Pythagorean Identity Application Quiz

10th Grade

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Thames

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Quizzes Created: 7387 | Total Attempts: 9,541,629

| Attempts: 11 | Questions: 20 | Updated: Dec 11, 2025

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1) If sin(θ) = −7/25 and θ in QIV, compute cos(θ).

−24/25

7/25

24/25

−7/25

Given: sinθ = −7/25, θ in QIV. Goal: Find cosθ.

Step 1: cos²θ = 1 − (−7/25)² = 1 − 49/625 = 576/625.

Step 2: cosθ = ±24/25.

Step 3: In QIV, cosine is positive cosθ = 24/25.

So final answer is 24/25.

Explanation

Submit

About This Quiz

Pythagorean Identity Application Quiz - Quiz

Ready to turn sin²θ + cos²θ = 1 into fast answers? In this quiz, you’ll use the identity plus quadrant signs to find missing sin, cos, and tan values from a single clue. Expect clean radicals, exact fractions, and lots of “which quadrant?” reasoning that builds accuracy and speed.

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2) If sin(θ) = 8/17 with θ in QI, what is cos²(θ)?

64/289

17/289

81/289

225/289

Given: sinθ = 8/17, θ in QI. Goal: Find cos²θ.

Step 1: cos²θ = 1 − (8/17)² = 1 − 64/289 = 225/289.

So final answer is 225/289.

Explanation

Given: sinθ = 8/17, θ in QI. Goal: Find cos²θ.

Step 1: cos²θ = 1 − (8/17)² = 1 − 64/289 = 225/289.

So final answer is 225/289.

Submit

3) If cos(θ) = −5/13 and θ in QII, find tan(θ).

12/5

−12/5

−5/12

5/12

Given: cosθ = −5/13, θ in QII. Goal: Find tanθ.

Step 1: sin²θ = 1 − (25/169) = 144/169 sinθ = ±12/13.

Step 2: QII ⇒ sine positive sinθ = 12/13.

Step 3: tanθ = sinθ / cosθ = (12/13)/(−5/13) = −12/5.

So final answer is −12/5.

Explanation

Submit

4) Given sin(θ) = 15/17 and θ in QII, find cos(θ).

8/17

15/17

−8/17

−15/17

Given: sinθ = 15/17, θ in QII. Goal: Find cosθ.

Step 1: cos²θ = 1 − (15/17)² = 1 − 225/289 = 64/289.

Step 2: QII ⇒ cosine negative cosθ = −8/17.

So final answer is −8/17.

Explanation

Given: sinθ = 15/17, θ in QII. Goal: Find cosθ.

Step 1: cos²θ = 1 − (15/17)² = 1 − 225/289 = 64/289.

Step 2: QII ⇒ cosine negative cosθ = −8/17.

So final answer is −8/17.

Submit

5) If cos(θ) = √2/2 and θ in QIV, find sin(θ).

−√2/2

√2/2

√3/2

−√3/2

Given: cosθ = √2/2, θ in QIV. Goal: Find sinθ.

Step 1: sin²θ = 1 − (1/2) = 1/2 sinθ = ±√2/2.

Step 2: QIV ⇒ sine negative sinθ = −√2/2.

So final answer is −√2/2.

Explanation

Given: cosθ = √2/2, θ in QIV. Goal: Find sinθ.

Step 1: sin²θ = 1 − (1/2) = 1/2 sinθ = ±√2/2.

Step 2: QIV ⇒ sine negative sinθ = −√2/2.

So final answer is −√2/2.

Submit

6) Given sin(θ) = t with θ in QI, express cos(θ).

1 − t²

T² − 1

√(1 − t²)

−√(1 − t²)

Given: sinθ = t, θ in QI. Goal: Express cosθ.

Step 1: cos²θ = 1 − t².

Step 2: QI ⇒ cosine positive cosθ = √(1 − t²).

So final answer is √(1 − t²).

Explanation

Given: sinθ = t, θ in QI. Goal: Express cosθ.

Step 1: cos²θ = 1 − t².

Step 2: QI ⇒ cosine positive cosθ = √(1 − t²).

So final answer is √(1 − t²).

Submit

7) Given cos(θ) = u with θ in QIII, express sin(θ).

√(1 − u²)

1 − u²

−√(1 − u²)

U² − 1

Given: cosθ = u, θ in QIII. Goal: Express sinθ.

Step 1: sin²θ = 1 − u².

Step 2: QIII ⇒ sine negative sinθ = −√(1 − u²).

So final answer is −√(1 − u²).

Explanation

Given: cosθ = u, θ in QIII. Goal: Express sinθ.

Step 1: sin²θ = 1 − u².

Step 2: QIII ⇒ sine negative sinθ = −√(1 − u²).

So final answer is −√(1 − u²).

Submit

8) If sin(θ) = −3/5 and θ in QIII, what is cos(θ)?

4/5

−4/5

−3/5

3/5

Given: sinθ = −3/5, θ in QIII. Goal: Find cosθ.

Step 1: cos²θ = 1 − (−3/5)² = 1 − 9/25 = 16/25.

Step 2: QIII ⇒ cosine negative cosθ = −4/5.

So final answer is −4/5.

Explanation

Given: sinθ = −3/5, θ in QIII. Goal: Find cosθ.

Step 1: cos²θ = 1 − (−3/5)² = 1 − 9/25 = 16/25.

Step 2: QIII ⇒ cosine negative cosθ = −4/5.

So final answer is −4/5.

Submit

9) Let sin(θ) = a with θ in QIII. What is cos(θ)?

√(1 − a²)

−(1 − a²)

−√(1 − a²)

1 − a²

Given: sinθ = a, θ in QIII. Goal: Find cosθ.

Step 1: cos²θ = 1 − a².

Step 2: QIII ⇒ cosine negative cosθ = −√(1 − a²).

So final answer is −√(1 − a²).

Explanation

Given: sinθ = a, θ in QIII. Goal: Find cosθ.

Step 1: cos²θ = 1 − a².

Step 2: QIII ⇒ cosine negative cosθ = −√(1 − a²).

So final answer is −√(1 − a²).

Submit

10) Find sin(θ) if cos(θ) = −√3/2 and θ in QIII.

1/2

−1/2

√3/2

−√3/2

Given: cosθ = −√3/2, θ in QIII. Goal: Find sinθ.

Step 1: sin²θ = 1 − (3/4) = 1/4 sinθ = ±1/2.

Step 2: QIII ⇒ sine negative sinθ = −1/2.

So final answer is −1/2.

Explanation

Given: cosθ = −√3/2, θ in QIII. Goal: Find sinθ.

Step 1: sin²θ = 1 − (3/4) = 1/4 sinθ = ±1/2.

Step 2: QIII ⇒ sine negative sinθ = −1/2.

So final answer is −1/2.

Submit

11) Given sin(θ) = t, express tan(θ) in terms of t for θ in QII.

T/√(1 − t²)

√(1 − t²)/t

−t/√(1 − t²)

−√(1 − t²)/t

Given: sinθ = t, θ in QII. Goal: Express tanθ.

Step 1: cosθ = ±√(1 − t²).

Step 2: QII ⇒ cosine negative cosθ = −√(1 − t²).

Step 3: tanθ = t / (−√(1 − t²)) = −t/√(1 − t²).

So final answer is −t/√(1 − t²).

Explanation

Submit

12) If sin(θ) = 20/29 and θ in QI, what is cos(θ)?

−21/29

9/29

21/29

−9/29

Given: sinθ = 20/29, θ in QI. Goal: Find cosθ.

Step 1: cos²θ = 1 − (400/841) = 441/841 cosθ = ±21/29.

Step 2: QI ⇒ cosine positive cosθ = 21/29.

So final answer is 21/29.

Explanation

Given: sinθ = 20/29, θ in QI. Goal: Find cosθ.

Step 1: cos²θ = 1 − (400/841) = 441/841 cosθ = ±21/29.

Step 2: QI ⇒ cosine positive cosθ = 21/29.

So final answer is 21/29.

Submit

13) If sin(θ) = 4/5 and θ in QI, which equation verifies sin²(θ) + cos²(θ) = 1 numerically?

(4/5)² + (4/5)² = 1

(4/5) + (3/5) = 1

(3/5)² + (3/5)² = 1

(4/5)² + (3/5)² = 1

Given: sinθ = 4/5, θ in QI. Goal: Verify identity.

Step 1: cos²θ = 1 − (4/5)² = 1 − 16/25 = 9/25 cosθ = 3/5.

Step 2: Substitute (4/5)² + (3/5)² = 16/25 + 9/25 = 25/25 = 1.

So final answer is (4/5)² + (3/5)² = 1.

Explanation

Submit

14) If sin²(θ) + cos²(θ) = 1 and sin(θ) = 3/5 with θ in QII, what is cos(θ)?

−4/5

4/5

3/5

−3/5

Given: sinθ = 3/5, θ in QII. Goal: Find cosθ.

Step 1: sin²θ + cos²θ = 1 cos²θ = 1 − (3/5)² = 1 − 9/25 = 16/25.

Step 2: cosθ = ±4/5.

Step 3 (Quadrant Rule): In QII, cosine is negative cosθ = −4/5.

So final answer is −4/5.

Explanation

Submit

15) Given cos(θ) = −12/13 and θ in QIII, find sin(θ).

12/13

−5/13

−12/13

5/13

Given: cosθ = −12/13, θ in QIII. Goal: Find sinθ.

Step 1: sin²θ = 1 − cos²θ = 1 − (144/169) = 25/169.

Step 2: sinθ = ±5/13.

Step 3: In QIII, sine is negative sinθ = −5/13.

So final answer is −5/13.

Explanation

Submit

16) Let cos(θ) = 2/3 with θ in QI. What is tan(θ)?

√5/3

2/√5

√5/2

√5/5

Given: cosθ = 2/3, θ in QI. Goal: Find tanθ.

Step 1: sin²θ = 1 − (2/3)² = 1 − 4/9 = 5/9 sinθ = √5/3.

Step 2: tanθ = sinθ / cosθ = (√5/3)/(2/3) = √5/2.

So final answer is √5/2.

Explanation

Given: cosθ = 2/3, θ in QI. Goal: Find tanθ.

Step 1: sin²θ = 1 − (2/3)² = 1 − 4/9 = 5/9 sinθ = √5/3.

Step 2: tanθ = sinθ / cosθ = (√5/3)/(2/3) = √5/2.

So final answer is √5/2.

Submit

17) If sin(θ) = √3/2 and θ in QII, find tan(θ).

√3

1/√3

−√3

−1/√3

Given: sinθ = √3/2, θ in QII. Goal: Find tanθ.

Step 1: cos²θ = 1 − (√3/2)² = 1 − 3/4 = 1/4 cosθ = ±1/2.

Step 2: QII ⇒ cosine negative cosθ = −1/2.

Step 3: tanθ = (√3/2)/(−1/2) = −√3.

So final answer is −√3.

Explanation

Submit

18) Suppose cos(θ) = −1/4 and θ in QII. Find sin(θ).

−√15/4

√3/4

−√3/4

√15/4

Given: cosθ = −1/4, θ in QII. Goal: Find sinθ.

Step 1: sin²θ = 1 − (1/4)² = 1 − 1/16 = 15/16.

Step 2: sinθ = ±√15/4.

Step 3: QII ⇒ sine positive sinθ = √15/4.

So final answer is √15/4.

Explanation

Submit

19) For θ in QIII with tan(θ) = 4/3, determine cos(θ).

−3/5

4/5

3/5

−4/5

Given: tanθ = 4/3, θ in QIII. Goal: Find cosθ.

Step 1: Reference triangle opp = 4, adj = 3, hyp = 5.

Step 2: In QIII, both sine and cosine are negative cosθ = −3/5.

So final answer is −3/5.

Explanation

Submit

20) If cos(θ) = 5/13 and θ in QIV, what is tan(θ)?

5/12

−5/12

12/5

−12/5

Given: cosθ = 5/13, θ in QIV. Goal: Find tanθ.

Step 1: sin²θ = 1 − (25/169) = 144/169 sinθ = ±12/13.

Step 2: QIV ⇒ sine negative sinθ = −12/13.

Step 3: tanθ = sinθ / cosθ = (−12/13)/(5/13) = −12/5.

So final answer is −12/5.

Explanation