the angle theta is in the second quadrant and cos theta = -2/√29 determine possible coordinates for point P on the terminal arm of theta a. (2,5) b. (-2,√29) c. (-5,2) d. (-2,5)

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.

the answer is a.

about 233,333 ft3

hope you got it