Sine, Cosine, and Tangent
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AP Precalculus › Sine, Cosine, and Tangent
A 6.5 m flagpole casts a shadow when the sun’s angle of elevation is 40°. Based on the scenario described, what is the shadow length (nearest tenth)?
$\dfrac{6.5}{\tan(50^\circ)}\approx 5.5\text{ m}$
$6.5\tan(40^\circ)\approx 5.5\text{ m}$
$\dfrac{6.5}{\tan(40^\circ)}\approx 7.7\text{ m}$
$\dfrac{6.5}{\tan(40)}\approx 1.7\text{ m}$
Explanation
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.
A 12.0 m light pole casts a shadow when the sun’s angle of elevation is 30°. Based on the scenario described, what is the shadow length (nearest tenth meter) using $\tan$?
$24.0\text{ m}$
$10.4\text{ m}$
$6.9\text{ m}$
$20.8\text{ m}$
Explanation
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.
A worker needs a ladder to reach a 6.0 m height, leaning at a 70° angle with the ground. Based on the scenario described, what ladder length is required (nearest tenth) using $\sin$?
$5.6\text{ m}$
$17.5\text{ m}$
$2.1\text{ m}$
$6.4\text{ m}$
Explanation
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.
A ship detects an object at a 12° angle of depression, with 250 m horizontal distance. Based on the scenario described, what is the object’s depth (nearest meter)?
$250\tan(78^\circ)\approx 1176\text{ m}$
$250\tan(12^\circ)\approx 53\text{ m}$
$250\tan(12)\approx 160\text{ m}$
$250\cos(12^\circ)\approx 244\text{ m}$
Explanation
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.
A projectile is launched at 50 m/s at a 40° angle above horizontal. Based on the scenario described, what is its initial vertical speed (nearest tenth)?
$50\cos(40^\circ)\approx 38.3\text{ m/s}$
$50\sin(40)\approx 37.3\text{ m/s}$
$50\sin(40^\circ)\approx 32.1\text{ m/s}$
$50\sin(50^\circ)\approx 38.3\text{ m/s}$
Explanation
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 projectile's velocity can be decomposed into components, where the vertical component is opposite to the launch angle. Choice B is correct because the vertical speed component equals the total speed multiplied by sin(40°): vertical speed = 50 × sin(40°) ≈ 32.1 m/s. Choice A incorrectly uses cosine, which would give the horizontal component instead of the vertical component. To help students: Use vector diagrams showing velocity decomposition, emphasize that sine gives the vertical (opposite) component while cosine gives the horizontal (adjacent) component, and practice identifying which component is requested. Watch for: confusion between horizontal and vertical components, and mixing up sine and cosine for different velocity components.
A ladder must reach 4.2 m up a wall while making a 60° angle with the ground. Based on the scenario described, what ladder length is needed (nearest tenth)?
$\dfrac{4.2}{\sin(30^\circ)}=8.4\text{ m}$
$\dfrac{4.2}{\sin(60^\circ)}\approx 4.8\text{ m}$
$4.2\sin(60^\circ)\approx 3.6\text{ m}$
$\dfrac{4.2}{\cos(60^\circ)}=8.4\text{ m}$
Explanation
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.
A projectile is launched at 20 m/s at a 40° angle above horizontal on level ground. Based on the scenario described, what horizontal component of velocity results (nearest tenth) using $\cos$?
$17.3\text{ m/s}$
$15.3\text{ m/s}$
$30.6\text{ m/s}$
$12.9\text{ m/s}$
Explanation
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 velocity components in projectile motion. In this scenario, a projectile launched at 20 m/s at a 40° angle requires finding the horizontal velocity component using the cosine function. Choice A is correct because it applies the cosine function correctly: horizontal component = initial velocity × cos(angle) = 20 × cos(40°) ≈ 20 × 0.7660 ≈ 15.3 m/s. Choice B is incorrect because it appears to use sine instead of cosine, giving the vertical component rather than horizontal. To help students: Emphasize understanding velocity vector decomposition, practice identifying when to use cosine (for horizontal) versus sine (for vertical) components, and reinforce the connection to right triangle trigonometry. Watch for: confusion between horizontal and vertical components, and mixing up sine and cosine functions in projectile problems.
A projectile is launched at 30 m/s at a 25° angle above horizontal. Using the information provided in the passage, what vertical component of velocity results (nearest tenth) using $\sin$?
$27.2\text{ m/s}$
$11.7\text{ m/s}$
$12.7\text{ m/s}$
$63.9\text{ m/s}$
Explanation
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 decomposing velocity vectors in projectile motion problems. In this scenario, a projectile launched at 30 m/s at a 25° angle requires finding the vertical velocity component using the sine function. Choice A is correct because it applies the sine function correctly: vertical component = initial velocity × sin(angle) = 30 × sin(25°) ≈ 30 × 0.4226 ≈ 12.7 m/s. Choice D is incorrect because it appears to use a slightly different calculation or rounding, possibly from calculator error or using the wrong function. To help students: Emphasize understanding velocity vector decomposition, practice using sine for vertical components and cosine for horizontal, and reinforce the physical meaning of these components. Watch for: confusion between sine and cosine usage, and errors in calculator degree mode settings.
A student is 25 m from a tower and measures a 48° angle of elevation to its top. Based on the scenario described, what is the tower’s height (nearest tenth)?
$25\tan(42^\circ)\approx 22.5\text{ m}$
$25\tan(48)\approx 28.9\text{ m}$
$25\tan(48^\circ)\approx 27.8\text{ m}$
$25\sin(48^\circ)\approx 18.6\text{ m}$
Explanation
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 elevation from the student to the tower top creates a right triangle where the distance is adjacent and the height is opposite to the angle. Choice A is correct because it applies the tangent function correctly: height = distance × tan(angle) = 25 × tan(48°) ≈ 27.8 m. Choice D is incorrect because it uses 48 radians instead of 48 degrees, which would result in a different calculation entirely. To help students: Draw clear diagrams showing the angle of elevation, identify the known and unknown sides relative to the angle, and ensure calculator mode matches the angle units given. Watch for: confusion between degrees and radians, and errors in identifying which trigonometric function relates the given information.