Math 2 Quiz: Right Triangle Trig Applications
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Right Triangle Trig ApplicationsQuestion 1 of 13

A ladder leans against a building. When the bottom of the ladder is 8 feet from the building, the ladder makes a 70° angle with the ground. If the ladder is pulled out so that it makes a 55° angle with the ground, how much farther from the building is the bottom of the ladder?

2.1 feet
3.8 feet
5.6 feet
7.3 feet
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Math 2 Quiz

Math 2 Quiz: Right Triangle Trig Applications

Practice Right Triangle Trig Applications in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Right Triangle Trig Applications, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A ladder leans against a building. When the bottom of the ladder is 8 feet from the building, the ladder makes a 70° angle with the ground. If the ladder is pulled out so that it makes a 55° angle with the ground, how much farther from the building is the bottom of the ladder?

  1. 2.1 feet
  2. 3.8 feet
  3. 5.6 feet (correct answer)
  4. 7.3 feet
Explanation: The ladder length stays constant. Initially: L=8cos⁡(70°)≈80.342≈23.4L = \frac{8}{\cos(70°)} ≈ \frac{8}{0.342} ≈ 23.4 feet. When at 55° angle: new distance from building = Lcos⁡(55°)≈23.4×0.574≈13.4L \cos(55°) ≈ 23.4 \times 0.574 ≈ 13.4 feet. The ladder moved 13.4−8=5.413.4 - 8 = 5.4 feet farther, closest to 5.6 feet. Choice A might result from calculation errors, B from using wrong trigonometric ratios, and D from using the new distance instead of the difference.

Question 2

A rescue helicopter is 300 feet directly above a stranded hiker. The pilot spots the rescue team at an angle of depression of 28° from the helicopter's position. If the rescue team is at the same elevation as the hiker, what is the straight-line distance the rescue team must travel to reach the hiker?

  1. 465 feet
  2. 512 feet
  3. 564 feet (correct answer)
  4. 618 feet
Explanation: The horizontal distance from the helicopter to the rescue team is d=300tan⁡(28°)≈3000.5317≈564d = \frac{300}{\tan(28°)} ≈ \frac{300}{0.5317} ≈ 564 feet. Since the hiker is directly below the helicopter, this horizontal distance is also the straight-line distance the rescue team must travel to reach the hiker. Choice A might result from using sin⁡(28°)\sin(28°), choice B from calculation errors, and choice D from using a slightly different angle or approach.

Question 3

A cable car travels up a mountain slope at a constant incline. The horizontal distance covered is 2400 meters, and the car rises 900 meters vertically. If a passenger drops a ball at the midpoint of the journey, and the ball rolls straight down the slope, what is the angle at which the ball rolls relative to the horizontal?

  1. 20.6° below horizontal (correct answer)
  2. 69.4° below horizontal
  3. 20.6° above horizontal
  4. 69.4° above horizontal
Explanation: The slope of the cable car track has rise = 900m and run = 2400m. The angle of incline is θ=arctan⁡(9002400)=arctan⁡(0.375)≈20.6°\theta = \arctan(\frac{900}{2400}) = \arctan(0.375) ≈ 20.6°. When the ball rolls down the slope, it moves at the same angle as the track, but downward relative to horizontal, so 20.6° below horizontal. Choice B uses the complement (90° - 20.6° = 69.4°). Choices C and D incorrectly suggest the ball rolls upward.

Question 4

A cable is stretched from the top of a 150-foot radio antenna to a point on the ground 200 feet from the base of the antenna. Due to wind, the cable makes a 3° angle with its straight-line path. What is the horizontal displacement of the attachment point on the antenna from its intended position?

  1. 11.8 feet
  2. 15.8 feet
  3. 14.5 feet
  4. 13.1 feet (correct answer)
Explanation: When you encounter a problem involving wind displacement or deviation from a straight path, you're working with vector components and trigonometry. The key insight is that the wind causes the cable to bow sideways, creating a horizontal displacement from where it should be. First, find the cable's intended length using the Pythagorean theorem: 1502+2002=62500=250\sqrt{150^2 + 200^2} = \sqrt{62500} = 250 feet. This would be the straight-line distance without wind. The wind causes the cable to deviate 3° from this straight path. Since the cable maintains the same length (250 feet), the horizontal displacement equals the horizontal component of this deviation. Using trigonometry, the displacement is: 250×sin⁡(3°)=250×0.0523=13.1250 \times \sin(3°) = 250 \times 0.0523 = 13.1 feet. Choice A (11.8 feet) likely results from using 150×sin⁡(3°)150 \times \sin(3°) instead of the full cable length. Choice B (15.8 feet) might come from incorrectly using 200×sin⁡(5°)200 \times \sin(5°) or similar calculation errors. Choice C (14.5 feet) could result from using an approximation like 250×0.058250 \times 0.058 instead of the precise sine value. The correct answer is D (13.1 feet) because it properly applies the sine function to the full cable length and the given deviation angle. Remember: in wind displacement problems, always identify the total length of the deflected object first, then use trigonometric functions with the deflection angle. The displacement is typically the sine component when measuring perpendicular deviation from the intended path.

Question 5

A loading ramp is built to reach a platform that is 4.5 feet high. Safety regulations require that the ramp's angle with the horizontal not exceed 12°. What is the minimum horizontal length the ramp must span?

  1. 18.2 feet
  2. 21.2 feet (correct answer)
  3. 24.8 feet
  4. 28.1 feet
Explanation: Using the angle constraint: tan⁡(12°)=4.5horizontal length\tan(12°) = \frac{4.5}{\text{horizontal length}}. Therefore, minimum horizontal length = 4.5tan⁡(12°)≈4.50.213≈21.1\frac{4.5}{\tan(12°)} ≈ \frac{4.5}{0.213} ≈ 21.1 feet. This matches choice B. Choice A might result from using sin⁡(12°)\sin(12°) instead of tan⁡(12°)\tan(12°), choice C from using a different angle, and choice D from calculation errors or using the ramp length instead of horizontal span.

Question 6

A surveyor needs to find the distance across a lake. She measures a baseline of 200 meters along the shore, then measures angles from each end of the baseline to a landmark on the opposite shore. From one end, the angle is 78°78° from the baseline direction, and from the other end, the angle is 65°65° from the baseline direction. What is the distance from the first measurement point to the landmark?

  1. 200sin⁡(65°)sin⁡(37°)\frac{200 \sin(65°)}{\sin(37°)} meters (correct answer)
  2. 200sin⁡(78°)sin⁡(37°)\frac{200 \sin(78°)}{\sin(37°)} meters
  3. 200sin⁡(65°)sin⁡(143°)\frac{200 \sin(65°)}{\sin(143°)} meters
  4. 200sin⁡(78°)sin⁡(143°)\frac{200 \sin(78°)}{\sin(143°)} meters
Explanation: This forms a triangle where the baseline is 200m, and the angles at the ends are 78°78° and 65°65° from the baseline. The angle at the landmark is 180°−78°−65°=37°180° - 78° - 65° = 37°. Using the Law of Sines: dsin⁡(65°)=200sin⁡(37°)\frac{d}{\sin(65°)} = \frac{200}{\sin(37°)}, where dd is the distance from the first point to the landmark. Therefore d=200sin⁡(65°)sin⁡(37°)d = \frac{200 \sin(65°)}{\sin(37°)}. Choice B uses the wrong angle (78° instead of 65°), while choices C and D incorrectly calculate the third angle as 143° instead of 37°.

Question 7

A photographer wants to capture a building in a single frame. The building is 180 feet tall, and the photographer's camera is 6 feet above ground level. If the camera's field of view has a vertical angle of 28°28°, what is the minimum horizontal distance the photographer must be from the building to capture it entirely?

  1. 1742tan⁡(14°)\frac{174}{2\tan(14°)} feet
  2. 1802tan⁡(14°)\frac{180}{2\tan(14°)} feet
  3. 174tan⁡(28°)\frac{174}{\tan(28°)} feet
  4. 87tan⁡(14°)\frac{87}{\tan(14°)} feet (correct answer)
Explanation: The camera at 6 feet height needs to capture from ground level (6 feet below) to the top of the 180-foot building (174 feet above camera level). The total vertical span is 174 feet. For optimal framing, the camera should point toward the middle of this span, which is 87 feet above the camera level. With a 28° field of view, the camera can see ±14° from its pointing direction. The distance required is d=87tan⁡(14°)d = \frac{87}{\tan(14°)} feet. Choice A divides by 2 unnecessarily, choice B uses the wrong height measurement, and choice C uses the full 28° angle instead of the half-angle.

Question 8

A mountain climber at base camp observes the summit at an angle of elevation of 28°28°. After hiking 2000 meters horizontally toward the mountain, the angle of elevation increases to 35°35°. What is the height of the summit above base camp?

  1. 2000tan⁡(28°)tan⁡(35°)tan⁡(35°)+tan⁡(28°)\frac{2000 \tan(28°) \tan(35°)}{\tan(35°) + \tan(28°)} meters
  2. 2000tan⁡(28°)tan⁡(35°)tan⁡(35°)−tan⁡(28°)\frac{2000 \tan(28°) \tan(35°)}{\tan(35°) - \tan(28°)} meters (correct answer)
  3. 2000tan⁡(35°)−2000tan⁡(28°)2000 \tan(35°) - 2000 \tan(28°) meters
  4. 2000(tan⁡(35°)−tan⁡(28°))tan⁡(35°)tan⁡(28°)\frac{2000(\tan(35°) - \tan(28°))}{\tan(35°) \tan(28°)} meters
Explanation: Let hh be the height and dd be the initial horizontal distance to the mountain base. From the first observation: tan⁡(28°)=hd\tan(28°) = \frac{h}{d}, so d=htan⁡(28°)d = \frac{h}{\tan(28°)}. From the second observation: tan⁡(35°)=hd−2000\tan(35°) = \frac{h}{d-2000}, so d−2000=htan⁡(35°)d - 2000 = \frac{h}{\tan(35°)}. Substituting the first equation into the second: htan⁡(28°)−2000=htan⁡(35°)\frac{h}{\tan(28°)} - 2000 = \frac{h}{\tan(35°)}. Rearranging: htan⁡(28°)−htan⁡(35°)=2000\frac{h}{\tan(28°)} - \frac{h}{\tan(35°)} = 2000, so h(1tan⁡(28°)−1tan⁡(35°))=2000h\left(\frac{1}{\tan(28°)} - \frac{1}{\tan(35°)}\right) = 2000. This gives h(tan⁡(35°)−tan⁡(28°)tan⁡(28°)tan⁡(35°))=2000h\left(\frac{\tan(35°) - \tan(28°)}{\tan(28°)\tan(35°)}\right) = 2000, so h=2000tan⁡(28°)tan⁡(35°)tan⁡(35°)−tan⁡(28°)h = \frac{2000\tan(28°)\tan(35°)}{\tan(35°) - \tan(28°)}. Choice A has the wrong sign in the denominator, choice C assumes the height difference equals the change in horizontal distance times the tangent functions (incorrect), and choice D has the tangent terms inverted in the fraction.

Question 9

A radio antenna on top of a building casts a shadow. From the end of the antenna's shadow, the angle of elevation to the top of the antenna is 52°52° and to the top of the building (base of antenna) is 37°37°. If the building is 80 feet tall, what is the length of the antenna?

  1. 80(tan⁡(52°)−tan⁡(37°))80(\tan(52°) - \tan(37°)) feet
  2. 80tan⁡(52°)tan⁡(37°)−80\frac{80\tan(52°)}{\tan(37°)} - 80 feet
  3. 80(tan⁡(52°)−tan⁡(37°))tan⁡(37°)\frac{80(\tan(52°) - \tan(37°))}{\tan(37°)} feet (correct answer)
  4. 80tan⁡(52°)−80tan⁡(37°)80\tan(52°) - 80\tan(37°) feet
Explanation: Let dd be the horizontal distance from the building base to the end of the shadow, and LL be the antenna length. From the angle to the building top: tan⁡(37°)=80d\tan(37°) = \frac{80}{d}, so d=80tan⁡(37°)d = \frac{80}{\tan(37°)}. From the angle to the antenna top: tan⁡(52°)=80+Ld\tan(52°) = \frac{80 + L}{d}. Substituting: tan⁡(52°)=80+L80/tan⁡(37°)=(80+L)tan⁡(37°)80\tan(52°) = \frac{80 + L}{80/\tan(37°)} = \frac{(80 + L)\tan(37°)}{80}. Solving: 80tan⁡(52°)=(80+L)tan⁡(37°)=80tan⁡(37°)+Ltan⁡(37°)80\tan(52°) = (80 + L)\tan(37°) = 80\tan(37°) + L\tan(37°). Therefore: Ltan⁡(37°)=80tan⁡(52°)−80tan⁡(37°)=80(tan⁡(52°)−tan⁡(37°))L\tan(37°) = 80\tan(52°) - 80\tan(37°) = 80(\tan(52°) - \tan(37°)), so L=80(tan⁡(52°)−tan⁡(37°))tan⁡(37°)L = \frac{80(\tan(52°) - \tan(37°))}{\tan(37°)}. Choice A omits the division by tan⁡(37°)\tan(37°), choice B uses an incorrect relationship, and choice D assumes the horizontal distance is 80 feet (confusing it with the building height).

Question 10

A security camera is mounted 20 feet high on a wall. The camera can tilt from 10°10° above horizontal to 40°40° below horizontal. A person 6 feet tall walks along a path parallel to the wall. What is the width of the zone where the person's head (6 feet above ground) can be monitored by the camera?

  1. 14tan⁡(10°)+14tan⁡(40°)\frac{14}{\tan(10°)} + \frac{14}{\tan(40°)} feet
  2. 20tan⁡(10°)+20tan⁡(40°)\frac{20}{\tan(10°)} + \frac{20}{\tan(40°)} feet
  3. 20(cot⁡(10°)+cot⁡(40°))20(\cot(10°) + \cot(40°)) feet
  4. 14(cot⁡(10°)+cot⁡(40°))14(\cot(10°) + \cot(40°)) feet (correct answer)
Explanation: The camera is at height 20 feet and the person's head is at height 6 feet, so the vertical distance is 20 - 6 = 14 feet. When the camera tilts 10° above horizontal, it can see the person's head at horizontal distance d1=14tan⁡(10°)=14cot⁡(10°)d_1 = \frac{14}{\tan(10°)} = 14\cot(10°) from the wall. When tilted 40° below horizontal, it can see the person's head at distance d2=14tan⁡(40°)=14cot⁡(40°)d_2 = \frac{14}{\tan(40°)} = 14\cot(40°). The total width of the monitored zone is d1+d2=14cot⁡(10°)+14cot⁡(40°)=14(cot⁡(10°)+cot⁡(40°))d_1 + d_2 = 14\cot(10°) + 14\cot(40°) = 14(\cot(10°) + \cot(40°)). Choice A uses the correct calculation but with tangent instead of cotangent, choice B incorrectly uses the full camera height instead of the height difference, and choice C uses the wrong height difference.

Question 11

A cell tower is located on a hill 150 feet above the surrounding terrain. From a point on the ground level, the angle of elevation to the top of the tower is 32°32°, and the angle of elevation to the base of the tower is 18°18°. What is the height of the tower itself?

  1. 150tan⁡(32°)−150tan⁡(18°)150 \tan(32°) - 150 \tan(18°) feet
  2. 150(tan⁡(32°)−tan⁡(18°))tan⁡(18°)\frac{150(\tan(32°) - \tan(18°))}{\tan(18°)} feet (correct answer)
  3. 150tan⁡(32°)tan⁡(18°)−150\frac{150 \tan(32°)}{\tan(18°)} - 150 feet
  4. 150(tan⁡(32°)−tan⁡(18°))150(\tan(32°) - \tan(18°)) feet
Explanation: Let dd be the horizontal distance from the observation point to the base of the hill. From the angle to the base: tan⁡(18°)=150d\tan(18°) = \frac{150}{d}, so d=150tan⁡(18°)d = \frac{150}{\tan(18°)}. From the angle to the top: tan⁡(32°)=150+hd\tan(32°) = \frac{150 + h}{d} where hh is the tower height. Substituting: tan⁡(32°)=150+h150/tan⁡(18°)=(150+h)tan⁡(18°)150\tan(32°) = \frac{150 + h}{150/\tan(18°)} = \frac{(150 + h)\tan(18°)}{150}. Solving: 150tan⁡(32°)=(150+h)tan⁡(18°)150\tan(32°) = (150 + h)\tan(18°), so h=150tan⁡(32°)tan⁡(18°)−150=150(tan⁡(32°)−tan⁡(18°))tan⁡(18°)h = \frac{150\tan(32°)}{\tan(18°)} - 150 = \frac{150(\tan(32°) - \tan(18°))}{\tan(18°)}. Choice A assumes the horizontal distances are the same as the height (150), choice C has the correct calculation but written differently, and choice D incorrectly multiplies by 150 instead of dividing by tan⁡(18°)\tan(18°).

Question 12

A zip line runs from a platform 45 feet high to a landing area. The zip line makes a 15°15° angle below horizontal. Due to safety regulations, the zip line must maintain at least 12 feet of clearance above the ground at all points. What is the maximum horizontal distance the landing area can be from the platform?

  1. 33tan⁡(15°)\frac{33}{\tan(15°)} feet (correct answer)
  2. 45tan⁡(15°)\frac{45}{\tan(15°)} feet
  3. 45−12tan⁡(15°)\frac{45 - 12}{\tan(15°)} feet
  4. 45sin⁡(15°)\frac{45}{\sin(15°)} feet
Explanation: The zip line starts at 45 feet and descends at a 15° angle below horizontal. It must stay at least 12 feet above ground, so it can descend at most 45 - 12 = 33 feet vertically. Using trigonometry with the 15° descent angle: tan⁡(15°)=vertical drophorizontal distance=33d\tan(15°) = \frac{\text{vertical drop}}{\text{horizontal distance}} = \frac{33}{d}. Therefore d=33tan⁡(15°)d = \frac{33}{\tan(15°)}. Choice B ignores the 12-foot clearance requirement, choice C has the right numbers but written as a single fraction, and choice D uses sine instead of tangent, which would give the zip line length, not horizontal distance.

Question 13

A rescue helicopter hovers 400 feet directly above a stranded hiker. The helicopter pilot spots a clearing that could serve as a landing site. The angle of depression from the helicopter to the clearing is 18°18°. After landing and walking in a straight line toward the hiker, the rescue team measures the angle of elevation to the hiker's position as 31°31°. How far did the rescue team walk from the clearing to reach the hiker?

  1. 400tan⁡(18°)−400tan⁡(31°)\frac{400}{\tan(18°)} - \frac{400}{\tan(31°)} feet
  2. 400(tan⁡(31°)−tan⁡(18°))tan⁡(18°)tan⁡(31°)\frac{400(\tan(31°) - \tan(18°))}{\tan(18°)\tan(31°)} feet
  3. 400(cot⁡(18°)−cot⁡(31°))400(\cot(18°) - \cot(31°)) feet (correct answer)
  4. 400tan⁡(31°)−400tan⁡(18°)\frac{400}{\tan(31°)} - \frac{400}{\tan(18°)} feet
Explanation: When you encounter angle of depression and elevation problems, you're working with two connected right triangles that share a common vertical height. The key insight is recognizing how these triangles relate to each other geometrically. Let's set up the problem systematically. The helicopter is 400 feet above the hiker, creating a vertical line. From the helicopter, the angle of depression to the clearing is 18°, which means the horizontal distance from the hiker to the clearing is 400tan⁡(18°)\frac{400}{\tan(18°)} (using the relationship: horizontal distance = height ÷ tan(angle)). When the rescue team reaches the hiker's base and looks up at a 31° angle of elevation, the horizontal distance from this ground position to directly below the helicopter is 400tan⁡(31°)\frac{400}{\tan(31°)}. The distance the rescue team walked is the difference between these two horizontal distances: 400tan⁡(18°)−400tan⁡(31°)\frac{400}{\tan(18°)} - \frac{400}{\tan(31°)}. Factoring out 400 gives us 400(1tan⁡(18°)−1tan⁡(31°))400\left(\frac{1}{\tan(18°)} - \frac{1}{\tan(31°)}\right), which equals 400(cot⁡(18°)−cot⁡(31°))400(\cot(18°) - \cot(31°)) since cotangent is the reciprocal of tangent. Choice A has the terms in the right order but isn't factored. Choice B incorrectly manipulates the algebraic expression. Choice D reverses the subtraction order, which would give a negative distance since tan⁡(31°)>tan⁡(18°)\tan(31°) > \tan(18°). The correct answer is C. Strategy tip: In angle problems involving two different viewpoints, always draw a diagram showing both triangles and identify what distances you need to find versus subtract.