = base^2

( 4x )^2 = ( base )^2

4x = base

Therefore, cosA = $\dfrac{base}{hypotenuse}$

                           =$\dfrac{4x}{5x}$

                           = $\bold{\dfrac{4}{5}}$

And, tanA = $\dfrac{height}{base}$

     

                 = $\dfrac{3x}{4x}$

                 = $\bold{\dfrac{3}{4}}$

                Method : 2 ( Identities )

We know, sin^2 A + cos^2 A = 1

           Substituting value,

( 3 / 5 )^2 + cos^2 A = 1

cos^2 A = 1 - 9 / 25

cos^2 A = ( 25 - 9 ) / 25

cos^2 A = 16 / 25

cos^2 A = ( 4 / 5 )^2

cos A = 4 / 5

tanA = sinA / cosA

        = ( 3 / 5 ) / (  4 / 5 )

        = 3 / 4