Geometry Quiz: Applying Laws Of Sines And Cosines
9 questions · exam conditions
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Applying Laws Of Sines And CosinesQuestion 1 of 9

Two ships leave port at the same time. Ship A travels at 15 mph in a direction N30°E, while Ship B travels at 20 mph in a direction S45°E. After 2 hours, what is the distance between the two ships to the nearest mile?

35 miles based on vector addition principles
70 miles using the Law of Cosines
52 miles using the Law of Sines
47 miles using direct distance calculation
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Geometry Quiz

Geometry Quiz: Applying Laws Of Sines And Cosines

Practice Applying Laws Of Sines And Cosines in Geometry 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 Applying Laws Of Sines And Cosines, giving you a quick way to practice the rules, question types, and explanations that matter most for Geometry.

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

Two ships leave port at the same time. Ship A travels at 15 mph in a direction N30°E, while Ship B travels at 20 mph in a direction S45°E. After 2 hours, what is the distance between the two ships to the nearest mile?

  1. 35 miles based on vector addition principles
  2. 70 miles using the Law of Cosines (correct answer)
  3. 52 miles using the Law of Sines
  4. 47 miles using direct distance calculation
Explanation: After 2 hours, Ship A has traveled 30 miles and Ship B has traveled 40 miles. The angle between their paths is 30° + 90° + 45° = 165°. Using the Law of Cosines to find the distance between ships: d² = 30² + 40² - 2(30)(40)cos(165°) = 900 + 1600 - 2400cos(165°) ≈ 2500 + 2320 = 4820, so d ≈ 70 miles. Choice A uses simple vector addition incorrectly. Choice C misapplies the Law of Sines. Choice D results from using the wrong angle between the paths.

Question 2

In triangle XYZXYZ, you know that x=20x = 20, y=16y = 16, and ∠Z=45°\angle Z = 45°. When solving for angle XX using the Law of Sines after finding side zz, which potential issue must you consider?

  1. The triangle might not exist due to impossible side lengths
  2. The Law of Sines cannot be applied in this configuration
  3. There could be two possible values for angle XX (ambiguous case) (correct answer)
  4. Angle XX must be obtuse based on the given information
Explanation: When you have two sides and an angle opposite one of them (SSA configuration), you're dealing with a classic ambiguous case scenario in triangle problems. Here, you know sides x=20x = 20 and y=16y = 16, plus ∠Z=45°\angle Z = 45° opposite the unknown side zz. After using the Law of Cosines to find side zz, you'll apply the Law of Sines: sin⁡Xx=sin⁡Zz\frac{\sin X}{x} = \frac{\sin Z}{z}. This gives you sin⁡X=xsin⁡Zz\sin X = \frac{x \sin Z}{z}. The critical issue is that the sine function can produce two different angles in the range 0°0° to 180°180° for the same sine value. If sin⁡X=k\sin X = k, then XX could be either arcsin⁡(k)\arcsin(k) or 180°−arcsin⁡(k)180° - \arcsin(k). Both angles have identical sine values but are supplementary to each other. Looking at the wrong answers: (A) is incorrect because with these side lengths and angle, a valid triangle will exist. (B) is wrong since the Law of Sines applies perfectly once you know all three sides and one angle. (D) is incorrect because there's nothing in the given information that forces angle XX to be obtuse—in fact, you might get both an acute and obtuse solution. The answer is (C) because the ambiguous case means two possible triangles could satisfy the given conditions, leading to two different values for angle XX. Study tip: Whenever you see SSA (two sides and a non-included angle), immediately think "ambiguous case." Always check if your sine calculation yields two possible angle measures.

Question 3

A pilot flies from airport A to airport B (150 miles away) and then to airport C. The angle at airport B is 125°125°, and the distance from B to C is 200 miles. To find the direct distance from A to C using the Law of Cosines, what is the correct setup?

  1. AC2=1502+2002−2(150)(200)cos⁡(125°)AC^2 = 150^2 + 200^2 - 2(150)(200)\cos(125°) (correct answer)
  2. AC2=1502+2002−2(150)(200)cos⁡(55°)AC^2 = 150^2 + 200^2 - 2(150)(200)\cos(55°)
  3. AC=150sin⁡(125°)sin⁡CAC = \frac{150 \sin(125°)}{\sin C} where angle C is unknown
  4. AC2=1502+2002+2(150)(200)cos⁡(125°)AC^2 = 150^2 + 200^2 + 2(150)(200)\cos(125°)
Explanation: The Law of Cosines states c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C, where CC is the included angle between sides aa and bb. Here, the angle at B (125°125°) is between the sides AB (150 miles) and BC (200 miles), so AC2=1502+2002−2(150)(200)cos⁡(125°)AC^2 = 150^2 + 200^2 - 2(150)(200)\cos(125°). Choice B incorrectly uses the supplement of 125°125°. Choice C attempts to use Law of Sines without sufficient information. Choice D has the wrong sign in the formula.

Question 4

A surveyor needs to find the distance across a lake. From point PP on one shore, she measures the distance to point QQ on the opposite shore as 250 meters, and to point RR (also on the opposite shore) as 180 meters. If ∠QPR=42°\angle QPR = 42°, what is the distance QRQR to the nearest meter?

  1. 168 meters using direct triangle measurement
  2. 172 meters using the Law of Cosines (correct answer)
  3. 185 meters using the Law of Sines
  4. 196 meters using trigonometric ratios
Explanation: Using the Law of Cosines: QR2=PQ2+PR2−2(PQ)(PR)cos⁡(∠QPR)=2502+1802−2(250)(180)cos⁡(42°)=62500+32400−90000cos⁡(42°)≈94900−66879=28021QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos(\angle QPR) = 250^2 + 180^2 - 2(250)(180)\cos(42°) = 62500 + 32400 - 90000\cos(42°) \approx 94900 - 66879 = 28021, so QR≈172QR \approx 172 meters. Choice A results from incorrect angle usage. Choice C comes from misapplying the Law of Sines when the Law of Cosines is needed. Choice D results from treating this as a right triangle problem.

Question 5

In triangle ABCABC with sides a=13a = 13, b=18b = 18, and c=25c = 25, a student calculates cos⁡C=132+182−2522(13)(18)=169+324−625468=−132468\cos C = \frac{13^2 + 18^2 - 25^2}{2(13)(18)} = \frac{169 + 324 - 625}{468} = \frac{-132}{468}. What can you conclude about angle CC?

  1. Angle CC is acute since the calculation shows positive cosine
  2. The calculation is incorrect; this triangle cannot exist
  3. Angle CC is obtuse since cos⁡C<0\cos C < 0, approximately 106°106° (correct answer)
  4. Angle CC equals exactly 90°90° based on these measurements
Explanation: When you encounter a triangle with three given side lengths, the Law of Cosines is your tool for finding angles. The student's calculation using cos⁡C=a2+b2−c22ab\cos C = \frac{a^2 + b^2 - c^2}{2ab} is mathematically correct and reveals important information about the triangle's geometry. Let's verify: cos⁡C=132+182−2522(13)(18)=169+324−625468=−132468≈−0.282\cos C = \frac{13^2 + 18^2 - 25^2}{2(13)(18)} = \frac{169 + 324 - 625}{468} = \frac{-132}{468} ≈ -0.282. Since cosine is negative, angle C must be obtuse (greater than 90°). Using inverse cosine: C=cos⁡−1(−0.282)≈106°C = \cos^{-1}(-0.282) ≈ 106°. This confirms answer C is correct. Now for the wrong answers: A incorrectly states the cosine is positive when the calculation clearly shows −132468\frac{-132}{468} is negative. B suggests the triangle cannot exist, but you can verify this triangle satisfies the triangle inequality (the sum of any two sides exceeds the third side). D claims the angle is exactly 90°, which would require cos⁡C=0\cos C = 0, not the negative value we calculated. The key insight is recognizing the connection between cosine signs and angle types: positive cosine means acute angles, zero cosine means right angles, and negative cosine means obtuse angles. When using the Law of Cosines, always check the sign of your result—it immediately tells you whether you're dealing with an acute or obtuse angle, which helps verify your final answer makes geometric sense.

Question 6

Refer to the triangle below. What is the measure of ∠A\angle A, to the nearest degree?

  1. 44°44° (correct answer)
  2. 71°71°
  3. 65°65°
  4. 52°52°
Explanation: Using the Law of Cosines to solve for ∠A\angle A (opposite side a=8a = 8): cos⁡A=b2+c2−a22bc=102+112−822(10)(11)=157220≈0.7136\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{10^2 + 11^2 - 8^2}{2(10)(11)} = \dfrac{157}{220} \approx 0.7136, so A≈44°A \approx 44°. Choice B solves for the angle opposite side 11. Choice C solves for the angle opposite side 10. Choice D reflects a sign error in the numerator.

Question 7

Refer to the parallelogram ABCDABCD below. What is the length of diagonal BDBD, to the nearest tenth?

  1. 11.011.0 (correct answer)
  2. 8.78.7
  3. 13.213.2
  4. 7.47.4
Explanation: In parallelogram ABCDABCD, consecutive angles are supplementary, so ∠A=180°−65°=115°\angle A = 180° - 65° = 115°. In △ABD\triangle ABD: BD2=AB2+AD2−2(AB)(AD)cos⁡A=72+62−2(7)(6)cos⁡115°=49+36−84(−0.4226)≈85+35.5=120.5BD^2 = AB^2 + AD^2 - 2(AB)(AD)\cos A = 7^2 + 6^2 - 2(7)(6)\cos 115° = 49 + 36 - 84(-0.4226) \approx 85 + 35.5 = 120.5, so BD≈11.0BD \approx 11.0. Choice B uses cos⁡65°\cos 65° directly (forgetting supplementary angle), giving about 8.7. Choice C is approximately diagonal ACAC. Choice D uses cos⁡65°\cos 65° with the wrong operations.

Question 8

In triangle PQR, p = 7, q = 9, and ∠R = 120°. When using the Law of Cosines to find side r, what is the intermediate step that shows r² before taking the square root?

  1. r² = 49 + 81 - 126cos(120°) = 193
  2. r² = 49 + 81 + 126cos(120°) = 67
  3. r² = 49 + 81 - 63 = 67
  4. r² = 49 + 81 + 63 = 193 (correct answer)
Explanation: When you encounter a triangle problem with two sides and the included angle, the Law of Cosines is your go-to tool. The formula is c2=a2+b2−2abcos⁡(C)c^2 = a^2 + b^2 - 2ab\cos(C), where C is the angle opposite side c. Here, you're finding side r (opposite angle R), so the formula becomes r2=p2+q2−2pqcos⁡(R)r^2 = p^2 + q^2 - 2pq\cos(R). Substituting the given values: r2=72+92−2(7)(9)cos⁡(120°)r^2 = 7^2 + 9^2 - 2(7)(9)\cos(120°). First, calculate the squares: 72=497^2 = 49 and 92=819^2 = 81. Next, find 2(7)(9)=1262(7)(9) = 126. Now you need cos⁡(120°)\cos(120°). Since 120° is in the second quadrant, cosine is negative: cos⁡(120°)=−12\cos(120°) = -\frac{1}{2}. Substituting everything: r2=49+81−126(−12)=49+81−(−63)=49+81+63=193r^2 = 49 + 81 - 126(-\frac{1}{2}) = 49 + 81 - (-63) = 49 + 81 + 63 = 193. This matches answer choice D. Answer A incorrectly shows −126cos⁡(120°)-126\cos(120°) as a single negative term, missing that cosine of 120° is already negative. Answer B makes the same error and also incorrectly changes the minus sign to plus in the Law of Cosines formula. Answer C shows the right final calculation (49 + 81 + 63) but gets the wrong sum. Remember: when the given angle is obtuse (greater than 90°), its cosine is negative. In the Law of Cosines, this creates a double negative that becomes positive, effectively adding rather than subtracting the final term.

Question 9

In triangle ABCABC, ∠A=35°\angle A = 35°, ∠B=68°\angle B = 68°, and side a=14a = 14. If you need to find side cc, which approach would give the most direct solution?

  1. Use Law of Sines: c=asin⁡Csin⁡Ac = \frac{a \sin C}{\sin A} where C=77°C = 77° (correct answer)
  2. Use Law of Cosines: c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C
  3. First find side bb, then use Law of Cosines
  4. Use Law of Sines: c=asin⁡Bsin⁡Cc = \frac{a \sin B}{\sin C} where C=77°C = 77°
Explanation: Since we know two angles and the side opposite one of them (ASA case), the Law of Sines provides the most direct approach. ∠C=180°−35°−68°=77°\angle C = 180° - 35° - 68° = 77°, then csin⁡C=asin⁡A\frac{c}{\sin C} = \frac{a}{\sin A}, so c=14sin⁡77°sin⁡35°≈23.8c = \frac{14 \sin 77°}{\sin 35°} \approx 23.8. Choice B requires finding side bb first, making it less direct. Choice C is unnecessarily complicated. Choice D sets up the Law of Sines incorrectly with wrong angle relationships.