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Let $0 < \theta < \pi$.

The area of the triangle in the plane formed by the vertices $(-1,0), (1,0), (\cos\theta, \sin\theta)$ is 

  1. not more than $1$
  2. can be more than $1$ but not more than $2$
  3. can be more than $2$ but not more than $\pi$
  4. can be more than $\pi$ but not more than $2\pi$
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Two coordinates of triangle lies on the x-axis, $(-1,0)$ and $(1,0)$.

so base of the triangle $= 2 $

height of triangle will be $=\sin \theta$

so area will be $\frac{1}{2} \times 2 \times \sin \theta = \sin \theta$

for $0<\theta<\pi$, maximum value of $\sin \theta$ will be $1$ only at $\theta = 90^{\circ}$.

Area of given triangle can be maximum $1$

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