Proving, Applying Laws of Sines/Cosines
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Geometry › Proving, Applying Laws of Sines/Cosines
A surveyor needs to find the area of a triangular plot of land. She measures two adjacent sides as $$45$$ meters and $$60$$ meters, with an included angle of $$120°$$. Using an auxiliary line from the vertex of the $$120°$$ angle perpendicular to the opposite side, what is the area of the triangular plot?
$$1350$$ square meters
$$675$$ square meters
$$675\sqrt{3}$$ square meters
$$1350\sqrt{3}$$ square meters
Explanation
Using the area formula $$A = \frac{1}{2}ab\sin(C)$$ where $$a = 45$$, $$b = 60$$, and $$C = 120°$$. The area is $$\frac{1}{2} \cdot 45 \cdot 60 \cdot \sin(120°) = \frac{1}{2} \cdot 45 \cdot 60 \cdot \frac{\sqrt{3}}{2} = \frac{2700\sqrt{3}}{4} = 675\sqrt{3}$$. Choice A incorrectly doubles the result. Choice C uses $$\sin(120°) = 1$$ instead of $$\frac{\sqrt{3}}{2}$$. Choice D omits the $$\sqrt{3}$$ factor entirely.
In triangle $$PQR$$, side $$PQ = 12$$, side $$QR = 15$$, and angle $$Q = 60°$$. An auxiliary line is drawn from vertex $$P$$ perpendicular to side $$QR$$. Which expression correctly represents the area of triangle $$PQR$$ using the derived formula?
$$\frac{1}{2} \cdot 12 \cdot 15 \cdot \cos(60°) = 45$$
$$\frac{1}{2} \cdot 12 \cdot 15 \cdot \sin(60°) = 45\sqrt{3}$$
$$\frac{1}{2} \cdot 12 \cdot 15 \cdot \sin(30°) = 45$$
$$\frac{1}{2} \cdot 12 \cdot 15 \cdot \tan(60°) = 90\sqrt{3}$$
Explanation
The formula $$A = \frac{1}{2}ab\sin(C)$$ uses the sine of the included angle between the two known sides. Here, sides $$PQ = 12$$ and $$QR = 15$$ form angle $$Q = 60°$$, so the area is $$\frac{1}{2} \cdot 12 \cdot 15 \cdot \sin(60°) = 90 \cdot \frac{\sqrt{3}}{2} = 45\sqrt{3}$$. 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.
Two triangles have the same area calculated using $$A = \frac{1}{2}ab\sin(C)$$. Triangle 1 has sides $$a_1 = 8$$, $$b_1 = 12$$, and included angle $$C_1 = 30°$$. Triangle 2 has sides $$a_2 = 6$$, $$b_2 = 16$$, and included angle $$C_2$$. What is the measure of angle $$C_2$$?
$$90°$$
$$60°$$
$$30°$$
$$45°$$
Explanation
When you encounter problems involving the area formula $$A = \frac{1}{2}ab\sin(C)$$, 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: $$A_1 = \frac{1}{2}(8)(12)\sin(30°) = \frac{1}{2}(96)(\frac{1}{2}) = 24$$. Since the triangles have equal areas, Triangle 2 also has area 24.
Now set up the equation for Triangle 2: $$24 = \frac{1}{2}(6)(16)\sin(C_2)$$. Simplifying: $$24 = 48\sin(C_2)$$, so $$\sin(C_2) = \frac{24}{48} = \frac{1}{2}$$. Therefore, $$C_2 = 30°$$.
Looking at the wrong answers: Choice (A) $$60°$$ would give $$\sin(60°) = \frac{\sqrt{3}}{2} \approx 0.866$$, resulting in an area of about 41.6, which is too large. Choice (B) $$45°$$ would yield $$\sin(45°) = \frac{\sqrt{2}}{2} \approx 0.707$$, giving an area of about 33.9, still too large. Choice (D) $$90°$$ would produce $$\sin(90°) = 1$$, resulting in an area of 48, which is exactly double what we need.
Remember that when using $$A = \frac{1}{2}ab\sin(C)$$, the angle C must be the included angle between sides a and b. Always double-check your sine values for common angles—knowing that $$\sin(30°) = \frac{1}{2}$$ is essential for quick problem-solving.
In the diagram, $ ABC$ is not a right triangle. The included angle at $A$ is $\theta$, with adjacent sides $AB=c$ and $AC=b$, and opposite side $BC=a$.
A student claims: “If $\theta=90^\circ$, then the Law of Cosines becomes the Pythagorean Theorem.” Which expression correctly shows this special case?
If $\theta=90^\circ$, then $a^2=b^2+c^2+2bc\cos90^\circ=b^2+c^2+2bc$.
If $\theta=90^\circ$, then $a=b+c$ because right triangles add sides.
If $\theta=90^\circ$, then $a^2=b^2+c^2-2bc\cos90^\circ=b^2+c^2$.
If $\theta=90^\circ$, then $\sin\theta=1$ so $\tfrac{a}{1}=\tfrac{b}{\sin B}$.
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.
In non-right triangle $ABC$, the included angle $\angle C$ is shown between sides $AC$ and $BC$. Given $AC=5$, $BC=9$, and $\angle C=120^\circ$, which reasoning correctly applies a law to find $AB$ and explains why this law still works when the angle is obtuse?
Use the Law of Cosines since it generalizes the Pythagorean Theorem: $AB^2=AC^2+BC^2-2(AC)(BC)\cos C$.
Use the Law of Sines with $\dfrac{AB}{\sin C}=\dfrac{AC}{\sin C}$ to conclude $AB=AC$ and then find $AB$.
Estimate $AB$ from the picture because $\angle C$ looks close to $180^\circ$ so $AB\approx AC+BC$.
Use the Pythagorean Theorem because any triangle with an obtuse angle can be split into two right triangles, so $AB^2=AC^2+BC^2$.
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.
In the diagram, $ ABC$ is not a right triangle. Segment $AD$ is an altitude to $BC$ (so $AD\perp BC$). Side $AB=c$, side $AC=b$, and angles at $B$ and $C$ are $\beta$ and $\gamma$.
A student wants to prove a Law of Sines relationship. Which statement correctly identifies the key equal quantities obtained from the right triangles $\triangle ABD$ and $\triangle ACD$?
$BD=c\sin\beta$ and $DC=b\sin\gamma$, so $BD=DC$.
$AD=c\cos\beta$ and $AD=b\cos\gamma$, so $\tfrac{c}{\cos\beta}=\tfrac{b}{\cos\gamma}$.
$AD=c\sin\beta$ and $AD=b\sin\gamma$, so $c\sin\beta=b\sin\gamma$.
Since $AD$ is an altitude, $\beta+\gamma=90^\circ$, so $\sin\beta=\cos\gamma$.
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.
A non-right triangle $\triangle XYZ$ has $XY=6$, $XZ=11$, and included angle $\angle YXZ=30^\circ$ emphasized at $X$.
Which setup correctly uses a justification aligned with the Law of Cosines (and its Pythagorean special case) to find $YZ$?
Use the Law of Sines: $\frac{YZ}{\sin 30^\circ}=\frac{11}{\sin Y}$ and choose $\angle Y$ by visual estimation.
Use $YZ=XY+XZ-\cos 30^\circ$ because the cosine term is subtracted from the sum of the sides.
Use $YZ^2=XY^2+XZ^2-2(XY)(XZ)\cos 30^\circ$, which reduces to the Pythagorean Theorem when the included angle is $90^\circ$.
Use $YZ^2=XY^2+XZ^2$ because the included angle is given, so the triangle can be treated as right at $X$.
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.
A non-right triangle $\triangle MNO$ has sides $MN=8$, $MO=13$, and $NO=17$. The included angle at $M$ is emphasized (between $MN$ and $MO$).
Which setup correctly finds $\angle M$ using a justification consistent with the Law of Cosines (as a Pythagorean generalization)?
Assume $\angle M=90^\circ$ because it is the included angle, then use $17^2=8^2+13^2$.
Write $\cos M=\frac{17}{8+13}$ since cosine compares the opposite side to the sum of adjacent sides.
Set $17^2=8^2+13^2-2(8)(13)\cos M$, so $\cos M=\frac{8^2+13^2-17^2}{2\cdot 8\cdot 13}$.
Set $\frac{17}{\sin M}=\frac{13}{\sin N}$ and choose $\angle N$ by estimating it from the drawing.
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.
A triangle has an area of $$30\sqrt{2}$$ square units. Two of its sides measure $$12$$ units and $$10$$ 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?
$$30°$$
$$45°$$
$$135°$$
$$60°$$
Explanation
When you encounter a triangle problem involving area, two sides, and an included angle, you're working with the area formula: $$A = \frac{1}{2}ab\sin C$$, where $$a$$ and $$b$$ are the two known sides and $$C$$ is the angle between them.
Given: area = $$30\sqrt{2}$$, sides = 12 and 10 units. Substituting into the formula:
$$30\sqrt{2} = \frac{1}{2} \cdot 12 \cdot 10 \cdot \sin C$$
$$30\sqrt{2} = 60\sin C$$
$$\sin C = \frac{30\sqrt{2}}{60} = \frac{\sqrt{2}}{2}$$
Since $$\sin C = \frac{\sqrt{2}}{2}$$, the angle $$C = 45°$$ (or $$135°$$). 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 $$C = 45°$$.
Answer A ($$60°$$) would give $$\sin 60° = \frac{\sqrt{3}}{2}$$, not $$\frac{\sqrt{2}}{2}$$. Answer B ($$30°$$) would give $$\sin 30° = \frac{1}{2}$$, which is too small. Answer D ($$135°$$) technically satisfies $$\sin 135° = \frac{\sqrt{2}}{2}$$, but the auxiliary line context indicates the acute angle.
Study tip: Always memorize the exact sine values for special angles ($$30°$$, $$45°$$, $$60°$$). When you see $$\frac{\sqrt{2}}{2}$$ as a sine value, immediately think $$45°$$. The area formula with sine is crucial for any triangle where you know two sides and need the included angle.
In the derivation of the area formula $$A = \frac{1}{2}ab\sin(C)$$, an auxiliary line creates a right triangle. If the original triangle has sides $$a = 15$$, $$b = 20$$, and the included angle $$C = 150°$$, what is the length of the auxiliary line (height)?
$$15\sqrt{3}$$ units
$$10\sqrt{3}$$ units
$$7.5$$ units
$$10$$ units
Explanation
When the auxiliary line is drawn from the vertex with angle $$C$$ perpendicular to side $$b$$, it creates a height $$h$$. In the right triangle formed, $$h = a\sin(C) = 15\sin(150°) = 15 \cdot \frac{1}{2} = 7.5$$ units (since $$\sin(150°) = \sin(30°) = \frac{1}{2}$$). Choice A would result from using $$b = 20$$ instead of $$a = 15$$. Choice C incorrectly uses $$\sin(150°) = \frac{\sqrt{3}}{2}$$. Choice D combines the wrong sine value with side $$b$$.