Proving, Applying Laws of Sines/Cosines
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Geometry › Proving, Applying Laws of Sines/Cosines
In the diagram, $ ABC$ is not a right triangle. The included angle at $A$ is $\theta$, with $AB=c$, $AC=b$, and $BC=a$.
Which setup correctly applies the Law of Cosines to solve for $\cos\theta$ in terms of $a$, $b$, and $c$?
Since the triangle is not right, $\cos\theta=0$.
$\cos\theta=\dfrac{a}{b+c}$.
$\cos\theta=\dfrac{a^2-b^2-c^2}{2bc}$.
$\cos\theta=\dfrac{b^2+c^2-a^2}{2bc}$.
Explanation
The skill is applying the Law of Cosines to solve for the cosine of an angle. The geometric setup involves triangle ABC with angle θ at A, sides AB = c, AC = b, BC = a. The derivation idea is to rearrange the standard formula to isolate cos θ. To apply the Law of Cosines, compute cos θ = (b² + c² - a²) / (2bc). This is justified as it derives from the projection and Pythagorean applications. A distractor negates terms incorrectly or assumes zero cosine. To transfer this strategy, ask why the formula isolates cosine before how to compute it.
In triangle $ABC$ (not right), an altitude $CD$ is drawn to $AB$. Given $AC=8$, $BC=11$, and $\angle A=35^\circ$, which plan correctly uses the altitude to justify a Law of Sines relationship and then find $\angle B$?
Use the Law of Cosines with $AB^2=AC^2+BC^2-2(AC)(BC)\cos C$, then set $C=90^\circ$ to find $B$.
Use $CD=AC\cos A$ and $CD=BC\cos B$ to get $AC\cos A=BC\cos B$, then solve for $B$.
Use $CD=AC\sin A$ and $CD=BC\sin B$ to get $AC\sin A=BC\sin B$, then solve for $\sin B$.
Apply the Law of Sines directly because it is a memorized formula; no altitude reasoning is needed.
Explanation
This problem demonstrates using an altitude to derive and apply the Law of Sines. The geometric setup has altitude CD to side AB, with AC = 8, BC = 11, and angle A = 35°. In right triangle ACD, sin A = CD/AC, giving CD = AC·sin A = 8·sin(35°), and in right triangle BCD, sin B = CD/BC, so CD = BC·sin B = 11·sin B. Setting these equal: 8·sin(35°) = 11·sin B, which solves for sin B and thus angle B. This approach justifies the Law of Sines relationship AC·sin A = BC·sin B through the common altitude CD. Option B incorrectly uses cosine instead of sine for the altitude relationship, while option C unnecessarily invokes the Law of Cosines. The altitude method shows why the Law of Sines works: it captures how angles relate to opposite sides through a common height.
In the diagram, $ ABC$ is not a right triangle. The included angle at $A$ is labeled $\theta$, with adjacent sides $AB=9$ and $AC=12$.
Which setup correctly finds the length of $BC$ when $\theta=120^\circ$, using reasoning consistent with the Law of Cosines (generalizing the Pythagorean Theorem to an obtuse included angle)?
Because the picture would look wide at $A$, estimate $BC\approx 3$.
Use $BC^2=9^2+12^2$ since the largest angle makes it a right triangle.
Use $BC^2=9^2+12^2-2(9)(12)\cos120^\circ$, then take $BC=\sqrt{9^2+12^2-2(9)(12)\cos120^\circ}$.
Use $\tfrac{BC}{\sin120^\circ}=\tfrac{12}{\sin120^\circ}$, so $BC=12$.
Explanation
The skill is applying the Law of Cosines for obtuse angles, generalizing Pythagorean. The geometric setup is triangle ABC with obtuse angle 120° at A, adjacent sides AB = 9, AC = 12, opposite BC. The derivation idea is to use negative cosine for obtuse angles, adding to the sum of squares. To apply the Law of Cosines, compute BC² = 9² + 12² - 2·9·12·cos 120°, with cos 120° = -0.5. This is justified as it accounts for the angle spreading the sides. A distractor ignores the cosine and assumes right triangle. To transfer this strategy, ask why obtuse angles increase the opposite side before how to calculate.
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 $\sin\theta=1$ so $\tfrac{a}{1}=\tfrac{b}{\sin B}$.
If $\theta=90^\circ$, then $a^2=b^2+c^2-2bc\cos90^\circ=b^2+c^2$.
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.
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 the diagram, $ ABC$ is not a right triangle. The included angle at $A$ is $\angle A=60^\circ$, with adjacent sides $AB=7$ and $AC=10$. The opposite side is $BC=a$.
Which setup both (i) uses the Law of Cosines as a generalization of the Pythagorean Theorem and (ii) correctly computes $a$?
Use $a^2=7^2+10^2-2(7)(10)\cos60^\circ$, so $a=\sqrt{79}$.
Assume a right triangle: $a^2=7^2+10^2$, so $a=\sqrt{149}$.
Use $\tfrac{a}{\sin60^\circ}=\tfrac{7}{\sin60^\circ}$, so $a=7$.
Since the diagram looks obtuse at $A$, take $a\approx 17$ by visual estimation.
Explanation
The skill is applying the Law of Cosines as a generalization of the Pythagorean theorem. The geometric setup is triangle ABC with angle 60° at A, adjacent sides AB = 7 and AC = 10, opposite side BC = a. The derivation idea is to incorporate the cosine of the included angle to adjust for non-right triangles. To apply the Law of Cosines, calculate a² = 7² + 10² - 2·7·10·cos 60°, resulting in a = √79. This is justified because cos 60° = 0.5 reduces the subtraction, yielding a value between the right-triangle case and others. A distractor assumes a right triangle, ignoring the angle adjustment. To transfer this strategy, ask why the law extends Pythagorean before how to plug in values.
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)?
Set $17^2=8^2+13^2-2(8)(13)\cos M$, so $\cos M=\frac{8^2+13^2-17^2}{2\cdot 8\cdot 13}$.
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 $\frac{17}{\sin M}=\frac{13}{\sin N}$ 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 non-right triangle $\triangle DEF$ has $DE=5$, $DF=12$, and $EF=13$. The included angle at $D$ is emphasized.
Which reasoning correctly decides whether $\angle D$ is acute, right, or obtuse using the Law of Cosines as a generalization of the Pythagorean Theorem?
Because the diagram is not drawn to scale, you cannot classify $\angle D$ even with the side lengths given.
Use the Law of Sines with $\frac{13}{\sin D}$ and infer $\angle D$ from how large 13 is compared to 12.
Since $5+12=13$, the triangle must be right at $D$, so $\angle D=90^\circ$ by inspection.
Compute $\cos D=\frac{DE^2+DF^2-EF^2}{2(DE)(DF)}$; since this value is $0$, conclude $\angle D$ is right.
Explanation
The skill involves proving and applying the Law of Cosines to classify angles. In triangle DEF, sides DE and DF enclose angle D, with opposite side EF. The derivation idea uses cosine sign to determine acute, right, or obtuse compared to Pythagorean. Apply the law by computing cos D = (5² + 12² - 13²)/(2·5·12), yielding 0 for right angle. This is correct because zero cosine confirms 90° precisely. A distractor like choice B misuses side sum instead of squares. To transfer this strategy, always ask why cosine sign indicates angle type before how to compute it.
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 $YZ^2=XY^2+XZ^2-2(XY)(XZ)\cos 30^\circ$, 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 $X$.
Use $YZ=XY+XZ-\cos 30^\circ$ because the cosine term is subtracted from the sum of the sides.
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 $\triangle XYZ$ 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 $YZ^2=XY^2+XZ^2-2(XY)(XZ)\cos 30^\circ$, 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 $X$.
Use $YZ=XY+XZ-\cos 30^\circ$ because the cosine term is subtracted from the sum of the sides.
Use the Law of Sines: $\frac{YZ}{\sin 30^\circ}=\frac{11}{\sin Y}$ and choose $\angle Y$ 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.
In triangle $ABC$ (not right), an altitude $CD$ is drawn to $AB$. Suppose you know $AC$ and $BC$ and you measure $\angle A$ and $\angle B$. Which reasoning correctly explains why $\dfrac{AC}{\sin B}=\dfrac{BC}{\sin A}$ must be true using the altitude (not memorized formulas)?
Because $CD=AC\sin A$ and $CD=BC\sin B$, so $AC\sin A=BC\sin B$ and the ratio follows by rearranging.
Because $\triangle ABC$ is a right triangle once the altitude is drawn, so the sine ratios are the same in both smaller triangles.
Because $CD=AC\sin B$ and $CD=BC\sin A$, so $AC\sin B=BC\sin A$ and the ratio follows.
Because the diagram appears symmetric about $CD$, so $AC=BC$ and therefore $\dfrac{AC}{\sin B}=\dfrac{BC}{\sin A}$.
Explanation
This problem asks for a geometric justification of the Law of Sines using an altitude. The setup has altitude CD from C to side AB, creating two right triangles where we can apply basic trigonometry. In right triangle ACD, sin A = CD/AC, which gives CD = AC·sin A, and in right triangle BCD, sin B = CD/BC, which gives CD = BC·sin B. Since both expressions equal the same altitude CD, we have AC·sin A = BC·sin B. Rearranging this equality by dividing both sides by sin A·sin B gives AC/sin B = BC/sin A, which is the Law of Sines relationship. Option B reverses the angles in the sine expressions, while option C incorrectly claims the original triangle becomes right. This altitude-based reasoning shows why the Law of Sines must be true: it expresses the invariant relationship between a triangle's sides and opposite angles through a common height.