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Precalculus Quiz

Precalculus Quiz: Solving Right Triangles Pythagorean Theorem Trigonometry

Practice Solving Right Triangles Pythagorean Theorem Trigonometry in Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

A right triangle has legs of length aaa and bbb, with hypotenuse ccc. If one acute angle measures θ\thetaθ, and sin⁡(θ)=35\sin(\theta) = \frac{3}{5}sin(θ)=53​, which expression correctly represents cos⁡(90°−θ)\cos(90° - \theta)cos(90°−θ) in terms of the triangle's sides?

Select an answer to continue

What this quiz covers

This quiz focuses on Solving Right Triangles Pythagorean Theorem Trigonometry, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

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.

All questions

Question 1

A right triangle has legs of length aaa and bbb, with hypotenuse ccc. If one acute angle measures θ\thetaθ, and sin⁡(θ)=35\sin(\theta) = \frac{3}{5}sin(θ)=53​, which expression correctly represents cos⁡(90°−θ)\cos(90° - \theta)cos(90°−θ) in terms of the triangle's sides?

  1. cos⁡(90°−θ)=adjacent to θhypotenuse=45\cos(90° - \theta) = \frac{\text{adjacent to } \theta}{\text{hypotenuse}} = \frac{4}{5}cos(90°−θ)=hypotenuseadjacent to θ​=54​
  2. cos⁡(90°−θ)=sin⁡(θ)=opposite to θhypotenuse=35\cos(90° - \theta) = \sin(\theta) = \frac{\text{opposite to } \theta}{\text{hypotenuse}} = \frac{3}{5}cos(90°−θ)=sin(θ)=hypotenuseopposite to θ​=53​ (correct answer)
  3. cos⁡(90°−θ)=1sin⁡(θ)=hypotenuseopposite to θ=53\cos(90° - \theta) = \frac{1}{\sin(\theta)} = \frac{\text{hypotenuse}}{\text{opposite to } \theta} = \frac{5}{3}cos(90°−θ)=sin(θ)1​=opposite to θhypotenuse​=35​
  4. cos⁡(90°−θ)=tan⁡(θ)=opposite to θadjacent to θ=34\cos(90° - \theta) = \tan(\theta) = \frac{\text{opposite to } \theta}{\text{adjacent to } \theta} = \frac{3}{4}cos(90°−θ)=tan(θ)=adjacent to θopposite to θ​=43​

Explanation: For complementary angles, cos(90° - θ) = sin(θ). Since sin(θ) = 3/5, we have cos(90° - θ) = 3/5. This represents the ratio of the side opposite to θ over the hypotenuse. Choice A incorrectly applies cos(θ) = 4/5 instead of the complementary relationship. Choice C confuses the relationship with the reciprocal (cosecant). Choice D incorrectly uses the tangent ratio instead of recognizing the complementary angle relationship.

Question 2

Two complementary angles have measures (3x−15)°(3x - 15)°(3x−15)° and (2x+25)°(2x + 25)°(2x+25)°. If sin⁡(3x−15)°=0.6\sin(3x - 15)° = 0.6sin(3x−15)°=0.6, what is the value of cos⁡(2x+25)°\cos(2x + 25)°cos(2x+25)°?

  1. cos⁡(2x+25)°=0.6\cos(2x + 25)° = 0.6cos(2x+25)°=0.6 because complementary angles have equal sines and cosines
  2. cos⁡(2x+25)°=0.8\cos(2x + 25)° = 0.8cos(2x+25)°=0.8 because cos⁡2(2x+25)°+sin⁡2(2x+25)°=1\cos^2(2x + 25)° + \sin^2(2x + 25)° = 1cos2(2x+25)°+sin2(2x+25)°=1
  3. cos⁡(2x+25)°=0.6\cos(2x + 25)° = 0.6cos(2x+25)°=0.6 because sin⁡(α)=cos⁡(90°−α)\sin(\alpha) = \cos(90° - \alpha)sin(α)=cos(90°−α) for complementary angles (correct answer)
  4. cos⁡(2x+25)°=0.4\cos(2x + 25)° = 0.4cos(2x+25)°=0.4 because cos⁡(θ)=1−sin⁡(90°−θ)\cos(\theta) = 1 - \sin(90° - \theta)cos(θ)=1−sin(90°−θ) for complementary angles

Explanation: Since the angles are complementary: (3x - 15) + (2x + 25) = 90. Solving: 5x + 10 = 90, so 5x = 80, giving x = 16. The angles are (3×16 - 15)° = 33° and (2×16 + 25)° = 57°. For complementary angles α and β, sin(α) = cos(β) and cos(α) = sin(β). Since sin(33°) = 0.6, we have cos(57°) = cos(2x + 25)° = 0.6. Choice A gives the correct answer but with incorrect reasoning about equal sines and cosines. Choice B incorrectly applies the Pythagorean identity to find cos(57°) when sin(57°) ≠ 0.6. Choice D uses a non-existent trigonometric relationship.

Question 3

In right triangle △JKL\triangle JKL△JKL, ∠L=90∘\angle L = 90^\circ∠L=90∘, JL=8JL = 8JL=8 in, and KL=15KL = 15KL=15 in. Side JKJKJK is the hypotenuse. Based on the triangle described, what is the exact length of side JKJKJK?

  1. 171717 in (correct answer)
  2. 232323 in
  3. 161\sqrt{161}161​ in
  4. 289\sqrt{289}289​ in

Explanation: This question tests the ability to solve right triangles using the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse equals the sum of the squares of the two legs: a² + b² = c². This is an 8-15-17 Pythagorean triple, so we can recognize immediately that the missing side is 17 without calculation. Choice A is correct because it shows correct substitution into the Pythagorean theorem: √(8² + 15²) = √(64 + 225) = √289 = 17 in. Choice B makes the error of adding the sides instead of adding their squares: 8 + 15 = 23 instead of √(64 + 225) = 17. Key to right triangle problems: first identify the right angle and hypotenuse (longest side, opposite the right angle), then decide whether you have enough information for Pythagorean theorem (two sides known) or need trigonometry (one side and one angle known). Recognize common Pythagorean triples (3-4-5, 5-12-13, 8-15-17) and their multiples to save time on calculations—if you see two sides of a triple, the third can be determined without calculation.

Question 4

In right triangle △ABC\triangle ABC△ABC, ∠C=90∘\angle C = 90^\circ∠C=90∘. If AC=8AC = 8AC=8 and AB=17AB = 17AB=17 (hypotenuse), what is the exact value of sin⁡(∠A)\sin(\angle A)sin(∠A)?

  1. 817\frac{8}{17}178​
  2. 1517\frac{15}{17}1715​ (correct answer)
  3. 1715\frac{17}{15}1517​
  4. 158\frac{15}{8}815​

Explanation: This question tests the ability to solve right triangles using trigonometric ratios. Trigonometric ratios relate the angles of a right triangle to the ratios of its sides: sin(θ) = opposite/hypotenuse, cos(θ) = adjacent/hypotenuse, and tan(θ) = opposite/adjacent. From the perspective of angle A, the side BC is the opposite side, and we need to find BC first using the Pythagorean theorem: AC² + BC² = AB², so 8² + BC² = 17², giving 64 + BC² = 289, thus BC² = 225 and BC = 15. Choice B is correct because sin(A) = opposite/hypotenuse = BC/AB = 15/17. Choice A incorrectly uses the adjacent side instead of the opposite side, calculating cos(A) = AC/AB = 8/17 instead of sin(A) = 15/17. Remember the SOH-CAH-TOA mnemonic for choosing the correct trig ratio: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent, where opposite and adjacent are always relative to the angle in question.

Question 5

A ladder is 101010 ft long and leans against a vertical wall. The bottom of the ladder is 666 ft from the wall, forming a right triangle with the ground and the wall. What is the height (in feet) the ladder reaches up the wall?

  1. 444 ft
  2. 161616 ft
  3. 888 ft (correct answer)
  4. 136\sqrt{136}136​ ft

Explanation: This question tests the ability to solve right triangles using the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse equals the sum of the squares of the two legs: a² + b² = c². The ladder forms the hypotenuse (10 ft), the distance from the wall is one leg (6 ft), and we need to find the height up the wall, which is the other leg, so we use: 6² + height² = 10², which gives 36 + height² = 100, so height² = 64, and taking the square root yields height = 8 ft. Choice C is correct because when we substitute the given values into the rearranged Pythagorean theorem, we get height = √(10² - 6²) = √(100 - 36) = √64 = 8 ft. Choice A incorrectly subtracts the sides instead of using the Pythagorean theorem: 10 - 6 = 4 instead of √(10² - 6²) = 8. This is a 6-8-10 triangle (a scaled 3-4-5 triple), so we can recognize immediately that the missing side is 8 without calculation. Recognize common Pythagorean triples (3-4-5, 5-12-13, 8-15-17) and their multiples to save time on calculations—if you see two sides of a triple, the third can be determined without calculation.

Question 6

In right triangle ABCABCABC, the right angle is at CCC. The legs are AC=9AC=9AC=9 cm and BC=12BC=12BC=12 cm, and the hypotenuse is AB=cAB=cAB=c. What is the exact length of side ccc?

  1. 151515 cm (correct answer)
  2. 212121 cm
  3. 333 cm
  4. 63 \sqrt{63}63​ cm

Explanation: This question tests the ability to solve right triangles using the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse equals the sum of the squares of the two legs: a² + b² = c². Given sides of length 9 and 12, we substitute into the Pythagorean theorem: 9² + 12² = c², which gives 81 + 144 = c², so c² = 225, and taking the square root yields c = 15. Choice A is correct because when we substitute the given values into the Pythagorean theorem, we get c = √(9² + 12²) = √(81 + 144) = √225 = 15 cm. Choice B incorrectly adds the sides instead of adding their squares: 9 + 12 = 21 instead of √(9² + 12²) = 15. Key to right triangle problems: first identify the right angle and hypotenuse (longest side, opposite the right angle), then decide whether you have enough information for Pythagorean theorem (two sides known) or need trigonometry (one side and one angle known). Recognize common Pythagorean triples (3-4-5, 5-12-13, 8-15-17) and their multiples to save time on calculations—if you see two sides of a triple, the third can be determined without calculation.

Question 7

A right triangle has one leg of length 7 and hypotenuse of length 25. An angle θ\thetaθ in this triangle satisfies cos⁡(θ)=725\cos(\theta) = \frac{7}{25}cos(θ)=257​. What is the value of sin⁡(90°−θ)+cos⁡(90°−θ)\sin(90° - \theta) + \cos(90° - \theta)sin(90°−θ)+cos(90°−θ)?

  1. sin⁡(90°−θ)+cos⁡(90°−θ)=725+2425=3125\sin(90° - \theta) + \cos(90° - \theta) = \frac{7}{25} + \frac{24}{25} = \frac{31}{25}sin(90°−θ)+cos(90°−θ)=257​+2524​=2531​ (correct answer)
  2. sin⁡(90°−θ)+cos⁡(90°−θ)=2425+725=3125\sin(90° - \theta) + \cos(90° - \theta) = \frac{24}{25} + \frac{7}{25} = \frac{31}{25}sin(90°−θ)+cos(90°−θ)=2524​+257​=2531​
  3. sin⁡(90°−θ)+cos⁡(90°−θ)=2425+2425=4825\sin(90° - \theta) + \cos(90° - \theta) = \frac{24}{25} + \frac{24}{25} = \frac{48}{25}sin(90°−θ)+cos(90°−θ)=2524​+2524​=2548​
  4. sin⁡(90°−θ)+cos⁡(90°−θ)=725+725=1425\sin(90° - \theta) + \cos(90° - \theta) = \frac{7}{25} + \frac{7}{25} = \frac{14}{25}sin(90°−θ)+cos(90°−θ)=257​+257​=2514​

Explanation: First, find the other leg using the Pythagorean theorem: other leg = √(25² - 7²) = √(625 - 49) = √576 = 24. Given cos(θ) = 7/25, we can find sin(θ) = 24/25. Using complementary angle relationships: sin(90° - θ) = cos(θ) = 7/25 and cos(90° - θ) = sin(θ) = 24/25. Therefore, sin(90° - θ) + cos(90° - θ) = 7/25 + 24/25 = 31/25. Choice B reverses the order but gets the same sum. Choice C incorrectly uses sin(θ) for both terms. Choice D incorrectly uses cos(θ) for both terms.

Question 8

In right triangle △MNO\triangle MNO△MNO, ∠O=90∘\angle O = 90^\circ∠O=90∘, the hypotenuse is MN=12MN = 12MN=12 cm, and ∠M=30∘\angle M = 30^\circ∠M=30∘. Using the given information, what is the exact length of side NONONO (the side opposite ∠M\angle M∠M)?

  1. 666 cm (correct answer)
  2. 636\sqrt{3}63​ cm
  3. 12312\sqrt{3}123​ cm
  4. 434\sqrt{3}43​ cm

Explanation: This question tests the ability to solve right triangles using trigonometric ratios. Trigonometric ratios relate the angles of a right triangle to the ratios of its sides: sin(θ) = opposite/hypotenuse, cos(θ) = adjacent/hypotenuse, and tan(θ) = opposite/adjacent. From the perspective of angle M (30°), the side NO is opposite, and MN=12 is the hypotenuse, so we use sin(M) = NO/12; since sin(30°)=1/2, NO=12 × (1/2)=6. Choice A is correct because it uses the sine ratio for the 30° angle: 12 sin(30°) = 12 × (1/2) = 6 cm. Choice B confuses the 30-60-90 triangle ratios with 45-45-90 triangle ratios, using √3 where it's not needed for the side opposite 30°. Remember the SOH-CAH-TOA mnemonic for choosing the correct trig ratio: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent, where opposite and adjacent are always relative to the angle in question. For complementary angles in a right triangle, sin(θ) = cos(90° - θ), which explains why the sine of one acute angle equals the cosine of the other.

Question 9

In right triangle △STU\triangle STU△STU, ∠U=90∘\angle U = 90^\circ∠U=90∘, SU=12SU = 12SU=12 (leg adjacent to ∠S\angle S∠S), and TU=5TU = 5TU=5 (leg opposite ∠S\angle S∠S). Using the given information, what is the value of tan⁡(∠S)\tan(\angle S)tan(∠S)?

  1. 125\frac{12}{5}512​
  2. 513\frac{5}{13}135​
  3. 512\frac{5}{12}125​ (correct answer)
  4. 135\frac{13}{5}513​

Explanation: This question tests the ability to solve right triangles using trigonometric ratios. Trigonometric ratios relate the angles of a right triangle to the ratios of its sides: sin(θ) = opposite/hypotenuse, cos(θ) = adjacent/hypotenuse, and tan(θ) = opposite/adjacent. From the perspective of angle S, the side TU=5 is opposite, and SU=12 is adjacent, so we use tan(S) = opposite/adjacent = 5/12. Choice C is correct because it uses the tangent ratio with the opposite and adjacent sides: tan(S) = 5/12. Choice A inverts the ratio, calculating adjacent/opposite = 12/5 instead of opposite/adjacent. Remember the SOH-CAH-TOA mnemonic for choosing the correct trig ratio: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent, where opposite and adjacent are always relative to the angle in question. Key to right triangle problems: first identify the right angle and hypotenuse (longest side, opposite the right angle), then decide whether you have enough information for Pythagorean theorem (two sides known) or need trigonometry (one side and one angle known).

Question 10

A ladder leans against a wall, making an angle α\alphaα with the ground. The ladder's length is 20 feet, and it reaches 16 feet up the wall. Using the relationship between sine and cosine of complementary angles, what is cos⁡(α)+sin⁡(90°−α)\cos(\alpha) + \sin(90° - \alpha)cos(α)+sin(90°−α)?

  1. cos⁡(α)+sin⁡(90°−α)=1620+1220=2820=1.4\cos(\alpha) + \sin(90° - \alpha) = \frac{16}{20} + \frac{12}{20} = \frac{28}{20} = 1.4cos(α)+sin(90°−α)=2016​+2012​=2028​=1.4
  2. cos⁡(α)+sin⁡(90°−α)=1620+1620=3220=1.6\cos(\alpha) + \sin(90° - \alpha) = \frac{16}{20} + \frac{16}{20} = \frac{32}{20} = 1.6cos(α)+sin(90°−α)=2016​+2016​=2032​=1.6
  3. cos⁡(α)+sin⁡(90°−α)=1220+1620=2820=1.4\cos(\alpha) + \sin(90° - \alpha) = \frac{12}{20} + \frac{16}{20} = \frac{28}{20} = 1.4cos(α)+sin(90°−α)=2012​+2016​=2028​=1.4
  4. cos⁡(α)+sin⁡(90°−α)=1220+1220=2420=1.2\cos(\alpha) + \sin(90° - \alpha) = \frac{12}{20} + \frac{12}{20} = \frac{24}{20} = 1.2cos(α)+sin(90°−α)=2012​+2012​=2024​=1.2 (correct answer)

Explanation: When you encounter a ladder problem with trigonometry, you're working with a right triangle where the ladder is the hypotenuse, the wall height is the opposite side to angle α, and the ground distance is the adjacent side to angle α. First, let's find the missing side using the Pythagorean theorem. With a 20-foot ladder reaching 16 feet up the wall, the ground distance is 202−162=400−256=144=12\sqrt{20^2 - 16^2} = \sqrt{400 - 256} = \sqrt{144} = 12202−162​=400−256​=144​=12 feet. Now we can find the trigonometric ratios: sin⁡(α)=oppositehypotenuse=1620\sin(\alpha) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{16}{20}sin(α)=hypotenuseopposite​=2016​ and cos⁡(α)=adjacenthypotenuse=1220\cos(\alpha) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{20}cos(α)=hypotenuseadjacent​=2012​. The key insight is understanding complementary angle relationships. When two angles are complementary (sum to 90°), the sine of one equals the cosine of the other. Therefore, sin⁡(90°−α)=cos⁡(α)=1220\sin(90° - \alpha) = \cos(\alpha) = \frac{12}{20}sin(90°−α)=cos(α)=2012​. So cos⁡(α)+sin⁡(90°−α)=1220+1220=2420=1.2\cos(\alpha) + \sin(90° - \alpha) = \frac{12}{20} + \frac{12}{20} = \frac{24}{20} = 1.2cos(α)+sin(90°−α)=2012​+2012​=2024​=1.2, which is answer D. Answer A incorrectly uses sin⁡(α)\sin(\alpha)sin(α) for cos⁡(α)\cos(\alpha)cos(α), confusing opposite and adjacent sides. Answer B makes both errors—wrong values for both terms. Answer C correctly identifies cos⁡(α)=1220\cos(\alpha) = \frac{12}{20}cos(α)=2012​ but incorrectly uses sin⁡(α)=1620\sin(\alpha) = \frac{16}{20}sin(α)=2016​ instead of applying the complementary angle relationship. Remember: for complementary angles, sin⁡(90°−θ)=cos⁡(θ)\sin(90° - \theta) = \cos(\theta)sin(90°−θ)=cos(θ) and cos⁡(90°−θ)=sin⁡(θ)\cos(90° - \theta) = \sin(\theta)cos(90°−θ)=sin(θ). This relationship is fundamental in trigonometry and appears frequently on exams.

Question 11

In right triangle MNOMNOMNO, the right angle is at NNN. The hypotenuse is MO=13MO = 13MO=13, and MN=5MN = 5MN=5. What is the length of NONONO?

  1. 888
  2. 101010
  3. 121212 (correct answer)
  4. 194\sqrt{194}194​

Explanation: This question tests the ability to solve right triangles using the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse equals the sum of the squares of the two legs: a² + b² = c². This is a 5-12-13 Pythagorean triple, so we can recognize immediately that the missing side is 12 without calculation. Choice C is correct because it shows correct substitution into the Pythagorean theorem, identifying the hypotenuse as 13 and subtracting the square of the known leg 5 to find the other leg as 12. Choice D incorrectly adds the squares instead of subtracting: √(169 + 25) = √194 instead of √(169 - 25) = 12. Key to right triangle problems: first identify the right angle and hypotenuse (longest side, opposite the right angle), then decide whether you have enough information for Pythagorean theorem (two sides known) or need trigonometry (one side and one angle known). Recognize common Pythagorean triples (3-4-5, 5-12-13, 8-15-17) and their multiples to save time on calculations—if you see two sides of a triple, the third can be determined without calculation.

Question 12

In right triangle JKLJKLJKL, the right angle is at KKK. If JK=9JK = 9JK=9 and KL=12KL = 12KL=12, what is the exact length of the hypotenuse JLJLJL?

  1. 151515 (correct answer)
  2. 212121
  3. 369\sqrt{369}369​
  4. 81+144\sqrt{81} + \sqrt{144}81​+144​

Explanation: This question tests the ability to solve right triangles using the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse equals the sum of the squares of the two legs: a² + b² = c². Given sides of length 9 and 12, we substitute into the Pythagorean theorem: 9² + 12² = c², which gives 81 + 144 = c², so c² = 225, and taking the square root yields c = 15. Choice A is correct because it shows correct substitution into the Pythagorean theorem with specific numbers 9 and 12, yielding √(81 + 144) = √225 = 15. Choice D makes the error of adding the square roots instead of taking the square root after adding the squares: √81 + √144 = 9 + 12 instead of √(81 + 144) = 15. Key to right triangle problems: first identify the right angle and hypotenuse (longest side, opposite the right angle), then decide whether you have enough information for Pythagorean theorem (two sides known) or need trigonometry (one side and one angle known). Recognize common Pythagorean triples (3-4-5, 5-12-13, 8-15-17) and their multiples to save time on calculations—if you see two sides of a triple, the third can be determined without calculation.

Question 13

In right triangle ABCABCABC, the right angle is at CCC. The hypotenuse is AB=13AB=13AB=13 m and one leg is BC=5BC=5BC=5 m. What is the exact length of the other leg ACACAC?

  1. 194 \sqrt{194}194​ m
  2. 121212 m (correct answer)
  3. 888 m
  4. 181818 m

Explanation: This question tests the ability to solve right triangles using the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse equals the sum of the squares of the two legs: a² + b² = c². Since we know the hypotenuse (13 m) and one leg (BC = 5 m), we need to find the other leg AC, so we rearrange: AC² + 5² = 13², which gives AC² + 25 = 169, so AC² = 144, and taking the square root yields AC = 12. Choice B is correct because this is a 5-12-13 Pythagorean triple, so we can recognize immediately that the missing side is 12 without calculation. Choice A incorrectly adds the squares: √(13² + 5²) = √(169 + 25) = √194, treating both given sides as legs instead of recognizing that 13 is the hypotenuse. When using the Pythagorean theorem, always check that you're putting the hypotenuse (the longest side) as c in the equation a² + b² = c², not one of the legs. Remember that in any right triangle, the hypotenuse is always the longest side, so if you're given two sides, the larger one is the hypotenuse if it's opposite the right angle.

Question 14

In right triangle △ABC\triangle ABC△ABC, ∠C=90∘\angle C = 90^\circ∠C=90∘ and ABABAB is the hypotenuse. If AB=13AB = 13AB=13 and AC=5AC = 5AC=5, what is the exact length of leg BCBCBC?

  1. 181818
  2. 121212 (correct answer)
  3. 194\sqrt{194}194​
  4. 888

Explanation: This question tests the ability to solve right triangles using the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse equals the sum of the squares of the two legs: a² + b² = c². Since we know the hypotenuse AB = 13 and one leg AC = 5, we need to find the other leg BC, so we rearrange: AC² + BC² = AB², which gives 5² + BC² = 13², so 25 + BC² = 169, therefore BC² = 144, and taking the square root yields BC = 12. Choice B is correct because when we substitute into the rearranged Pythagorean theorem, we get BC = √(13² - 5²) = √(169 - 25) = √144 = 12. Choice C incorrectly adds the squares instead of subtracting: √(169 + 25) = √194 instead of √(169 - 25) = 12. This is a 5-12-13 Pythagorean triple, so we can recognize immediately that the missing side is 12 without calculation. When using the Pythagorean theorem, always check that you're putting the hypotenuse (the longest side) as c in the equation a² + b² = c², not one of the legs.

Question 15

In right triangle △ABC\triangle ABC△ABC, ∠C=90∘\angle C = 90^\circ∠C=90∘ and ABABAB is the hypotenuse. If BC=12BC = 12BC=12 and AB=13AB = 13AB=13, what is the value of cos⁡(∠B)\cos(\angle B)cos(∠B)?

  1. 1312\frac{13}{12}1213​
  2. 513\frac{5}{13}135​
  3. 1213\frac{12}{13}1312​ (correct answer)
  4. 125\frac{12}{5}512​

Explanation: This question tests the ability to solve right triangles using trigonometric ratios. Trigonometric ratios relate the angles of a right triangle to the ratios of its sides: sin(θ) = opposite/hypotenuse, cos(θ) = adjacent/hypotenuse, and tan(θ) = opposite/adjacent. From the perspective of angle B, the side BC is the adjacent side (it touches angle B), and AB is the hypotenuse, so we use cos(B) = adjacent/hypotenuse = BC/AB = 12/13. Choice C is correct because it correctly identifies BC as the adjacent side to angle B and uses cos(B) = 12/13. Choice B incorrectly identifies the sides, using AC (which we'd need to calculate as 5) as if it were adjacent to angle B, when actually AC is opposite to angle B. This is a 5-12-13 Pythagorean triple, confirming our work since AC would be 5. Remember the SOH-CAH-TOA mnemonic for choosing the correct trig ratio: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent, where opposite and adjacent are always relative to the angle in question.

Question 16

In right triangle △ABC\triangle ABC△ABC, ∠C=90∘\angle C = 90^\circ∠C=90∘. If AC=12AC = 12AC=12 and BC=5BC = 5BC=5, what is the exact value of tan⁡(∠A)\tan(\angle A)tan(∠A)?

  1. 125\frac{12}{5}512​
  2. 513\frac{5}{13}135​
  3. 1213\frac{12}{13}1312​
  4. 512\frac{5}{12}125​ (correct answer)

Explanation: This question tests the ability to solve right triangles using trigonometric ratios. Trigonometric ratios relate the angles of a right triangle to the ratios of its sides: sin(θ) = opposite/hypotenuse, cos(θ) = adjacent/hypotenuse, and tan(θ) = opposite/adjacent. From the perspective of angle A, the side BC = 5 is the opposite side and AC = 12 is the adjacent side, so we use tan(A) = opposite/adjacent = BC/AC = 5/12. Choice D is correct because it correctly identifies BC as opposite to angle A and AC as adjacent to angle A, giving tan(A) = 5/12. Choice A inverts the ratio, calculating adjacent/opposite = 12/5 instead of opposite/adjacent = 5/12. This is a 5-12-13 Pythagorean triple, which we can verify by checking that 5² + 12² = 25 + 144 = 169 = 13². Remember the SOH-CAH-TOA mnemonic for choosing the correct trig ratio: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent, where opposite and adjacent are always relative to the angle in question.

Question 17

In right triangle STUSTUSTU, the right angle is at TTT. If ST=8ST = 8ST=8 and TU=15TU = 15TU=15, what is the exact value of tan⁡(∠S)\tan(\angle S)tan(∠S)?

  1. 815\frac{8}{15}158​
  2. 1517\frac{15}{17}1715​
  3. 158\frac{15}{8}815​ (correct answer)
  4. 178\frac{17}{8}817​

Explanation: This question tests the ability to solve right triangles using trigonometric ratios. Trigonometric ratios relate the angles of a right triangle to the ratios of its sides: sin(θ) = opposite/hypotenuse, cos(θ) = adjacent/hypotenuse, and tan(θ) = opposite/adjacent. From the perspective of angle S, the side of length 15 is the opposite, and the side of length 8 is the adjacent, so we use tan(S) = opposite/adjacent = 15/8 to find the ratio. Choice C is correct because it uses the tangent ratio with opposite side 15 and adjacent side 8 from the perspective of angle S. Choice A inverts the ratio, calculating adjacent/opposite = 8/15 instead of opposite/adjacent = 15/8. Remember the SOH-CAH-TOA mnemonic for choosing the correct trig ratio: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent, where opposite and adjacent are always relative to the angle in question. Key to right triangle problems: first identify the right angle and hypotenuse (longest side, opposite the right angle), then decide whether you have enough information for Pythagorean theorem (two sides known) or need trigonometry (one side and one angle known).

Question 18

In right triangle ABCABCABC, the right angle is at CCC. The legs are AC=12AC=12AC=12 and BC=5BC=5BC=5, and the hypotenuse is AB=13AB=13AB=13. What is the value of cos⁡(∠A)\cos(\angle A)cos(∠A)?

  1. 513\dfrac{5}{13}135​
  2. 1213\dfrac{12}{13}1312​ (correct answer)
  3. 1312\dfrac{13}{12}1213​
  4. 125\dfrac{12}{5}512​

Explanation: This question tests the ability to solve right triangles using trigonometric ratios. Trigonometric ratios relate the angles of a right triangle to the ratios of its sides: sin(θ) = opposite/hypotenuse, cos(θ) = adjacent/hypotenuse, and tan(θ) = opposite/adjacent. From the perspective of angle A, the side AC = 12 is adjacent to angle A (it touches angle A but isn't the hypotenuse), and AB = 13 is the hypotenuse, so we use cos(A) = adjacent/hypotenuse = 12/13. Choice B is correct because cos(A) = AC/AB = 12/13, correctly identifying AC as the adjacent side to angle A. Choice A inverts the ratio, calculating 12/5 which would be tan(A) = opposite/adjacent, confusing which ratio relates the given sides. Remember the SOH-CAH-TOA mnemonic for choosing the correct trig ratio: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent, where opposite and adjacent are always relative to the angle in question. For complementary angles in a right triangle, sin(θ) = cos(90° - θ), which explains why the sine of one acute angle equals the cosine of the other.

Question 19

A surveyor stands 15 m from the base of a tower on level ground. The angle of elevation to the top of the tower is 45∘45^\circ45∘. What is the height of the tower?

  1. 151515 m (correct answer)
  2. 15215\sqrt{2}152​ m
  3. 152\dfrac{15}{\sqrt{2}}2​15​ m
  4. 303030 m

Explanation: This question tests the ability to solve right triangles using trigonometric ratios. Trigonometric ratios relate the angles of a right triangle to the ratios of its sides: sin(θ) = opposite/hypotenuse, cos(θ) = adjacent/hypotenuse, and tan(θ) = opposite/adjacent. From the surveyor's position, the horizontal distance (15 m) is adjacent to the angle of elevation, and the tower height is opposite, so we use tan(45°) = height/15, solving to get height = 15 × tan(45°) = 15 × 1 = 15. Choice A is correct because tan(45°) = 1, so height = 15 × tan(45°) = 15 × 1 = 15 m. Choice B incorrectly treats this as a 45-45-90 triangle where the hypotenuse would be 15√2, but 15 m is the adjacent side (horizontal distance), not the hypotenuse. In a 45-45-90 triangle, the two legs are equal, so when the angle of elevation is 45°, the height equals the horizontal distance. Remember that for angle of elevation problems, use tangent when you know the horizontal distance and need the height, since tan(elevation angle) = height/horizontal distance.

Question 20

In right triangle △DEF\triangle DEF△DEF, ∠F=90∘\angle F = 90^\circ∠F=90∘. The hypotenuse is DE=13DE = 13DE=13 ft and one leg is DF=5DF = 5DF=5 ft. Based on the triangle described, what is the length of side EFEFEF?

  1. 194\sqrt{194}194​ ft
  2. 121212 ft (correct answer)
  3. 181818 ft
  4. 888 ft

Explanation: This question tests the ability to solve right triangles using the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse equals the sum of the squares of the two legs: a² + b² = c². Given the hypotenuse of length 13 and one leg of length 5, we rearrange the Pythagorean theorem to find the other leg: EF = √(13² - 5²) = √(169 - 25) = √144 = 12. Choice B is correct because it shows correct substitution into the Pythagorean theorem with the hypotenuse as c and solving for the missing leg: √(169 - 25) = 12 ft. Choice A incorrectly treats 5 as the hypotenuse, using √(13² + 5²) = √194 instead of correctly identifying that 13 is the hypotenuse. Key to right triangle problems: first identify the right angle and hypotenuse (longest side, opposite the right angle), then decide whether you have enough information for Pythagorean theorem (two sides known) or need trigonometry (one side and one angle known). When using the Pythagorean theorem, always check that you're putting the hypotenuse (the longest side) as c in the equation a² + b² = c², not one of the legs.