Applying Laws of Sines and Cosines - Precalculus

A triangle has sides of length $a$ , $b$ , and $c$ , where $a < b < c$ . If the triangle satisfies $a^2 + b^2 = c^2$ , which statement about using the Laws of Sines and Cosines is most accurate?
Answer: The Law of Cosines will give $\cos C = 0$ , confirming the right angle at C 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.
In triangle $ABC$ , $a=9$ , $b=12$ , and $c=15$ are given. What is the measure of angle $C$ (in degrees)?
Answer: $90^\circ$ 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 D is correct because it applies the correct law with proper substitution of values and accurate arithmetic, recognizing that this is a 3-4-5 right triangle scaled by 3. Choice C incorrectly gives 60°, which would require cos(C) = 0.5, but our calculation clearly shows cos(C) = 0. 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°.
In triangle $ABC$ , $a=9$ , $b=12$ , and the included angle $C=120^\circ$ are given. What is the length of side $c$ ?
Answer: $\sqrt{333}$ 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.
In triangle $ABC$ , sides $a=8$ and $b=10$ are known, and the included angle $C=60^\circ$ is known. What is the length of side $c$ (opposite $\angle C$ )?
Answer: $\sqrt{84}$ 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.
In triangle $ABC$ , sides $a=8$ and $b=15$ and the included angle $C=90^\circ$ are given. Which law should be used first to solve this triangle for side $c$ ?
Answer: Law of Cosines 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.
In triangle $ABC$ , $\angle A=30^\circ$ , side $a=10$ , and side $b=12$ are given (SSA, ambiguous case). How many triangles satisfy the given conditions?
Answer: 2 triangles 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.
In triangle $ABC$ , $A=45^\circ$ , $B=45^\circ$ , and $c=10$ (opposite $\angle C$ ). What is the length of side $a$ (opposite $\angle A$ )?
Answer: $5\sqrt{2}$ 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.
In triangle $ABC$ , sides $a=8$ and $b=10$ are known, and the included angle is $\angle C=60^\circ$ . What is the length of side $c$ (opposite $\angle C$ )?
Answer: $\sqrt{84}$ 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 uses the Pythagorean theorem 8² + 10² = c², which only works for right triangles, but this triangle has angle C = 60° (not 90°). 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.
In triangle $ABC$ , $\angle A=45^\circ$ , $\angle B=60^\circ$ , and side $a=12$ . What is the length of side $b$ ?
Answer: $6\sqrt{6}$ 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 = 60°, and side a = 12, we use the Law of Sines: a/sin(A) = b/sin(B), so b = a·sin(B)/sin(A) = 12·sin(60°)/sin(45°) = 12·(√3/2)/(√2/2) = 12·√3/√2 = 12√3/√2 · √2/√2 = 12√6/2 = 6√6. Choice A is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice B gives 6√3, which would result from incorrectly simplifying √3/√2 as √3 instead of rationalizing to get √6/2. 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.
In triangle $ABC$ , sides $a=11$ and $b=7$ are known and $C=60^\circ$ is known. What is the value of side $c$ (opposite $\angle C$ )?
Answer: $\sqrt{93}$ 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 = 11, b = 7, and angle C = 60°, we apply the Law of Cosines: c² = 11² + 7² - 2(11)(7)·cos(60°) = 121 + 49 - 154·(0.5) = 170 - 77 = 93, so c = √93. Choice B is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice D 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.
In triangle $ABC$ , side $a=9$ , side $b=12$ , and side $c=15$ . What is the measure of $\angle C$ (in degrees)?
Answer: $90^\circ$ 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°.
In triangle ABC, $a = 12$ , $b = 15$ , and $\angle C = 60°$ . What is the area of triangle ABC?
Answer: $45\sqrt{3}$ 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 $45\sqrt{3}$ .
In triangle $ABC$ , sides $a=11$ and $b=13$ and the included angle $C=60^\circ$ are given. Which equation correctly applies the Law of Cosines to find side $c$ ?
Answer: $c^2=11^2+13^2-2(11)(13)\cos(60^\circ)$ 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 sides and the included angle (SAS), this is an SAS configuration, which requires the Law of Cosines because we have two sides and the included angle to find the opposite side. Choice A is correct because it correctly identifies which law is needed for this configuration and sets up the formula with the proper sign. Choice C uses the Law of Sines when the Law of Cosines is needed—we have two sides and the included angle (SAS), not angle-side opposite pairs. 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.
In triangle $ABC$ , sides $a=9$ and $b=12$ are known and angle $C=60^\circ$ is known. Which equation correctly applies the Law of Cosines to find side $c$ ?
Answer: $c^2=9^2+12^2-2(9)(12)\cos(60^\circ)$ 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 sides and the included angle (SAS with a=9, b=12, C=60°), this is an SAS configuration, which requires the Law of Cosines because it involves the included angle between the known sides. Choice A is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice B uses the wrong sign in the Law of Cosines, adding 2ab·cos(C) instead of subtracting it. 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.
In triangle DEF, $d = 15$ , $e = 20$ , and $\angle F = 120°$ . After finding side $f$ using the Law of Cosines, which method would be most accurate for finding $\angle D$ ?
Answer: Use Law of Cosines: $\cos D = \frac{e^2 + f^2 - d^2}{2ef}$ 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.
In triangle ABC, $a = 7$ , $b = 9$ , and $c = 12$ . A student claims that $\cos A = \frac{7^2 + 9^2 - 12^2}{2(7)(9)}$ . What is wrong with this calculation?
Answer: The formula should use $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ 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.
A surveyor measures a triangular plot of land. From point A, the distance to point B is 200 meters and to point C is 150 meters. The angle at A is $75°$ . If the surveyor needs to find the distance from B to C, which calculation should be used?
Answer: $BC = \sqrt{200^2 + 150^2 - 2(200)(150)\cos(75°)}$ 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.
In triangle $ABC$ , $a=8$ , $b=10$ , and angle $A=30^\circ$ are given (SSA, ambiguous case). How many triangles satisfy the given conditions?
Answer: 2 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.
In triangle $ABC$ , sides $a=8$ , $b=10$ , and included angle $\angle C=60^\circ$ are given. To find side $c$ first, which approach is most appropriate?
Answer: Use the Law of Cosines because two sides and the included angle are known 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 sides and the included angle (SAS), this is an SAS configuration, which requires the Law of Cosines because the Law of Sines needs angle-side opposite pairs, not an included angle. Choice C is correct because it correctly identifies which law is needed for this configuration. Choice B incorrectly uses the Pythagorean theorem a² + b² = c², which only works for right triangles, but this triangle has angle C = 60°≠90°. 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.
In triangle $ABC$ , sides $a=8$ and $b=10$ are known and the included angle $C=60^\circ$ is known. What is the length of side $c$ (opposite $\angle C$ )?
Answer: $\sqrt{84}$ 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 uses the Pythagorean theorem 8² + 10² = c², which only works for right triangles, but this triangle has angle C = 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.

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