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$A (2, 0)$ is a fixed point on the plane while $P (sin (2 θ - 60^{\circ}), cos (2 θ - 60^{\circ}))$ is moving point .The position of the point P at $θ = 15^{0}$ is B and the position of the point P at $θ = 45^{0}$ is C then the area of the triangle ABC is $\frac{\sqrt{3}}{k} s q u a r e u n i t s t h e n t h e v a l u e o f k i s$

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Detailed Solution

A(2,0),

$θ = 15 ° \Rightarrow B (sin (2 θ - 60^{\circ}), cos (2 θ - 60^{\circ})) = (\frac{- 1}{2}, \frac{\sqrt{3}}{2})$

$θ = 45 ° \Rightarrow C (sin (2 θ - 60^{\circ}), cos (2 θ - 60^{\circ})) = (\frac{1}{2}, \frac{\sqrt{3}}{2})$

Area of triangle ABC= $= \frac{\sqrt{3}}{4}$

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