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Can someone help me with a two step equatio that is x=2
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and $\theta$ lies in $2^{\text{th}}$ quadrant.
where, $x-$ coordinate is negative, and $y-$ coordinate is positive
so it can't a.
now, cosine means, side adjacent over the hypotenuse, in Cartesian plane, that will be $x-$ coordinate over the distance from origin.
Assume the triangle , with base $2$ units and hypotenuse $\sqrt{29}$ and it's in second quadrant. (so )
now, the leftmost point on $x-$ axis is , obviously $(-2,0)$
and by Pythagoras theorem, we can find the perpendicular side, that will be $y^2=(\sqrt{29})^2-(2)^2\implies y=5$
so the coordinates of the upper vertex is $(-2,5)$, each point lying on this "ray" should have equal ratio of respective coordinates. i.e. $\frac25=\left|\frac xy\right| $
and it should lie on second quadrant, so $x<0 \, y>0$
Option d satisfies this.