Let P (sinθ , cosθ) (0 ≤ θ ≤ 2π) be a point and let OAB be a triangle with vertices (0, 0) , [√(3/2),0] and [0 ,√(3/2)] . Find θ if P lies inside the ΔOAB.

Equations of lines along OA , OB and AB are y = 0, x = 0, and x+y = √(3/2) respectively.

Now P and B will lie on the same side of y = 0 if cosθ > 0.

Similarly P and A will lie on the same side of x = 0 if sin θ > 0 and P and O will lie on the same side of x + y = √(3/2)  if sin θ + cosθ √(3/2) .

Hence P will lie inside the Δ ABC, if sinθ > 0, cosθ > 0 and sinθ + cosθ √(3/2) .

Now sinθ + cosθ √(3/2)

⇒ sin ( θ + π /4)

i.e. 0

or, 2π/3

Since sinθ >0 and cos θ > 0, so

0

Regards

Arun (askIITians forum expert)