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.

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.

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). In this SSA configuration with a=7, b=10, and angle A=30°, we first find the height h = b·sin(A) =10·sin(30°)=10·0.5=5; since angle A is acute, a>h (7>5), and a<b (7<10), there are two possible triangles. Choice C is correct because it correctly identifies 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. 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.

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=9, and angle C=120°, we apply the Law of Cosines: c²=11² +9² -2(11)(9)·cos(120°)=121+81-198·(-0.5)=202+99=301, so c=√301. Choice A is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice D uses the wrong sign in the Law of Cosines, subtracting 2ab·cos(C) instead of adding it since cos(120°)<0 makes -2ab·cos(C) positive. 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.

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.

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.

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) = (225 - 225)/216 = 0/216 = 0, so C = arccos(0) = 90°. Choice B is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice C makes an arithmetic error in calculating (a² + b² - c²), getting a non-zero value instead of 0. 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. 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°.

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 = 14, and angle C = 60°, this is an SAS configuration, which requires the Law of Cosines because we have two sides and the included angle. 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.

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.

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.