Precalculus Quiz: Applying Laws Of Sines And Cosines

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 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.

Question 1

1. $\frac{20\sin(70°)}{\sin(45°)}$
2. $\frac{20\sin(65°)}{\sin(45°)}$
3. $\frac{20\sin(45°)}{\sin(70°)}$
4. $\frac{20\sin(45°)}{\sin(65°)}$ (correct answer)

Explanation: When you encounter a triangle with two angles and one side given, you're dealing with a Law of Sines problem. This law states that in any triangle, the ratio of each side to the sine of its opposite angle is constant: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. First, find the missing angle. Since angles in a triangle sum to 180°, angle R = 180° - 45° - 70° = 65°. Now you have all three angles: P = 45°, Q = 70°, R = 65°, and side r = 20. To find side p (opposite angle P), set up the Law of Sines proportion: $\frac{p}{\sin P} = \frac{r}{\sin R}$. Substituting the known values: $\frac{p}{\sin 45°} = \frac{20}{\sin 65°}$. Solving for p: $p = \frac{20\sin 45°}{\sin 65°}$. Let's examine why the other answers are incorrect. Answer A uses $\sin 70°$ in the numerator and $\sin 45°$ in the denominator, which would actually give you the ratio $\frac{r}{p}$ rather than p itself. Answer B incorrectly uses $\sin 70°$ (angle Q) instead of $\sin 45°$ (angle P) in the numerator. Answer C places $\sin 70°$ in the denominator instead of $\sin 65°$, which would be trying to relate sides p and q instead of p and r. Study tip: Always identify all three angles first, then carefully match each side with its opposite angle in your Law of Sines setup. Double-check that you're solving for the correct variable by ensuring the unknown appears in the numerator of your final expression.

Question 2

1. Law of Sines
2. Law of Cosines (correct answer)
3. Pythagorean theorem only (Law of Sines/Cosines cannot be used)
4. Neither law; there is not enough information

Explanation: This question tests understanding of when to apply the Law of Cosines for solving non-right triangles. To decide which law to use: Law of Sines when you have angle-side opposite pairs to work with, Law of Cosines when you have two sides and the included angle or three sides with no angles. Since we know two sides (a = 8, b = 15) and the included angle (C = 90°), this is a SAS configuration, which requires the Law of Cosines because we have the two sides and the angle between them. Choice B is correct because it correctly identifies that the Law of Cosines should be used for this SAS configuration. Choice C incorrectly suggests using only the Pythagorean theorem and claims the Laws cannot be used, but the Law of Cosines actually reduces to the Pythagorean theorem when C = 90° since cos(90°) = 0, making c² = a² + b² - 2ab·cos(90°) = a² + b² - 0 = a² + b². The Law of Cosines reduces to the Pythagorean theorem when the angle is 90° (since cos(90°) = 0, the -2ab·cos(C) term vanishes), making it a generalization that works for all triangles.

Question 3

1. $\sqrt{117}$
2. $\sqrt{333}$ (correct answer)
3. $15$
4. $\sqrt{225}$

Explanation: This question tests understanding of the Law of Cosines for solving non-right triangles. The Law of Cosines states that c² = a² + b² - 2ab·cos(C) for any triangle (with analogous formulas for the other sides), generalizing the Pythagorean theorem, and is used when you know two sides and the included angle (SAS) or all three sides (SSS). Given sides a = 9, b = 12, and angle C = 120°, we apply the Law of Cosines: c² = a² + b² - 2ab·cos(C) = 9² + 12² - 2(9)(12)·cos(120°) = 81 + 144 - 216·(-1/2) = 225 - (-108) = 225 + 108 = 333, so c = √333. Choice B is correct because it applies the correct law with proper substitution of values and accurate arithmetic, remembering that cos(120°) = -1/2. Choice D gives the intermediate result c² = 225 instead of including the -2ab·cos(C) term, forgetting that cos(120°) is negative. Remember that the Law of Cosines generalizes the Pythagorean theorem: when the angle is 90°, cos(90°) = 0 makes the formula reduce to a² + b² = c², but for any other angle, you must include the -2ab·cos(C) term.

Question 4

1. Since $\sin K = 0.8$, then $\angle K = \arcsin(0.8) \approx 53.1°$
2. There are two possible values: $\angle K \approx 53.1°$ or $\angle K \approx 126.9°$ (correct answer)
3. The calculation is impossible because $\sin K > \frac{1}{2}$
4. The triangle is invalid because $j < k$ but $\angle J = 30°$ is too small

Explanation: This is the ambiguous case (SSA) of the Law of Sines. When 0 < sin K < 1, there are potentially two angles: K₁ = arcsin(0.8) ≈ 53.1° and K₂ = 180° - 53.1° = 126.9°. The student must check which value(s) create valid triangles by ensuring the sum of angles equals 180°. Choice A ignores the ambiguous case. Choice C incorrectly suggests the calculation is impossible. Choice D makes an invalid conclusion about the triangle's validity.

Question 5

1. $\sqrt{183}$ (correct answer)
2. $\sqrt{379}$
3. $\sqrt{365}$
4. $\sqrt{169}$

Explanation: This question tests understanding of the Law of Cosines for solving non-right triangles. The Law of Cosines states that c² = a² + b² - 2ab·cos(C) for any triangle (with analogous formulas for the other sides), generalizing the Pythagorean theorem, and is used when you know two sides and the included angle (SAS) or all three sides (SSS). Given sides a=13, b=14, and angle C=60°, we apply the Law of Cosines: c² =13² +14² -2(13)(14)·cos(60°)=169+196-364·(0.5)=365-182=183, so c=√183. Choice A is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice C forgets the crucial -2ab·cos(C) term in the Law of Cosines, computing only 13² +14²=365 when the full formula is c²=13² +14² -2(13)(14)·cos(60°). Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines. Remember that the Law of Cosines generalizes the Pythagorean theorem: when the angle is 90°, cos(90°) = 0 makes the formula reduce to a² + b² = c², but for any other angle, you must include the -2ab·cos(C) term.

Question 6

1. $\sqrt{244}$
2. $\sqrt{84}$ (correct answer)
3. $\sqrt{164}$
4. $\sqrt{36}$

Explanation: This question tests understanding of the Law of Cosines for solving non-right triangles. The Law of Cosines states that c² = a² + b² - 2ab·cos(C) for any triangle (with analogous formulas for the other sides), generalizing the Pythagorean theorem, and is used when you know two sides and the included angle (SAS) or all three sides (SSS). Given sides a = 8, b = 10, and angle C = 60°, we apply the Law of Cosines: c² = 8² + 10² - 2(8)(10)·cos(60°) = 64 + 100 - 160·(0.5) = 164 - 80 = 84, so c = √84. Choice B is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice C forgets the crucial -2ab·cos(C) term in the Law of Cosines, computing only a² + b² when the full formula is c² = a² + b² - 2ab·cos(C). Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines. Remember that the Law of Cosines generalizes the Pythagorean theorem: when the angle is 90°, cos(90°) = 0 makes the formula reduce to a² + b² = c², but for any other angle, you must include the -2ab·cos(C) term.

Question 7

1. $60^\circ$
2. $45^\circ$
3. $90^\circ$ (correct answer)
4. $120^\circ$

Explanation: This question tests understanding of the Law of Cosines for solving non-right triangles. The Law of Cosines states that c² = a² + b² - 2ab·cos(C) for any triangle (with analogous formulas for the other sides), generalizing the Pythagorean theorem, and is used when you know two sides and the included angle (SAS) or all three sides (SSS). With sides a = 9, b = 12, c = 15, we rearrange the Law of Cosines to solve for the angle: cos(C) = (a² + b² - c²)/(2ab) = (81 + 144 - 225)/(2·9·12) = 0/216 = 0, so C = arccos(0) = 90°. Choice C is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice D might result from a sign error, computing cos(C) = -0.5 instead of 0, which would give 120°. When using the Law of Cosines to find an angle, rearrange to cos(C) = (a² + b² - c²)/(2ab), compute the right side, then use arccos to find the angle, checking that the result is between 0° and 180°.

Question 8

1. 0
2. 1
3. 2 (correct answer)
4. Infinitely many

Explanation: This question tests understanding of when to apply the Law of Sines for solving non-right triangles, specifically the ambiguous SSA case. The Law of Sines states that in any triangle, the ratio of each side to the sine of its opposite angle is constant: a/sin(A) = b/sin(B) = c/sin(C), and is used when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA). Since we know a = 8, b = 10, and angle A = 30° (SSA configuration), we first find sin(B) using the Law of Sines: sin(B) = b·sin(A)/a = 10·sin(30°)/8 = 10·(0.5)/8 = 5/8 = 0.625. Since sin(B) = 0.625 < 1 and B could be either arcsin(0.625) ≈ 38.7° or 180° - 38.7° ≈ 141.3°, and both give valid triangles (since A + B < 180° in both cases), there are two possible triangles. Choice C is correct because it correctly identifies that the SSA configuration with these specific values yields two valid triangles. Choice B incorrectly suggests only one triangle exists, missing the second solution in the ambiguous case. In the SSA case (two sides and non-included angle), be alert for the ambiguous case: there might be two possible triangles, one triangle, or no triangle, depending on the specific values—always check if a second solution exists.

Question 9

1. $10$
2. $5\sqrt{2}$ (correct answer)
3. $10\sqrt{2}$
4. $5$

Explanation: This question tests understanding of the Law of Sines for solving non-right triangles. The Law of Sines states that in any triangle, the ratio of each side to the sine of its opposite angle is constant: a/sin(A) = b/sin(B) = c/sin(C), and is used when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA). Knowing angle A = 45°, angle B = 45°, and side c = 10, we first find C = 180° - 45° - 45° = 90°, then use the Law of Sines: a/sin(A) = c/sin(C), so a = 10·sin(45°)/sin(90°) = 10·(√2/2)/1 = 5√2. Choice B is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice C uses the wrong angle-side pairing in the Law of Sines, matching angle A with side c instead of its opposite side a. Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines. For the Law of Sines, always match each angle with its opposite side: angle A with side a, angle B with side b, angle C with side c—using the wrong pairing will give an incorrect answer.

Question 10

1. $45\sqrt{3}$ (correct answer)
2. $\frac{45\sqrt{3}}{2}$
3. $90$
4. $\frac{90\sqrt{3}}{3}$

Explanation: First, find side c using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C = 144 + 225 - 2(12)(15)\cos(60°) = 369 - 360(\frac{1}{2}) = 189$. So $c = \sqrt{189} = 3\sqrt{21}$. The area is $\frac{1}{2}ab\sin C = \frac{1}{2}(12)(15)\sin(60°) = 90 \cdot \frac{\sqrt{3}}{2} = 45\sqrt{3}$. Choice B uses the wrong area formula. Choice C forgets the $\sin(60°)$ factor. Choice D incorrectly simplifies $45\sqrt{3}$.

Question 11

1. Both laws apply equally well, but the Law of Cosines is more efficient
2. The Law of Sines cannot be applied because one angle is $90°$
3. The Law of Cosines will give $\cos C = 0$, confirming the right angle at C (correct answer)
4. The Law of Sines will give undefined results due to the right angle

Explanation: When you encounter a triangle where $a^2 + b^2 = c^2$, you're looking at a right triangle by the converse of the Pythagorean theorem. Since $c$ is the longest side, the right angle is opposite to side $c$, meaning angle $C = 90°$. Both the Law of Sines and Law of Cosines work perfectly with right triangles, but they reveal the right angle in different ways. Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C$. Since we know $c^2 = a^2 + b^2$, we can substitute: $a^2 + b^2 = a^2 + b^2 - 2ab\cos C$. This simplifies to $0 = -2ab\cos C$, which means $\cos C = 0$. Since $\cos 90° = 0$, this confirms that angle $C = 90°$. This makes option C correct. Option A is wrong because while both laws apply, the Law of Cosines isn't necessarily more "efficient" — it depends on what you're solving for. Option B contains a fundamental misconception: the Law of Sines works perfectly fine with right triangles. In fact, $\frac{\sin 90°}{c} = \frac{1}{c}$ is well-defined since $\sin 90° = 1$. Option D is similarly incorrect — the Law of Sines gives perfectly defined results with right triangles, not undefined ones. Study tip: Remember that both trigonometric laws work with all triangles, including right triangles. The Law of Cosines is particularly useful for confirming right angles because $\cos 90° = 0$ creates a clean algebraic relationship.

Question 12

1. $0$
2. $1$
3. $2$ (correct answer)
4. $3$

Explanation: This question tests understanding of the Law of Sines and the ambiguous case for solving non-right triangles. The Law of Sines states that in any triangle, the ratio of each side to the sine of its opposite angle is constant: $a / \sin(A) = b / \sin(B) = c / \sin(C)$, and is used when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA). Given $A = 30^\circ$, $a = 8$, $b = 10$, this is an SSA configuration, which requires the Law of Sines because we have two sides and a non-included angle; since A is acute, $a > b \sin A = 5$, and $a < b$, there are two possible triangles. Choice C is correct because it correctly identifies the two solutions in the SSA ambiguous case. Choice B identifies only one solution in the SSA ambiguous case when actually two triangles satisfy the given conditions. In the SSA case (two sides and non-included angle), be alert for the ambiguous case: there might be two possible triangles, one triangle, or no triangle, depending on the specific values—always check if a second solution exists. Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines.

Question 13

1. The denominator should be $2(7)(12)$ instead of $2(7)(9)$
2. The formula should use addition instead of subtraction in the numerator
3. The formula should use $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ (correct answer)
4. The formula is correct, but the student should find $\sin A$ instead

Explanation: When you encounter a triangle with three known sides, you're working with the Law of Cosines, which relates the sides and angles of any triangle. The key is understanding which form of the formula to use based on what you're trying to find. The Law of Cosines states that for any triangle with sides a, b, c opposite to angles A, B, C respectively: $c^2 = a^2 + b^2 - 2ab\cos C$. When you want to find an angle (like angle A), you need to rearrange this formula to solve for the cosine of that angle. To find $\cos A$, you use the version: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$. Notice that side a (opposite to angle A) appears as the subtracted term in the numerator, while the other two sides b and c are in the denominator. This gives us $\cos A = \frac{9^2 + 12^2 - 7^2}{2(9)(12)}$. The student's error is using the wrong form of the Law of Cosines. Choice C correctly identifies this mistake. Choice A is wrong because changing the denominator to $2(7)(12)$ would be finding $\cos B$, not $\cos A$. Choice B is incorrect because the Law of Cosines requires subtraction, not addition, in the numerator. Choice D is wrong because the formula itself is flawed, regardless of whether we want sine or cosine. Remember: in the Law of Cosines, the side opposite to the angle you're finding is always the one being subtracted in the numerator, and the other two sides appear in the denominator.

Question 14

1. Use the Law of Cosines first because two sides and the included angle are known.
2. Use the Law of Sines first because an angle-side opposite pair is known. (correct answer)
3. Use the Pythagorean theorem because the triangle must be right.
4. Use the Law of Cosines first because all three sides are known.

Explanation: This question tests understanding of when to apply each law for solving non-right triangles. To decide which law to use: Law of Sines when you have angle-side opposite pairs to work with, Law of Cosines when you have two sides and the included angle or three sides with no angles. Since we know two angles and one side opposite one of them (AAS), this is an AAS configuration, which requires the Law of Sines because we have angle-side opposite pairs to work with. Choice B is correct because it correctly identifies which law is needed for this configuration. Choice A incorrectly assumes we have two sides and the included angle known, but only one side is given. Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines. For the Law of Sines, always match each angle with its opposite side: angle A with side a, angle B with side b, angle C with side c—using the wrong pairing will give an incorrect answer.

Question 15

1. Law of Sines, because an opposite side-angle pair is given.
2. Law of Cosines, because two sides and the included angle are given (SAS). (correct answer)
3. Pythagorean Theorem, because the triangle is right.
4. Neither; there is not enough information to start.

Explanation: This question tests understanding of when to apply the Law of Cosines versus the Law of Sines for solving triangles. To decide which law to use: Law of Sines when you have angle-side opposite pairs to work with, Law of Cosines when you have two sides and the included angle or three sides with no angles. Since we know two sides (a = 8 and b = 13) and the included angle (C = 120°), this is a SAS configuration, which requires the Law of Cosines because we can directly compute the third side using c² = a² + b² - 2ab·cos(C). Choice B is correct because it correctly identifies the SAS configuration and prescribes the appropriate law. Choice A incorrectly suggests using the Law of Sines, but we don't have an angle-side opposite pair to start with—we'd need to find side c first using the Law of Cosines. Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines.

Question 16

1. 0 triangles
2. 1 triangle
3. 2 triangles (correct answer)
4. Infinitely many triangles

Explanation: This question tests understanding of the Law of Sines in the ambiguous SSA case for solving non-right triangles. The Law of Sines states that in any triangle, the ratio of each side to the sine of its opposite angle is constant: a/sin(A) = b/sin(B) = c/sin(C), and is used when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA). Given angle A = 30°, side a = 10, side b = 12, this is an SSA configuration, which requires the Law of Sines because we have two sides and a non-included angle; the height h = b sinA =12 sin(30°) =12·(0.5)=6, and since 6 <10 <12 and A is acute, there are two triangles. Choice C is correct because it correctly identifies the ambiguous case with two solutions. Choice B identifies only one solution in the SSA ambiguous case when actually two triangles satisfy the given conditions. Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines. In the SSA case (two sides and non-included angle), be alert for the ambiguous case: there might be two possible triangles, one triangle, or no triangle, depending on the specific values—always check if a second solution exists.

Question 17

1. $\sqrt{67}$
2. $\sqrt{193}$ (correct answer)
3. $\sqrt{130}$
4. $16$

Explanation: This question tests understanding of the Law of Cosines for solving non-right triangles. The Law of Cosines states that c² = a² + b² - 2ab·cos(C) for any triangle (with analogous formulas for the other sides), generalizing the Pythagorean theorem, and is used when you know two sides and the included angle (SAS) or all three sides (SSS). Given sides a = 7, b = 9, and angle C = 120°, we apply the Law of Cosines: c² = 7² + 9² - 2(7)(9)·cos(120°) = 49 + 81 - 126·(-0.5) = 130 + 63 = 193, so c = √193. Choice B is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice A makes an arithmetic error in handling the negative cosine, subtracting instead of adding the term after the sign adjustment. Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines. Remember that the Law of Cosines generalizes the Pythagorean theorem: when the angle is 90°, cos(90°) = 0 makes the formula reduce to a² + b² = c², but for any other angle, you must include the -2ab·cos(C) term.

Question 18

1. Use Law of Sines: $\sin D = \frac{d \sin F}{f}$
2. Use Law of Cosines: $\cos D = \frac{e^2 + f^2 - d^2}{2ef}$ (correct answer)
3. Use the fact that $\angle D = 180° - \angle E - \angle F$
4. Use Law of Sines: $\sin D = \frac{f \sin F}{d}$

Explanation: While both Law of Sines and Law of Cosines could work, the Law of Cosines is more accurate when we know all three sides because it avoids the ambiguous case and rounding errors that can occur with inverse sine. The Law of Cosines gives a unique answer for the angle. Choice A uses the correct Law of Sines formula but is less accurate. Choice C requires finding ∠E first. Choice D has the Law of Sines formula backwards.

Question 19

1. $BC = \sqrt{200^2 + 150^2 - 2(200)(150)\cos(75°)}$ (correct answer)
2. $BC = \sqrt{200^2 + 150^2 + 2(200)(150)\cos(75°)}$
3. $\frac{BC}{\sin(75°)} = \frac{200}{\sin C}$
4. $BC = \frac{200 \cdot 150 \cdot \sin(75°)}{200 + 150}$

Explanation: This is a direct application of the Law of Cosines. Given two sides (AB = 200, AC = 150) and the included angle (∠A = 75°), we find the third side BC using $c^2 = a^2 + b^2 - 2ab\cos C$. Choice B incorrectly uses addition instead of subtraction. Choice C attempts the Law of Sines but we don't know angle C. Choice D uses an invalid formula that resembles an area calculation.

Question 20

1. $4\sqrt{2}$
2. $8\sqrt{3}$
3. $4\sqrt{6}$ (correct answer)
4. $8\sqrt{2}$

Explanation: This question tests understanding of the Law of Sines for solving non-right triangles. The Law of Sines states that in any triangle, the ratio of each side to the sine of its opposite angle is constant: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$, and is used when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA). Knowing $\angle A=60^\circ$, $\angle B=45^\circ$, and side $b=8$, we use the Law of Sines: $\frac{a}{\sin 60^\circ} = \frac{b}{\sin 45^\circ}$, so $a = 8 \cdot \frac{\sin 60^\circ}{\sin 45^\circ} = 8 \cdot \frac{\sqrt{3}/2}{\sqrt{2}/2} = 8 \cdot \frac{\sqrt{3}}{\sqrt{2}} = 8 \cdot \sqrt{\frac{3}{2}} = 4\sqrt{6}$. Choice C is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice A makes an arithmetic error in calculating $\sin 60^\circ / \sin 45^\circ$, getting a factor of $\sqrt{2}$ instead of $\sqrt{3/2}$. For the Law of Sines, always match each angle with its opposite side: angle A with side a, angle B with side b, angle C with side c—using the wrong pairing will give an incorrect answer. Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines.