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Geometry

Geometry Help: Proving Applying Laws Of Sines Cosines

Review real example questions for Proving Applying Laws Of Sines Cosines in Geometry.

Question 1

A surveyor needs to find the area of a triangular plot of land. She measures two adjacent sides as 454545 meters and 606060 meters, with an included angle of 120°120°120°. Using an auxiliary line from the vertex of the 120°120°120° angle perpendicular to the opposite side, what is the area of the triangular plot?

  1. 135031350\sqrt{3}13503​ square meters
  2. 6753675\sqrt{3}6753​ square meters
  3. 135013501350 square meters
  4. 675675675 square meters
Explanation: Using the area formula A=12absin⁡(C)A = \frac{1}{2}ab\sin(C)A=21​absin(C) where a=45a = 45a=45, b=60b = 60b=60, and C=120°C = 120°C=120°. The area is 12⋅45⋅60⋅sin⁡(120°)=12⋅45⋅60⋅32=270034=6753\frac{1}{2} \cdot 45 \cdot 60 \cdot \sin(120°) = \frac{1}{2} \cdot 45 \cdot 60 \cdot \frac{\sqrt{3}}{2} = \frac{2700\sqrt{3}}{4} = 675\sqrt{3}21​⋅45⋅60⋅sin(120°)=21​⋅45⋅60⋅23​​=427003​​=6753​. Choice A incorrectly doubles the result. Choice C uses sin⁡(120°)=1\sin(120°) = 1sin(120°)=1 instead of 32\frac{\sqrt{3}}{2}23​​. Choice D omits the 3\sqrt{3}3​ factor entirely.

Question 2

In triangle PQRPQRPQR, side PQ=12PQ = 12PQ=12, side QR=15QR = 15QR=15, and angle Q=60°Q = 60°Q=60°. An auxiliary line is drawn from vertex PPP perpendicular to side QRQRQR. Which expression correctly represents the area of triangle PQRPQRPQR using the derived formula?

  1. 12⋅12⋅15⋅sin⁡(60°)=453\frac{1}{2} \cdot 12 \cdot 15 \cdot \sin(60°) = 45\sqrt{3}21​⋅12⋅15⋅sin(60°)=453​
  2. 12⋅12⋅15⋅cos⁡(60°)=45\frac{1}{2} \cdot 12 \cdot 15 \cdot \cos(60°) = 4521​⋅12⋅15⋅cos(60°)=45
  3. 12⋅12⋅15⋅sin⁡(30°)=45\frac{1}{2} \cdot 12 \cdot 15 \cdot \sin(30°) = 4521​⋅12⋅15⋅sin(30°)=45
  4. 12⋅12⋅15⋅tan⁡(60°)=903\frac{1}{2} \cdot 12 \cdot 15 \cdot \tan(60°) = 90\sqrt{3}21​⋅12⋅15⋅tan(60°)=903​
Explanation: The formula A=12absin⁡(C)A = \frac{1}{2}ab\sin(C)A=21​absin(C) uses the sine of the included angle between the two known sides. Here, sides PQ=12PQ = 12PQ=12 and QR=15QR = 15QR=15 form angle Q=60°Q = 60°Q=60°, so the area is 12⋅12⋅15⋅sin⁡(60°)=90⋅32=453\frac{1}{2} \cdot 12 \cdot 15 \cdot \sin(60°) = 90 \cdot \frac{\sqrt{3}}{2} = 45\sqrt{3}21​⋅12⋅15⋅sin(60°)=90⋅23​​=453​. Choice B incorrectly uses cosine instead of sine. Choice C uses the wrong angle (30° instead of 60°). Choice D incorrectly uses tangent instead of sine.

Question 3

Triangle XYZXYZXYZ has sides XY=10XY = 10XY=10, YZ=14YZ = 14YZ=14, and XZ=16XZ = 16XZ=16. To find the area using the formula A=12absin⁡(C)A = \frac{1}{2}ab\sin(C)A=21​absin(C), which of the following approaches requires drawing an auxiliary line and is mathematically sound?

  1. Find the altitude to side XZXZXZ, then use A=12⋅16⋅hA = \frac{1}{2} \cdot 16 \cdot hA=21​⋅16⋅h
  2. Find angle XXX using the Law of Sines, then calculate A=12⋅14⋅16⋅cos⁡(X)A = \frac{1}{2} \cdot 14 \cdot 16 \cdot \cos(X)A=21​⋅14⋅16⋅cos(X)
  3. Use Heron's formula directly since all three sides are known
  4. Find angle YYY using the Law of Cosines, then calculate A=12⋅10⋅14⋅sin⁡(Y)A = \frac{1}{2} \cdot 10 \cdot 14 \cdot \sin(Y)A=21​⋅10⋅14⋅sin(Y)
Explanation: When you're asked to find a triangle's area using A=12absin⁡(C)A = \frac{1}{2}ab\sin(C)A=21​absin(C), you need two sides and the included angle between them. Since you're only given the three side lengths (10, 14, and 16), you must first find one of the angles. Option D correctly identifies this approach: use the Law of Cosines to find angle YYY, then apply the area formula. The Law of Cosines states c2=a2+b2−2abcos⁡(C)c^2 = a^2 + b^2 - 2ab\cos(C)c2=a2+b2−2abcos(C). For angle YYY, you'd write 162=102+142−2(10)(14)cos⁡(Y)16^2 = 10^2 + 14^2 - 2(10)(14)\cos(Y)162=102+142−2(10)(14)cos(Y), solve for cos⁡(Y)\cos(Y)cos(Y), then find YYY. Finally, calculate A=12⋅10⋅14⋅sin⁡(Y)A = \frac{1}{2} \cdot 10 \cdot 14 \cdot \sin(Y)A=21​⋅10⋅14⋅sin(Y) using the two sides that form angle YYY. Option A uses the altitude formula A=12bhA = \frac{1}{2}bhA=21​bh, which is valid but doesn't use the required A=12absin⁡(C)A = \frac{1}{2}ab\sin(C)A=21​absin(C) formula specified in the question. Option B contains a critical error: it uses cos⁡(X)\cos(X)cos(X) instead of sin⁡(X)\sin(X)sin(X) in the area formula. The correct formula always uses sine, not cosine. Option C suggests Heron's formula, which works perfectly for this triangle but again doesn't use the specified A=12absin⁡(C)A = \frac{1}{2}ab\sin(C)A=21​absin(C) format. Remember: when you have three sides and need to use A=12absin⁡(C)A = \frac{1}{2}ab\sin(C)A=21​absin(C), always use the Law of Cosines first to find an angle, then apply the area formula with the two sides that form that angle.

Question 4

Two triangles have the same area calculated using A=12absin⁡(C)A = \frac{1}{2}ab\sin(C)A=21​absin(C). Triangle 1 has sides a1=8a_1 = 8a1​=8, b1=12b_1 = 12b1​=12, and included angle C1=30°C_1 = 30°C1​=30°. Triangle 2 has sides a2=6a_2 = 6a2​=6, b2=16b_2 = 16b2​=16, and included angle C2C_2C2​. What is the measure of angle C2C_2C2​?

  1. 60°60°60°
  2. 45°45°45°
  3. 30°30°30°
  4. 90°90°90°
Explanation: When you encounter problems involving the area formula A=12absin⁡(C)A = \frac{1}{2}ab\sin(C)A=21​absin(C), you're working with the relationship between two sides of a triangle and their included angle. Since both triangles have equal areas, you can set up an equation to find the unknown angle. First, calculate the area of Triangle 1: A1=12(8)(12)sin⁡(30°)=12(96)(12)=24A_1 = \frac{1}{2}(8)(12)\sin(30°) = \frac{1}{2}(96)(\frac{1}{2}) = 24A1​=21​(8)(12)sin(30°)=21​(96)(21​)=24. Since the triangles have equal areas, Triangle 2 also has area 24. Now set up the equation for Triangle 2: 24=12(6)(16)sin⁡(C2)24 = \frac{1}{2}(6)(16)\sin(C_2)24=21​(6)(16)sin(C2​). Simplifying: 24=48sin⁡(C2)24 = 48\sin(C_2)24=48sin(C2​), so sin⁡(C2)=2448=12\sin(C_2) = \frac{24}{48} = \frac{1}{2}sin(C2​)=4824​=21​. Therefore, C2=30°C_2 = 30°C2​=30°. Looking at the wrong answers: Choice (A) 60°60°60° would give sin⁡(60°)=32≈0.866\sin(60°) = \frac{\sqrt{3}}{2} \approx 0.866sin(60°)=23​​≈0.866, resulting in an area of about 41.6, which is too large. Choice (B) 45°45°45° would yield sin⁡(45°)=22≈0.707\sin(45°) = \frac{\sqrt{2}}{2} \approx 0.707sin(45°)=22​​≈0.707, giving an area of about 33.9, still too large. Choice (D) 90°90°90° would produce sin⁡(90°)=1\sin(90°) = 1sin(90°)=1, resulting in an area of 48, which is exactly double what we need. Remember that when using A=12absin⁡(C)A = \frac{1}{2}ab\sin(C)A=21​absin(C), the angle C must be the included angle between sides a and b. Always double-check your sine values for common angles—knowing that sin⁡(30°)=12\sin(30°) = \frac{1}{2}sin(30°)=21​ is essential for quick problem-solving.

Question 5

In the diagram, ABC is not a right triangle. The included angle at AAA is θ\thetaθ, with adjacent sides AB=cAB=cAB=c and AC=bAC=bAC=b, and opposite side BC=aBC=aBC=a.

A student claims: “If θ=90∘\theta=90^\circθ=90∘, then the Law of Cosines becomes the Pythagorean Theorem.” Which expression correctly shows this special case?

  1. If θ=90∘\theta=90^\circθ=90∘, then a2=b2+c2−2bccos⁡90∘=b2+c2a^2=b^2+c^2-2bc\cos90^\circ=b^2+c^2a2=b2+c2−2bccos90∘=b2+c2.
  2. If θ=90∘\theta=90^\circθ=90∘, then a=b+ca=b+ca=b+c because right triangles add sides.
  3. If θ=90∘\theta=90^\circθ=90∘, then sin⁡θ=1\sin\theta=1sinθ=1 so a1=bsin⁡B\tfrac{a}{1}=\tfrac{b}{\sin B}1a​=sinBb​.
  4. If θ=90∘\theta=90^\circθ=90∘, then a2=b2+c2+2bccos⁡90∘=b2+c2+2bca^2=b^2+c^2+2bc\cos90^\circ=b^2+c^2+2bca2=b2+c2+2bccos90∘=b2+c2+2bc.
Explanation: The skill is applying the Law of Cosines and verifying its special case as the Pythagorean theorem. The geometric setup is triangle ABC with included angle θ at A, sides b and c adjacent, a opposite. The derivation idea is to substitute θ = 90° where cos 90° = 0, simplifying the formula. To apply the Law of Cosines, get a² = b² + c² - 2bc · 0 = b² + c². This is justified because it matches the Pythagorean theorem for right angles at A. A distractor adds unnecessary terms or misapplies sine instead. To transfer this strategy, ask why the cosine vanishes at 90° before how to simplify.

Question 6

In non-right triangle ABCABCABC, the included angle ∠C\angle C∠C is shown between sides ACACAC and BCBCBC. Given AC=5AC=5AC=5, BC=9BC=9BC=9, and ∠C=120∘\angle C=120^\circ∠C=120∘, which reasoning correctly applies a law to find ABABAB and explains why this law still works when the angle is obtuse?

  1. Use the Law of Cosines since it generalizes the Pythagorean Theorem: AB2=AC2+BC2−2(AC)(BC)cos⁡CAB^2=AC^2+BC^2-2(AC)(BC)\cos CAB2=AC2+BC2−2(AC)(BC)cosC.
  2. Use the Pythagorean Theorem because any triangle with an obtuse angle can be split into two right triangles, so AB2=AC2+BC2AB^2=AC^2+BC^2AB2=AC2+BC2.
  3. Use the Law of Sines with ABsin⁡C=ACsin⁡C\dfrac{AB}{\sin C}=\dfrac{AC}{\sin C}sinCAB​=sinCAC​ to conclude AB=ACAB=ACAB=AC and then find ABABAB.
  4. Estimate ABABAB from the picture because ∠C\angle C∠C looks close to 180∘180^\circ180∘ so AB≈AC+BCAB\approx AC+BCAB≈AC+BC.
Explanation: This problem involves finding side AB when given two sides and their included obtuse angle. The geometric setup provides AC = 5, BC = 9, and the included angle C = 120°, making this a Side-Angle-Side (SAS) case with an obtuse angle. The Law of Cosines generalizes the Pythagorean theorem to any triangle: AB² = AC² + BC² - 2(AC)(BC)cos C. Applying this: AB² = 25 + 81 - 2(5)(9)cos(120°) = 106 - 90(-0.5) = 106 + 45 = 151. The law works for obtuse angles because cos(120°) = -0.5, making the correction term positive, which increases AB² beyond what the Pythagorean theorem would give. Option B wrongly claims the Pythagorean theorem applies to obtuse triangles, while option D relies on visual estimation. When dealing with obtuse angles, the negative cosine in the Law of Cosines accounts for the triangle "opening up" more than a right triangle.

Question 7

In the diagram, ABC is not a right triangle. Segment ADADAD is an altitude to BCBCBC (so AD⊥BCAD\perp BCAD⊥BC). Side AB=cAB=cAB=c, side AC=bAC=bAC=b, and angles at BBB and CCC are β\betaβ and γ\gammaγ.

A student wants to prove a Law of Sines relationship. Which statement correctly identifies the key equal quantities obtained from the right triangles △ABD\triangle ABD△ABD and △ACD\triangle ACD△ACD?

  1. AD=csin⁡βAD=c\sin\betaAD=csinβ and AD=bsin⁡γAD=b\sin\gammaAD=bsinγ, so csin⁡β=bsin⁡γc\sin\beta=b\sin\gammacsinβ=bsinγ.
  2. AD=ccos⁡βAD=c\cos\betaAD=ccosβ and AD=bcos⁡γAD=b\cos\gammaAD=bcosγ, so ccos⁡β=bcos⁡γ\tfrac{c}{\cos\beta}=\tfrac{b}{\cos\gamma}cosβc​=cosγb​.
  3. BD=csin⁡βBD=c\sin\betaBD=csinβ and DC=bsin⁡γDC=b\sin\gammaDC=bsinγ, so BD=DCBD=DCBD=DC.
  4. Since ADADAD is an altitude, β+γ=90∘\beta+\gamma=90^\circβ+γ=90∘, so sin⁡β=cos⁡γ\sin\beta=\cos\gammasinβ=cosγ.
Explanation: The skill is proving the Law of Sines using properties of right triangles formed by an altitude. The geometric setup features triangle ABC with altitude AD to BC, sides AB = c, AC = b, and angles β at B and γ at C. The derivation idea is to identify equal expressions for the altitude AD from trigonometric definitions in triangles ABD and ACD. To apply the Law of Sines, recognize AD = c sin β and AD = b sin γ, equating them for the relationship. This is correct as it directly leads to the proportional sides and sines after algebraic manipulation. A distractor incorrectly uses cosines instead of sines for the opposite side relations. To transfer this strategy, ask why the altitude is shared before how to express the sines.

Question 8

A non-right triangle △XYZ\triangle XYZ△XYZ has XY=6XY=6XY=6, XZ=11XZ=11XZ=11, and included angle ∠YXZ=30∘\angle YXZ=30^\circ∠YXZ=30∘ emphasized at XXX.

Which setup correctly uses a justification aligned with the Law of Cosines (and its Pythagorean special case) to find YZYZYZ?​

  1. Use YZ2=XY2+XZ2−2(XY)(XZ)cos⁡30∘YZ^2=XY^2+XZ^2-2(XY)(XZ)\cos 30^\circYZ2=XY2+XZ2−2(XY)(XZ)cos30∘, which reduces to the Pythagorean Theorem when the included angle is 90∘90^\circ90∘.
  2. Use YZ=XY+XZ−cos⁡30∘YZ=XY+XZ-\cos 30^\circYZ=XY+XZ−cos30∘ because the cosine term is subtracted from the sum of the sides.
  3. Use YZ2=XY2+XZ2YZ^2=XY^2+XZ^2YZ2=XY2+XZ2 because the included angle is given, so the triangle can be treated as right at XXX.
  4. Use the Law of Sines: YZsin⁡30∘=11sin⁡Y\frac{YZ}{\sin 30^\circ}=\frac{11}{\sin Y}sin30∘YZ​=sinY11​ and choose ∠Y\angle Y∠Y by visual estimation.
Explanation: The skill involves proving and applying the Law of Cosines for sides. In triangle XYZ, sides XY and XZ enclose acute angle YXZ at 30 degrees. The derivation idea adds a cosine term to generalize Pythagorean for any angle. Apply the law using YZ² = 6² + 11² - 2(6)(11)cos 30°, reducing to Pythagorean at 90°. This is correct because it connects to the special case effectively. A distractor like choice C treats it as right-angled without basis. To transfer this strategy, always ask why the formula generalizes Pythagorean before how to calculate the side.

Question 9

A non-right triangle △MNO\triangle MNO△MNO has sides MN=8MN=8MN=8, MO=13MO=13MO=13, and NO=17NO=17NO=17. The included angle at MMM is emphasized (between MNMNMN and MOMOMO).

Which setup correctly finds ∠M\angle M∠M using a justification consistent with the Law of Cosines (as a Pythagorean generalization)?​

  1. Set 172=82+132−2(8)(13)cos⁡M17^2=8^2+13^2-2(8)(13)\cos M172=82+132−2(8)(13)cosM, so cos⁡M=82+132−1722⋅8⋅13\cos M=\frac{8^2+13^2-17^2}{2\cdot 8\cdot 13}cosM=2⋅8⋅1382+132−172​.
  2. Set 17sin⁡M=13sin⁡N\frac{17}{\sin M}=\frac{13}{\sin N}sinM17​=sinN13​ and choose ∠N\angle N∠N by estimating it from the drawing.
  3. Assume ∠M=90∘\angle M=90^\circ∠M=90∘ because it is the included angle, then use 172=82+13217^2=8^2+13^2172=82+132.
  4. Write cos⁡M=178+13\cos M=\frac{17}{8+13}cosM=8+1317​ since cosine compares the opposite side to the sum of adjacent sides.
Explanation: The skill involves proving and applying the Law of Cosines to find angles in non-right triangles. In triangle MNO, sides MN and MO enclose angle M, with opposite side NO given. The derivation idea generalizes the Pythagorean theorem with a cosine adjustment for the included angle. Apply the law by setting 17² = 8² + 13² - 2(8)(13)cos M, solving for cos M. This is correct because it accurately isolates the angle using the formula's structure. A distractor like choice C assumes a right angle at M without justification. To transfer this strategy, always ask why the formula accounts for the angle before how to solve for cosine.

Question 10

A triangle has an area of 30230\sqrt{2}302​ square units. Two of its sides measure 121212 units and 101010 units respectively. If an auxiliary line is drawn from the vertex between these sides perpendicular to the opposite side, what is the measure of the included angle between the two known sides?

  1. 60°60°60°
  2. 30°30°30°
  3. 45°45°45°
  4. 135°135°135°
Explanation: When you encounter a triangle problem involving area, two sides, and an included angle, you're working with the area formula: A=12absin⁡CA = \frac{1}{2}ab\sin CA=21​absinC, where aaa and bbb are the two known sides and CCC is the angle between them. Given: area = 30230\sqrt{2}302​, sides = 12 and 10 units. Substituting into the formula: 302=12⋅12⋅10⋅sin⁡C30\sqrt{2} = \frac{1}{2} \cdot 12 \cdot 10 \cdot \sin C302​=21​⋅12⋅10⋅sinC 302=60sin⁡C30\sqrt{2} = 60\sin C302​=60sinC sin⁡C=30260=22\sin C = \frac{30\sqrt{2}}{60} = \frac{\sqrt{2}}{2}sinC=60302​​=22​​ Since sin⁡C=22\sin C = \frac{\sqrt{2}}{2}sinC=22​​, the angle C=45°C = 45°C=45° (or 135°135°135°). To determine which, consider that the auxiliary line mentioned is the altitude from the vertex between the known sides. This detail suggests we're working with the acute angle case, making C=45°C = 45°C=45°. Answer A (60°60°60°) would give sin⁡60°=32\sin 60° = \frac{\sqrt{3}}{2}sin60°=23​​, not 22\frac{\sqrt{2}}{2}22​​. Answer B (30°30°30°) would give sin⁡30°=12\sin 30° = \frac{1}{2}sin30°=21​, which is too small. Answer D (135°135°135°) technically satisfies sin⁡135°=22\sin 135° = \frac{\sqrt{2}}{2}sin135°=22​​, but the auxiliary line context indicates the acute angle. Study tip: Always memorize the exact sine values for special angles (30°30°30°, 45°45°45°, 60°60°60°). When you see 22\frac{\sqrt{2}}{2}22​​ as a sine value, immediately think 45°45°45°. The area formula with sine is crucial for any triangle where you know two sides and need the included angle.