Let P (sinθ , cosθ) (0 ≤ θ ≤ 2π) be a point and let OAB be a triangle with vertices ….

Q: 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.

Sol. Equations of lines along OA, OB and AB are y = 0, x = 0, and $x + y = \sqrt{\frac{3}{2}}$ 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 $x + y = \sqrt{\frac{3}{2}}$ if $sin\theta + cos\theta < \sqrt{\frac{3}{2}}$ .

Hence P will lie inside the ΔABC, if sinθ > 0, cosθ > 0 and $sin\theta + cos\theta < \sqrt{\frac{3}{2}}$

Now $sin\theta + cos\theta < \sqrt{\frac{3}{2}}$

$\displaystyle sin(\theta + \pi/4) \sqrt{\frac{3}{2}}$

i.e. 0 < θ + π/4 < π/3

or, 2π/3 < θ + π/4 < π

Since sinθ >0 and cos θ > 0 , so

$\displaystyle 0 < \theta < \frac{\pi}{12}$

$\displaystyle \frac{5 \pi}{12} < \theta < \frac{\pi}{2}$