A triangle was drawn on a Cartesian plane such that there exists a point for which the distance is equal to 4. And two of the vertices of this triangle have coordinates and .
Given that the sine of one of the interior angles of this triangle must always be a constant. Find this constant.
Give your answer to 3 decimal places.
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Point O appears to be the circumcentre, and the triangle P Q R has circumradius 4. Applying the sine rule, we find that sin P p = 2 r ⇒ sin P 5 = 8 ⇒ sin P = 8 5 = 0 . 6 2 5