Answer
The coordinates of P is $(1,\sqrt 3)$
Work Step by Step
Draw a perpendicular line from P to x-axis. It cuts the x-axis at Q. O is the origin.
OQP is a right triangle, right-angled at Q.
$\angle POQ = 60^{\circ} (Given)$
$OP = 2 (Given)$
$\sin\theta = \frac{Opposite Side}{Hypotenuse}$
Using trignometric ratio for $60^{\circ}$
$\sin60^{\circ} = \frac{PQ}{OP}$
Substituting the values for OP and $sin60^{\circ} = \frac{\sqrt 3}{2} $ [Trigonometric Function Value for $sin60^{\circ}$ ]
$\frac{\sqrt 3}{2} = \frac{PQ}{2}$
$PQ = \sqrt3$
By Pythagorean Theorem,
$OP^{2} = OQ^{2}+PQ^{2}$
$2^{2}= OQ^{2}+(\sqrt 3)^{2}$
$4= OQ^{2}+3$
$ OQ^{2}=4-3$
$ OQ^{2}=1$
$ OQ=1$
The coordinates of P =P(OQ,PQ) = $P(1,\sqrt 3)$
