1. The problem states the inequality: $$4 \tan t + 5 \sin \theta \geq 20.$$ 2. To understand this inequality, note that $$\tan t$$ and $$\sin \theta$$ are trigonometric functions with certain ranges. Specifically, $$\sin \theta$$ is bounded between $$-1$$ and $$1$$. However, $$\tan t$$ can take any real value except where it is undefined (odd multiples of $$\frac{\pi}{2}$$). 3. Rearranging the inequality, we get $$4 \tan t \geq 20 - 5 \sin \theta.$$ 4. For a given $$\theta$$, since $$\sin \theta \leq 1$$, the smallest right side is $$20 - 5 \times 1 = 15$$. For $$\sin \theta = -1$$, the right side becomes $$20 - (-5) = 25$$. 5. Therefore, $$4 \tan t \geq $$ a value between $$15$$ and $$25$$ depending on $$\sin \theta$$. Dividing both sides by 4, $$\tan t \geq \frac{20 - 5 \sin \theta}{4}.$$ 6. The inequality has solutions where $$\tan t$$ is sufficiently large depending on $$\sin \theta$$. Since $$\tan t$$ ranges over all real numbers, $$t$$ must be chosen so $$\tan t$$ satisfies this bound, considering the periodicity and undefined points. 7. Without extra constraints on $$t$$ or $$\theta$$, this inequality represents the region where the sum of $$4 \tan t$$ and $$5 \sin \theta$$ is at least $$20$$. Final answer: Values of $$t$$ and $$\theta$$ must satisfy $$4 \tan t + 5 \sin \theta \geq 20$$, equivalently $$\tan t \geq \frac{20 - 5 \sin \theta}{4}$$.