The complex number $(3 \operatorname{cis} 18^\circ)(-2\operatorname{cis} 37^\circ)$ is expressed in polar form as $r \operatorname{cis} \theta,$ where $r > 0$ and $0^\circ \le \theta < 360^\circ.$  Enter the ordered pair $(r, \theta).$
We can write
\[(3 \operatorname{cis} 18^\circ)(-2\operatorname{cis} 37^\circ) = (3)(-2) \operatorname{cis}(18^\circ + 37^\circ) = -6 \operatorname{cis} 55^\circ.\]Since we want $r > 0,$ we can write $-6 \operatorname{cis} 55^\circ = 6 \operatorname{cis} (55^\circ + 180^\circ) = 6 \operatorname{cis} 235^\circ.$  Hence, $(r,\theta) = \boxed{(6,235^\circ)}.$
(6,235)
Level 4
Precalculus
Problem: The complex number $(3 \operatorname{cis} 18^\circ)(-2\operatorname{cis} 37^\circ)$ is expressed in polar form as $r \operatorname{cis} \theta,$ where $r > 0$ and $0^\circ \le \theta < 360^\circ.$  Enter the ordered pair $(r, \theta).$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the product of the complex numbers in polar form, we first identify the magnitudes and angles of each complex number. The first complex number is given as $3 \operatorname{cis} 18^\circ$, where the magnitude is 3 and the angle is 18 degrees. The second complex number is $-2 \operatorname{cis} 37^\circ$. The negative sign indicates a rotation of 180 degrees, so we can express it as $2 \operatorname{cis}(37^\circ + 180^\circ) = 2 \operatorname{cis} 217^\circ$. 

Next, we calculate the magnitude of the product. The magnitudes multiply as follows: 
3 (from the first complex number) multiplied by 2 (from the second complex number) gives us a magnitude of 6.

Now, we add the angles to find the angle of the product:
18 degrees (from the first complex number) plus 217 degrees (from the adjusted second complex number) equals 235 degrees.

Thus, we have derived the polar form of the product as having a magnitude of 6 and an angle of 235 degrees. The ordered pair representing the polar form is therefore $(6, 235)$.<|END_OF_LATENT|> The final answer is $(6,235)$.