Question

If tan θ = √3, then value of sin θ is :

(A) 1/√3

(B) √3/2

(C) 2/√3

(D) 1

Answer 1

Answer is (B) √3/2

tan θ = AB/BC = √3 = √3/1

Let AB = √3 k

and BC = k
In right angled ∆ABC, by pythagorus theorem,

AC² =AB² + BC²

= (√3k)² + (k)²

= 3k² + k² = 4k²

∴ AC = 2k

Now,