Law of Sines and Cosines — Free Trigonometry Review Games.

Law of Sines: a/sin(A) = b/sin(B) = c/sin(C).

Law of Cosines applies to SAS (two sides and included angle) or SSS cases.

c^2 = a^2 + b^2 - 2ab*cos(C).

SSA with the Law of Sines can produce 0, 1, or 2 triangles.

cos(90)=0, so c^2=a^2+b^2, the Pythagorean theorem.

a/sin(A)=b/sin(B), so b=a*sin(B)/sin(A)=10*sin(45)/sin(30).

c^2 = 49+100-2(7)(10)cos(60) = 149-140(0.5) = 149-70 = 79.

SSA can produce 0, 1, or 2 possible triangles.

The largest angle is always opposite the longest side.

Area = (1/2)*a*b*sin(C) when you know two sides and the included angle.

Area = (1/2)(8)(6)sin(30) = (1/2)(48)(0.5) = 12.

Since a > b and A is acute, exactly one triangle is possible (no ambiguous case when the side opposite the given angle is longer).

49 = 9+25-30*cos(C), 49=34-30*cos(C), cos(C)=-15/30=-0.5, C=120 degrees.

\(s = (3+4+5)/2 = 12/2 = 6\).

\(s=6\). Area = \(\sqrt{6*3*2*1} = \sqrt{36} = 6\).

When two angles and a non-included side are known, the third angle follows from the angle sum property, and the Law of Sines then directly relates the known side to any other side. The choice 'Three sides (SSS)' is wrong because with no angles known there is no ratio to set equal, so the Law of Cosines must be used first. The general rule to remember is that Law of Sines requires at least one complete angle-side pair to form a usable ratio.

Rearranging \(a^2=b^2+c^2-2bc\cos(A)\) algebraically isolates \(\cos(A)\) as \(\frac{b^2+c^2-a^2}{2bc}\), matching the side opposite the desired angle. The option \(\frac{a^2+b^2-c^2}{2ab}\) is wrong because it actually solves for angle \(C\), not angle \(A\), since it isolates the side opposite \(C\). Always match the squared side being subtracted to the angle you are solving for.

If the given side opposite the known angle is too short to reach the base, no triangle can close, so zero triangles is possible, making it the minimum. The choice '2' is wrong as a minimum because two solutions only occurs in a specific mid-range condition, not as a guaranteed lower bound. Students should remember that SSA can yield 0, 1, or 2 triangles depending on how the given side compares to the triangle's height.

The extended Law of Sines states \(\frac{a}{\sin(A)} = 2R\), where \(R\) is the circumradius, so the ratio equals the diameter of the circumscribed circle. The distractor 'radius of the inscribed circle' is wrong because the inradius relates to area and semiperimeter, not to the sine ratio. This extended identity is a useful shortcut for finding circumradius problems in trigonometry.

SSA is ambiguous because the same two sides and non-included angle can be consistent with zero, one, or two valid triangle configurations depending on the geometry. The option 'always produces exactly two triangles' is wrong because many SSA setups produce only one triangle or none at all. Recognizing this ambiguity before solving prevents students from missing a second valid solution or asserting an impossible one.

AAS determines a unique triangle because once two angles are fixed, the third is forced, and the given side then fixes the triangle's exact size via the Law of Sines. The choice 'SSA' is wrong because that configuration is the classic ambiguous case that can yield multiple solutions. Students should memorize that ASA and AAS are always unique, while SSA requires careful checking.

With two sides and the included angle known, the Law of Cosines directly computes the third side because it incorporates the angle between the two known sides. The Law of Sines is wrong to use first here because it requires a known angle-side pair, which is not yet available before finding the third side. This ordering rule—Cosines first for SAS—is essential for efficient triangle solving.

With three sides and no angles, the Law of Cosines is the only tool that can produce an angle from side lengths alone, since it directly relates all three sides to one included angle. The Law of Sines is wrong to start with because it needs at least one known angle to form a valid ratio. This is why SSS problems always begin with the Law of Cosines before any sine ratios can be used.

SSA is ambiguous because the non-included angle does not pin down a single triangle shape, allowing the third side to swing into two different positions in some cases. The option 'Two sides and the included angle (SAS)' is wrong because the included angle fixes the triangle uniquely, leaving no ambiguity. Remembering that only SSA is ambiguous helps students quickly identify when to check for multiple solutions.

The Law of Cosines states \(a^2=b^2+c^2-2bc\cos(A)\), where the angle used is the one opposite the side being solved for, here angle \(A\) opposite side \(a\). The option using \(\sin(A)\) instead of \(\cos(A)\) is wrong because the Law of Cosines specifically uses cosine to generalize the Pythagorean theorem for non-right triangles. Always match the angle in the formula to the side you are computing.

The correct form pairs side \(b\) with angle \(B\), the angle directly opposite it, giving \(b^2=a^2+c^2-2ac\cos(B)\). The option using \(\cos(A)\) instead of \(\cos(B)\) is wrong because the angle in the formula must be opposite the side being solved for, not an arbitrary other angle. This side-angle matching rule applies to every version of the Law of Cosines.

If \(a\) is shorter than the height \(h=b\sin(A)\), side \(a\) cannot reach the base line to close the triangle, so no valid triangle exists. The option '1' is wrong because a single triangle only occurs when \(a\) equals the height exactly (forming a right triangle) or when \(a \geq b\). Sketching the height comparison is the fastest way to determine the number of solutions in SSA problems.

Since every triangle's interior angles sum to \(180^\circ\), subtracting the two known angles from \(180^\circ\) directly yields the third angle with no trigonometric law required. The 'Law of Cosines' option is wrong here because that law relates sides and one angle, not simply two known angles to a third. This angle-sum shortcut is often the first step before applying the Law of Sines in AAS or ASA problems.

Using the Law of Sines, \(\sin(B) = \frac{b\sin(A)}{a} = \frac{6\sin(50^\circ)}{8} \approx 0.5745\), giving \(B \approx 35.1^\circ\) after taking the inverse sine. The option \(\approx 44.9^\circ\) is wrong because it does not match the arcsine of the correctly computed ratio. Since \(a > b\) here, only the acute solution is valid, so no ambiguous second triangle needs to be checked.

The height for the ambiguous case test is \(h=b\sin(A)=8\sin(30^\circ)=8(0.5)=4\), which is then compared to the given side \(a\) to determine the number of triangles. The option '8' is wrong because that simply restates \(b\) without multiplying by \(\sin(A)\). This height calculation is the essential first step whenever solving an SSA triangle problem.

Since angle \(A\) is obtuse and the side opposite it, \(a=10\), is longer than \(b=8\), exactly one valid triangle exists because an obtuse angle can only pair with the larger side in a consistent triangle. The option '2' is wrong because two solutions never occur when the given angle is obtuse. Whenever the given angle is obtuse, students should immediately check that \(a>b\) rather than searching for a second solution.

By the Law of Cosines, \(c^2=9^2+12^2-2(9)(12)\cos(50^\circ) \approx 225-138.85 \approx 86.15\), so \(c \approx 9.28\). The option \(\approx 8.10\) is wrong because it does not match the square root of the correctly computed value \(86.15\). This SAS setup always requires the Law of Cosines first, since no angle-side pair is initially available for the Law of Sines.

Since \(6^2+8^2=36+64=100=10^2\), the triangle satisfies the Pythagorean theorem, so the angle opposite the longest side, \(10\), must be exactly \(90^\circ\). The option \(53.1^\circ\) is wrong because that value corresponds to the angle opposite the side of length \(8\), not \(10\). Recognizing Pythagorean triples lets students bypass the full Law of Cosines calculation for right triangles.

By the Law of Sines, \(a = \frac{b\sin(A)}{\sin(B)} = \frac{12\sin(40^\circ)}{\sin(60^\circ)} \approx \frac{7.71}{0.866} \approx 8.91\). The option \(\approx 9.80\) is wrong because it does not result from correctly dividing \(12\sin(40^\circ)\) by \(\sin(60^\circ)\). This AAS setup is unambiguous, so only one triangle and one value of \(a\) exist.

With \(s=\frac{9+10+11}{2}=15\), Heron's formula gives \(\text{Area}=\sqrt{15(15-9)(15-10)(15-11)}=\sqrt{15\cdot6\cdot5\cdot4}=\sqrt{1800}\approx 42.43\). The option \(\approx 38.10\) is wrong because it does not equal the square root of \(1800\). Heron's formula is especially useful for SSS triangles where no angle is known to use the \(\frac{1}{2}ab\sin(C)\) area formula.

By the Law of Cosines, \(\cos(C)=\frac{5^2+6^2-9^2}{2(5)(6)}=\frac{-20}{60}\approx -0.333\), so \(C=\arccos(-0.333)\approx 109.5^\circ\). The option \(\approx 88.2^\circ\) is wrong because it corresponds to a positive cosine value, not the negative value found here. A negative cosine result always signals that the angle is obtuse, which students should recognize immediately.

Since \(a=10\) is greater than \(b=6\), side \(a\) is long enough to guarantee it swings past \(b\) to close only one triangle, regardless of the height comparison. The option '2' is wrong because two solutions require \(a\) to be strictly between the height and \(b\), which cannot happen when \(a\) already exceeds \(b\). A quick rule: whenever \(a \geq b\) in SSA with an acute angle, exactly one triangle results.

Since the interior angles of a triangle sum to \(180^\circ\), \(C=180^\circ-70^\circ-45^\circ=65^\circ\). The option \(55^\circ\) is wrong because it does not correctly subtract both given angles from \(180^\circ\). This basic angle-sum step is often required before applying the Law of Sines to find remaining sides.

By the extended Law of Sines, \(2R=\frac{a}{\sin(A)}=\frac{10}{\sin(30^\circ)}=\frac{10}{0.5}=20\). The option '10' is wrong because it forgets to divide by \(\sin(30^\circ)=0.5\), which doubles the correct result. This extended identity connects triangle side-angle ratios directly to the circumscribed circle's size.

By the Law of Cosines, \(c^2=7^2+9^2-2(7)(9)\cos(120^\circ)=130-126(-0.5)=130+63=193\), so \(c=\sqrt{193}\approx 13.89\). The option \(\approx 12.20\) is wrong because it does not match the square root of the correctly computed value \(193\). Note that because \(C\) is obtuse, \(\cos(C)\) is negative, which increases \(c^2\) beyond \(a^2+b^2\).

By the Law of Cosines, \(\cos(A)=\frac{20^2+18^2-14^2}{2(20)(18)}=\frac{528}{720}\approx 0.733\), so \(A=\arccos(0.733)\approx 42.8^\circ\). The option \(\approx 60.0^\circ\) is wrong because it does not equal the arccosine of the correctly computed ratio \(0.733\). In SSS problems, always find the angle opposite the side you are solving for by placing that side's square in the numerator with a negative sign.

The height is \(h=b\sin(A)=9\sin(35^\circ)\approx 5.16\); since \(h<a<b\) (that is, \(5.16<6<9\)), two distinct triangles satisfy the given conditions. The option '1' is wrong because a single triangle only occurs when \(a=h\) or when \(a\geq b\), neither of which applies here. This three-way comparison of \(h\), \(a\), and \(b\) is the standard test for counting SSA solutions.

By the Law of Sines, \(\sin(B)=\frac{b\sin(A)}{a}=\frac{9\sin(25^\circ)}{12}\approx 0.317\), giving \(B\approx 18.5^\circ\). The option \(\approx 161.5^\circ\) is wrong because although it is a valid arcsine supplement mathematically, it is rejected here since \(a>b\) guarantees only one triangle exists. Always check whether \(a\geq b\) before considering the obtuse supplementary angle as a second solution.

The SAS area formula gives \(\text{Area}=\frac{1}{2}ab\sin(C)=\frac{1}{2}(10)(14)\sin(45^\circ)=70(0.7071)\approx 49.5\). The option \(\approx 70.0\) is wrong because it omits multiplying by \(\sin(45^\circ)\), simply using half the product of the two sides. This formula only requires two sides and the included angle, making it faster than Heron's formula when an angle is already known.

By the Law of Cosines, \(\cos(C)=\frac{4^2+5^2-8^2}{2(4)(5)}=\frac{-23}{40}\approx -0.575\), so \(C=\arccos(-0.575)\approx 125.1^\circ\). The option \(\approx 96.4^\circ\) is wrong because it does not match the arccosine of the correctly computed negative ratio. Whenever the square of the longest side exceeds the sum of the squares of the other two, the opposite angle will be obtuse.

The height is \(h=b\sin(A)=10\sin(50^\circ)\approx 7.66\), which is greater than the given side \(a=5\), meaning side \(a\) is too short to ever reach the base and close a triangle. The option 'Exactly one triangle exists' is wrong because that outcome only occurs when \(a\geq h\), which is not satisfied here. This comparison of \(a\) to \(h\) is critical for correctly identifying impossible SSA configurations before attempting further calculations.

Because angle \(A\) is obtuse and the side opposite it, \(a=20\), is greater than \(b=15\), the geometry allows exactly one valid triangle, since an obtuse angle can only be paired with the longer of the two given sides. The option '2' is wrong because two solutions are impossible when the given angle is obtuse, regardless of the specific side lengths. This obtuse-angle shortcut lets students skip the full height comparison used for acute-angle ambiguous cases.

Using the Law of Cosines, \(\cos(C)=\frac{13^2+14^2-15^2}{2(13)(14)}=\frac{140}{364}\approx 0.385\), giving \(C\approx 67.4^\circ\), which is larger than the other two computed angles of \(\approx 59.5^\circ\) and \(\approx 53.1^\circ\). The option \(\approx 53.1^\circ\) (opposite side \(13\)) is wrong because that angle is opposite the shortest side and is therefore the smallest, not the largest, angle. In any triangle, the largest angle is always opposite the longest side, a principle useful for quickly identifying which angle to compute first.

The height is \(h=b\sin(A)=10(0.5)=5\), so two triangles exist only when \(5<a<10\); since \(6\) falls strictly within this range, it produces two solutions. The option '\(4\)' is wrong because \(4<5=h\), meaning the side is too short to reach the base at all, producing zero triangles instead. Memorizing the inequality \(h<a<b\) as the exact window for two solutions is essential for correctly counting ambiguous case outcomes.

Comparing \(12^2=144\) to \(7^2+8^2=49+64=113\), since \(144>113\), the Law of Cosines gives a negative cosine for the angle opposite side \(12\), confirming the triangle is obtuse. The option 'Right, since \(12^2=7^2+8^2\)' is wrong because \(144 \neq 113\), so the Pythagorean equality does not hold. This side-square comparison is a fast way to classify a triangle's largest angle without fully computing it.

The height is \(h=b\sin(A)=9\sin(60^\circ)\approx 7.79\), which exceeds the given side \(a=4\), so side \(a\) cannot reach far enough to close any triangle. The option '1, a unique triangle' is wrong because a single solution requires \(a\geq h\), a condition clearly violated here since \(4<7.79\). This scenario illustrates why checking the height first can save time by ruling out impossible SSA configurations immediately.

With \(s=\frac{8+15+17}{2}=20\), Heron's formula gives \(\text{Area}=\sqrt{20(20-8)(20-15)(20-17)}=\sqrt{20\cdot12\cdot5\cdot3}=\sqrt{3600}=60\). The option \(52.5\) is wrong because it does not equal the square root of the correctly computed product \(3600\). Since \(8^2+15^2=17^2\), this triangle is actually a right triangle, so the area could also be quickly confirmed with \(\frac{1}{2}(8)(15)=60\).

The area formula for SAS gives \(\text{Area}=\frac{1}{2}ab\sin(C)=\frac{1}{2}(10)(14)\sin(60^\circ)=70(0.866)\approx 60.6\). The option \(\approx 70.0\) is wrong because it neglects the \(\sin(60^\circ)\) factor and simply uses half the product of the two sides. This two-step reasoning—first identifying the included angle, then applying \(\frac{1}{2}ab\sin(C)\)—avoids the need to first solve for the third side.

The height is \(h=b\sin(A)=12\sin(35^\circ)\approx 6.88\); since \(6.88<8<12\), side \(a\) falls strictly between the height and \(b\), producing exactly two valid triangles. The option '1' is wrong because a single triangle requires either \(a=h\) exactly or \(a\geq b\), neither of which is true here. This is a classic example of the ambiguous case requiring the full \(h<a<b\) inequality check.

Two angles and the included side form an ASA configuration, and once the third angle is found by subtracting from \(180^\circ\), the Law of Sines can directly relate the known side to the unknown sides. The option 'Law of Cosines, since two angles are known' is wrong because the Law of Cosines requires two sides and an included angle, not two angles and one side, to solve for a missing part. Recognizing ASA versus SAS versus SSS at the start of a problem determines which law is appropriate to apply first.

Since cosine is negative only for angles between \(90^\circ\) and \(180^\circ\) on the unit circle, a computed value of \(\cos(C)=-0.42\) guarantees that angle \(C\) is obtuse. The option 'Angle \(C\) is acute' is wrong because acute angles between \(0^\circ\) and \(90^\circ\) always yield a positive cosine value. Checking the sign of the cosine result is a quick way to verify whether a solved angle should be greater than or less than \(90^\circ\) before finishing a problem.

By the Law of Cosines, \(x^2=5^2+12^2-2(5)(12)\cos(120^\circ)=169-120(-0.5)=169+60=229\), so \(x=\sqrt{229}\approx 15.13\). The option \(\approx 13.00\) is wrong because it matches the Pythagorean result for a right angle rather than accounting for the obtuse \(120^\circ\) angle, which increases the third side's length. Because the included angle here is obtuse, the negative cosine term adds to rather than subtracts from the sum of squares, producing a longer third side than in a right triangle.

When \(A=90^\circ\), solving \(\sin(B)=\frac{b\sin(A)}{a}\) gives one acute angle \(B\), but its supplement \(180^\circ-B\) combined with \(A=90^\circ\) would exceed \(180^\circ\), so that second solution is always invalid. The option 'Always exactly two triangles' is wrong because the supplementary angle case is automatically eliminated whenever the given angle is already \(90^\circ\) or greater. This shows why true ambiguity (two solutions) can only occur when the given angle is strictly acute.

First, the Law of Cosines gives \(c^2=9^2+6^2-2(9)(6)\cos(40^\circ)\approx 117-82.7\approx 34.3\), so \(c\approx 5.86\); then the Law of Sines gives \(\sin(A)=\frac{a\sin(C)}{c}=\frac{9\sin(40^\circ)}{5.86}\approx 0.988\), so \(A\approx 81.1^\circ\). The option \(c\approx 6.50\), then \(A\approx 75.4^\circ\) is wrong because it does not follow from the correctly computed value of \(c^2\approx 34.3\). This two-law sequence—Cosines to find the missing side, then Sines to find a remaining angle—is a standard synthesis strategy for SAS triangle problems.

Since the sum of any two angles in a triangle must be less than \(180^\circ\), the student must add each candidate value of \(B\) to the known \(A=35^\circ\) and discard any solution where the sum reaches or exceeds \(180^\circ\). The option 'Because both values are always valid answers regardless of the angle sum' is wrong because accepting an invalid supplementary angle would produce an impossible triangle with angles summing to more than \(180^\circ\). This angle-sum check is the final verification step required in every ambiguous-case problem before declaring one or two valid triangles.