In figure , AD = DB and ∠B = 90° then find the values of the following:

i. sin θ

ii. cos θ

iii. tan θ

In ΔABC, By Pythagoras theorem

Hypotenuse² = Perpendicular² + Base²

⇒ AC² = AB² + BC²

⇒ b² = a² + BC²

⇒ b² – a² = BC²

Now,

AD + DB = AB

⇒ DB + DB = AB

⇒ 2DB = AB

⇒ 2DB = a

In ΔBDC, By Baudhayan theorem

Hypotenuse² = Perpendicular² + Base²

⇒ CD² = DB² + BC²

⇒