Question

If tan@ = 1/2 then find the value of sin 20​

Answer 1

Step-by-step explanation:

Let theta = x

tan x=1/2 = prep(AB)/base(BC)

Let in right angled triangle B=90° ,prep.(AB) =k unit. then base (BC)= 2k units.

AC^2=AB^2+BC^2

AC^2= k^2 + 4k^2=5k^2

AC (hypotenuse) =k.√5.units

sin x= prep.(AB)/hypo.(AC)

sin x=k units/k.√5.units.

sin x= 1/√5 . , Answer.

Second-Method :-

Formula : sec^2 x = 1+ tan^2 x.

sec^2 x = 1 +(1/2)^2 = 5/4.

or, sec^2 x = 5/4.

or, 1/cos^2 x = 5/4.

or, cos^2 x = 4/5.

or, 1 - sin^2 x = 4/5.

or, sin^2 x = 1 -4/5 = 1/5.

or, sin x = 1/√5. Answer.

Third - Method:-

If tan x= 1/2 . , sin x = ?

we know that:-

Formula:

sin x = tan x/√(1+tan^2 x). , putting tan x = 1/2.

sin x= 1/2/√(1+1/4)= (1/2)/(√5/2) = 1/√5. Answer.