Applying Laws of Sines and Cosines - Geometry

1. In triangle $XYZ$, you know that $x = 20$, $y = 16$, and $\angle Z = 45°$. When solving for angle $X$ using the Law of Sines after finding side $z$, which potential issue must you consider?
   Answer: There could be two possible values for angle $X$ (ambiguous case) When you have two sides and an angle opposite one of them (SSA configuration), you're dealing with a classic ambiguous case scenario in triangle problems. Here, you know sides $x = 20$ and $y = 16$, plus $\angle Z = 45°$ opposite the unknown side $z$. After using the Law of Cosines to find side $z$, you'll apply the Law of Sines: $\frac{\sin X}{x} = \frac{\sin Z}{z}$. This gives you $\sin X = \frac{x \sin Z}{z}$. The critical issue is that the sine function can produce two different angles in the range $0°$ to $180°$ for the same sine value. If $\sin X = k$, then $X$ could be either $\arcsin(k)$ or $180° - \arcsin(k)$. Both angles have identical sine values but are supplementary to each other. Looking at the wrong answers: (A) is incorrect because with these side lengths and angle, a valid triangle will exist. (B) is wrong since the Law of Sines applies perfectly once you know all three sides and one angle. (D) is incorrect because there's nothing in the given information that forces angle $X$ to be obtuse—in fact, you might get both an acute and obtuse solution. The answer is (C) because the ambiguous case means two possible triangles could satisfy the given conditions, leading to two different values for angle $X$. Study tip: Whenever you see SSA (two sides and a non-included angle), immediately think "ambiguous case." Always check if your sine calculation yields two possible angle measures.
2. A pilot flies from airport A to airport B (150 miles away) and then to airport C. The angle at airport B is $125°$, and the distance from B to C is 200 miles. To find the direct distance from A to C using the Law of Cosines, what is the correct setup?
   Answer: $AC^2 = 150^2 + 200^2 - 2(150)(200)\cos(125°)$ The Law of Cosines states $c^2 = a^2 + b^2 - 2ab\cos C$, where $C$ is the included angle between sides $a$ and $b$. Here, the angle at B ($125°$) is between the sides AB (150 miles) and BC (200 miles), so $AC^2 = 150^2 + 200^2 - 2(150)(200)\cos(125°)$. Choice B incorrectly uses the supplement of $125°$. Choice C attempts to use Law of Sines without sufficient information. Choice D has the wrong sign in the formula.
3. In triangle $ABC$, $\angle A = 35°$, $\angle B = 68°$, and side $a = 14$. If you need to find side $c$, which approach would give the most direct solution?
   Answer: Use Law of Sines: $c = \frac{a \sin C}{\sin A}$ where $C = 77°$ Since we know two angles and the side opposite one of them (ASA case), the Law of Sines provides the most direct approach. $\angle C = 180° - 35° - 68° = 77°$, then $\frac{c}{\sin C} = \frac{a}{\sin A}$, so $c = \frac{14 \sin 77°}{\sin 35°} \approx 23.8$. Choice B requires finding side $b$ first, making it less direct. Choice C is unnecessarily complicated. Choice D sets up the Law of Sines incorrectly with wrong angle relationships.
4. A surveyor needs to find the distance across a lake. From point $P$ on one shore, she measures the distance to point $Q$ on the opposite shore as 250 meters, and to point $R$ (also on the opposite shore) as 180 meters. If $\angle QPR = 42°$, what is the distance $QR$ to the nearest meter?
   Answer: 172 meters using the Law of Cosines Using the Law of Cosines: $QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos(\angle QPR) = 250^2 + 180^2 - 2(250)(180)\cos(42°) = 62500 + 32400 - 90000\cos(42°) \approx 94900 - 66879 = 28021$, so $QR \approx 172$ meters. Choice A results from incorrect angle usage. Choice C comes from misapplying the Law of Sines when the Law of Cosines is needed. Choice D results from treating this as a right triangle problem.
5. In triangle PQR, p = 7, q = 9, and ∠R = 120°. When using the Law of Cosines to find side r, what is the intermediate step that shows r² before taking the square root?
   Answer: r² = 49 + 81 + 63 = 193 When you encounter a triangle problem with two sides and the included angle, the Law of Cosines is your go-to tool. The formula is $c^2 = a^2 + b^2 - 2ab\cos(C)$, where C is the angle opposite side c. Here, you're finding side r (opposite angle R), so the formula becomes $r^2 = p^2 + q^2 - 2pq\cos(R)$. Substituting the given values: $r^2 = 7^2 + 9^2 - 2(7)(9)\cos(120°)$. First, calculate the squares: $7^2 = 49$ and $9^2 = 81$. Next, find $2(7)(9) = 126$. Now you need $\cos(120°)$. Since 120° is in the second quadrant, cosine is negative: $\cos(120°) = -\frac{1}{2}$. Substituting everything: $r^2 = 49 + 81 - 126(-\frac{1}{2}) = 49 + 81 - (-63) = 49 + 81 + 63 = 193$. This matches answer choice D. Answer A incorrectly shows $-126\cos(120°)$ as a single negative term, missing that cosine of 120° is already negative. Answer B makes the same error and also incorrectly changes the minus sign to plus in the Law of Cosines formula. Answer C shows the right final calculation (49 + 81 + 63) but gets the wrong sum. Remember: when the given angle is obtuse (greater than 90°), its cosine is negative. In the Law of Cosines, this creates a double negative that becomes positive, effectively adding rather than subtracting the final term.
6. Two ships leave port at the same time. Ship A travels at 15 mph in a direction N30°E, while Ship B travels at 20 mph in a direction S45°E. After 2 hours, what is the distance between the two ships to the nearest mile?
   Answer: 70 miles using the Law of Cosines After 2 hours, Ship A has traveled 30 miles and Ship B has traveled 40 miles. The angle between their paths is 30° + 90° + 45° = 165°. Using the Law of Cosines to find the distance between ships: d² = 30² + 40² - 2(30)(40)cos(165°) = 900 + 1600 - 2400cos(165°) ≈ 2500 + 2320 = 4820, so d ≈ 70 miles. Choice A uses simple vector addition incorrectly. Choice C misapplies the Law of Sines. Choice D results from using the wrong angle between the paths.
7. Refer to the triangle below. What is the measure of $\angle A$, to the nearest degree?
   Answer: $44°$ Using the Law of Cosines to solve for $\angle A$ (opposite side $a = 8$): $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{10^2 + 11^2 - 8^2}{2(10)(11)} = \dfrac{157}{220} \approx 0.7136$, so $A \approx 44°$. Choice B solves for the angle opposite side 11. Choice C solves for the angle opposite side 10. Choice D reflects a sign error in the numerator.
8. In triangle $ABC$ with sides $a = 13$, $b = 18$, and $c = 25$, a student calculates $\cos C = \frac{13^2 + 18^2 - 25^2}{2(13)(18)} = \frac{169 + 324 - 625}{468} = \frac{-132}{468}$. What can you conclude about angle $C$?
   Answer: Angle $C$ is obtuse since $\cos C < 0$, approximately $106°$ When you encounter a triangle with three given side lengths, the Law of Cosines is your tool for finding angles. The student's calculation using $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ is mathematically correct and reveals important information about the triangle's geometry. Let's verify: $\cos C = \frac{13^2 + 18^2 - 25^2}{2(13)(18)} = \frac{169 + 324 - 625}{468} = \frac{-132}{468} ≈ -0.282$. Since cosine is negative, angle C must be obtuse (greater than 90°). Using inverse cosine: $C = \cos^{-1}(-0.282) ≈ 106°$. This confirms answer C is correct. Now for the wrong answers: A incorrectly states the cosine is positive when the calculation clearly shows $\frac{-132}{468}$ is negative. B suggests the triangle cannot exist, but you can verify this triangle satisfies the triangle inequality (the sum of any two sides exceeds the third side). D claims the angle is exactly 90°, which would require $\cos C = 0$, not the negative value we calculated. The key insight is recognizing the connection between cosine signs and angle types: positive cosine means acute angles, zero cosine means right angles, and negative cosine means obtuse angles. When using the Law of Cosines, always check the sign of your result—it immediately tells you whether you're dealing with an acute or obtuse angle, which helps verify your final answer makes geometric sense.
9. Refer to the parallelogram $ABCD$ below. What is the length of diagonal $BD$, to the nearest tenth?
   Answer: $11.0$ In parallelogram $ABCD$, consecutive angles are supplementary, so $\angle A = 180° - 65° = 115°$. In $\triangle ABD$: $BD^2 = AB^2 + AD^2 - 2(AB)(AD)\cos A = 7^2 + 6^2 - 2(7)(6)\cos 115° = 49 + 36 - 84(-0.4226) \approx 85 + 35.5 = 120.5$, so $BD \approx 11.0$. Choice B uses $\cos 65°$ directly (forgetting supplementary angle), giving about 8.7. Choice C is approximately diagonal $AC$. Choice D uses $\cos 65°$ with the wrong operations.

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