This interval represents Quadrant III. At $$\theta=\pi$$, the point on the unit circle is $$(-1, 0)$$, so $$\cos\pi=-1$$ and $$\sin\pi=0$$. At $$\theta=\frac{3\pi}{2}$$, the point is $$(0, -1)$$, so $$\cos(\frac{3\pi}{2})=0$$ and $$\sin(\frac{3\pi}{2})=-1$$. As $$\theta$$ goes from $$\pi$$ to $$\frac{3\pi}{2}$$, the y-coordinate ($$\sin\theta$$) goes from 0 to -1 (a decrease), and the x-coordinate ($$\cos\theta$$) goes from -1 to 0 (an increase).

The equation of the circle is $$x^2 + y^2 = 4^2 = 16$$. Given $$y=-2$$, we have $$x^2 + (-2)^2 = 16$$, so $$x^2 + 4 = 16$$, and $$x^2 = 12$$. Thus, $$x = \pm\sqrt{12} = \pm 2\sqrt{3}$$. Since the point is in Quadrant III, the x-coordinate must be negative, so $$x = -2\sqrt{3}$$. The value of $$\cos\theta$$ is $$\frac{x}{r} = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}$$.

By convention, a negative angle measure indicates a clockwise rotation from the initial side (the positive x-axis). The magnitude of the rotation is $$300^\circ$$. Therefore, an angle of $$-300^\circ$$ is formed by a clockwise rotation of $$300^\circ$$.

This question tests AP Precalculus skills involving trigonometric functions (sine, cosine, and tangent) in applied contexts. Trigonometric functions relate angles to side lengths in right triangles, useful for real-world applications involving heights, distances, and angles. In this scenario, the ladder forms a right triangle where the ladder is the hypotenuse, the wall height (4.2 m) is opposite to the ground angle (60°), and we need to find the ladder length. Choice C is correct because sin(60°) = opposite/hypotenuse = 4.2/ladder length, so ladder length = 4.2/sin(60°) ≈ 4.8 m. Choice B incorrectly uses cosine, which would relate the adjacent side (ground distance) to the hypotenuse, not the opposite side (wall height). To help students: Draw the triangle clearly labeling all parts, identify which trigonometric function relates the known side to the unknown side, and practice solving for the hypotenuse. Watch for: confusion between sine and cosine based on which sides are given, and errors in algebraic manipulation.

The line $$y=-x$$ with $$x>0$$ lies in Quadrant IV. This line makes a $$45^\circ$$ angle with the negative y-axis and the positive x-axis. The angle in standard position is $$315^\circ$$ or $$\frac{7\pi}{4}$$ radians. A point on this ray could be $$(1, -1)$$. The distance to the origin is $$r = \sqrt{1^2 + (-1)^2} = \sqrt{2}$$. Therefore, $$\sin\theta = \frac{y}{r} = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$$.

The angle $$\theta$$ is in Quadrant III, where both sine and cosine are negative. Using the Pythagorean identity $$\sin^2\theta + \cos^2\theta = 1$$, we have $$(-\frac{5}{13})^2 + \cos^2\theta = 1$$, which gives $$\frac{25}{169} + \cos^2\theta = 1$$. Thus, $$\cos^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$$, so $$\cos\theta = \pm\frac{12}{13}$$. Since $$\theta$$ is in Quadrant III, $$\cos\theta = -\frac{12}{13}$$. Then, $$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-5/13}{-12/13} = \frac{5}{12}$$.

This question tests AP Precalculus skills involving trigonometric functions (sine, cosine, and tangent) in applied contexts. Trigonometric functions relate angles to side lengths in right triangles, useful for real-world applications involving heights, distances, and angles. In this scenario, the sun's angle of elevation creates a right triangle where the flagpole is the opposite side and the shadow is the adjacent side to the angle. Choice B is correct because tan(40°) = opposite/adjacent = 6.5/shadow length, so shadow length = 6.5/tan(40°) ≈ 7.7 m. Choice A incorrectly multiplies instead of dividing, which would give the height if we knew the shadow length, not the shadow length from the height. To help students: Emphasize the relationship between the sun's angle and the resulting shadow, practice setting up the tangent ratio correctly, and reinforce when to multiply versus divide. Watch for: confusion about which quantity is known versus unknown, and errors in algebraic manipulation of the tangent equation.

This question tests AP Precalculus skills involving trigonometric functions (sine, cosine, and tangent) in applied contexts. Trigonometric functions relate angles to side lengths in right triangles, useful for real-world applications involving heights, distances, and angles. In this scenario, a 12.0 m light pole casts a shadow when the sun's angle of elevation is 30°, requiring the tangent function to find shadow length. Choice B is correct because it applies the tangent function correctly: tan(30°) = opposite/adjacent = 12.0/shadow length, so shadow length = 12.0/tan(30°) ≈ 12.0/0.5774 ≈ 20.8 m. Choice A is incorrect because it appears to multiply 12.0 by tan(30°) instead of dividing, giving a much smaller value. To help students: Emphasize drawing clear diagrams with the sun's rays, vertical pole, and horizontal shadow, practice setting up ratios correctly, and remember that lower sun angles create longer shadows. Watch for: confusion about which way to set up the division, and forgetting that tan(30°) = 1/√3 ≈ 0.577.

This question tests AP Precalculus skills involving trigonometric functions (sine, cosine, and tangent) in applied contexts. Trigonometric functions relate angles to side lengths in right triangles, useful for real-world applications involving heights, distances, and angles. In this scenario, a ladder leaning at a 70° angle with the ground must reach a 6.0 m height, and we need to find the ladder length using the sine function. Choice B is correct because it applies the sine function correctly: sin(70°) = opposite/hypotenuse = 6.0/ladder length, so ladder length = 6.0/sin(70°) ≈ 6.0/0.9397 ≈ 6.4 m. Choice A is incorrect because it appears to multiply 6.0 by sin(70°) instead of dividing, giving a much smaller value. To help students: Emphasize identifying which side is the hypotenuse (the ladder), practice recognizing when to use sine versus other functions, and always check that answers make physical sense. Watch for: confusion about which angle to use and mixing up multiplication versus division when solving for the unknown.

This question tests AP Precalculus skills involving trigonometric functions (sine, cosine, and tangent) in applied contexts. Trigonometric functions relate angles to side lengths in right triangles, useful for real-world applications involving heights, distances, and angles. In this scenario, the angle of depression from a ship to an object below creates a right triangle where the horizontal distance is adjacent and the depth is opposite to the angle. Choice A is correct because it applies the tangent function correctly: depth = horizontal distance × tan(angle of depression) = 250 × tan(12°) ≈ 53 m. Choice D is incorrect because it uses 12 radians instead of 12 degrees, which would give an entirely different result. To help students: Emphasize that angles of depression are measured from the horizontal downward, practice drawing diagrams to visualize the problem setup, and reinforce the importance of using degree mode on calculators. Watch for: confusion between angles of elevation and depression, and mixing up degree and radian measures.