Solving Right Triangles: Pythagorean Theorem, Trigonometry - Geometry

1. In the right triangle shown, $\angle B$ is a right angle. The hypotenuse is $\overline{AC}$. If $AC=15$ and $\angle A=40^\circ$, what is the length of $BC$?
   Answer: $15\sin(40^\circ)$ This problem requires finding the side opposite to a given angle using trigonometry. We have hypotenuse AC = 15, angle A = 40°, and angle B is the right angle. Since BC is opposite to angle A and AC is the hypotenuse, we use sine: sin(A) = opposite/hypotenuse = BC/AC. Rearranging: BC = AC × sin(A) = 15sin(40°). The answer is 15sin(40°) because sine relates the opposite side to the hypotenuse. Students might incorrectly use cosine (15cos(40°)) if they confuse which side is opposite versus adjacent to angle A. When using trigonometry, always draw and label your triangle to clearly identify which sides are opposite, adjacent, and hypotenuse relative to your angle.
2. A right triangle $\triangle DEF$ is shown in the plane with a right angle at $E$. The hypotenuse is $DF=20$. The acute angle at $D$ is $35^\circ$. What is the length of leg $DE$?
   Answer: $20\cos(35^\circ)$ This problem asks for a leg of a right triangle given the hypotenuse and an acute angle. We have a right angle at E, hypotenuse DF = 20, and angle D = 35°. Since we know an angle and the hypotenuse, we use trigonometry rather than the Pythagorean theorem. To find leg DE (adjacent to angle D), we use cosine: cos(35°) = adjacent/hypotenuse = DE/20. Solving for DE: DE = 20·cos(35°). This is the correct expression because cosine relates the adjacent side to the hypotenuse. A common error would be using sine (which gives the opposite side) or dividing by cosine. When given an angle and the hypotenuse, multiply by the appropriate trig function to find a leg.
3. In the diagram, $\triangle MNO$ is a right triangle with the right angle explicitly marked at $N$. The hypotenuse is $\overline{MO}$. The side $NO$ is labeled $7$, and the acute angle at $O$ is labeled $28^\circ$. Which expression represents the correct setup to find the length of the hypotenuse $\overline{MO}$? (Diagram is not drawn to scale; no other angles are marked.)
   Answer: $\cos 28^\circ=\dfrac{7}{MO}$ Solving right triangles involves finding unknown sides or angles using the Pythagorean theorem or trigonometric ratios. In this problem, the adjacent side is 7, and the acute angle is 28°. Cosine applies because it relates the adjacent side to the hypotenuse for the given angle. The equation is cos(28°) = 7 / MO. This matches choice B, as it sets up correctly for solving MO. A common misconception is using sine for the adjacent side, leading to sin(28°) = 7 / MO in choice A. To transfer this strategy, always choose between the Pythagorean theorem or trigonometry based on whether sides or angles are provided before computing.
4. In the coordinate plane, triangle $\triangle ABC$ is a right triangle with the right angle at $B$. The legs are $AB=6$ and $BC=8$. What is the length of hypotenuse $AC$? Give your answer in simplest form.
   Answer: $10$ This problem asks us to find the hypotenuse of a right triangle using the Pythagorean theorem. We're given that triangle ABC has a right angle at B, with legs AB = 6 and BC = 8. Since we know both legs of the right triangle, we can apply the Pythagorean theorem: a² + b² = c², where c is the hypotenuse. Setting up the equation: AC² = AB² + BC² = 6² + 8² = 36 + 64 = 100, so AC = √100 = 10. The answer is 10 because we take the positive square root when finding a length. A common error would be adding the legs directly (6 + 8 = 14) instead of using the Pythagorean theorem. When solving right triangles, identify what information you have first—if you have two sides, use the Pythagorean theorem; if you have an angle and a side, use trigonometry.
5. In the right triangle shown, $\angle A$ is a right angle. The hypotenuse is $\overline{BC}$. If $AB=8$ and $BC=17$, what is the length of $AC$?
   Answer: $15$ This problem asks us to find a leg of a right triangle using the Pythagorean theorem. We're given leg AB = 8 and hypotenuse BC = 17, with angle A being the right angle. Since we need to find a leg when given the other leg and hypotenuse, we use: a² = c² - b². Setting up: AC² = BC² - AB² = 17² - 8² = 289 - 64 = 225, so AC = √225 = 15. The answer is 15, not √225, because we must simplify the square root. A student might add the squares instead (getting √353) by incorrectly treating this as finding a hypotenuse. Always check: is the unknown side the hypotenuse (longest side) or a leg before deciding whether to add or subtract.
6. In right triangle $\triangle PQR$, the right angle is at $Q$. The hypotenuse is $PR=13$, and one leg is $QR=5$. What is the length of $PQ$?
   Answer: $12$ This problem requires finding a leg of a right triangle when given the hypotenuse and the other leg. We have a right angle at Q, hypotenuse PR = 13, and leg QR = 5. The Pythagorean theorem applies because we know two sides: PR² = PQ² + QR². Rearranging to solve for the unknown leg: PQ² = PR² - QR² = 13² - 5² = 169 - 25 = 144, so PQ = √144 = 12. The answer is 12 because we solved correctly for the missing leg. A common mistake would be to add the squares instead of subtracting (√194 comes from incorrectly computing 13² + 5²). Remember: when finding a leg, subtract the known leg squared from the hypotenuse squared.
7. In right triangle ABC with right angle at C, if $\frac{\sin(A)}{\cos(A)} = \frac{4}{3}$, what is the value of $\sin(B) \cdot \cos(B)$?
   Answer: $\frac{12}{25}$ When you encounter trigonometric ratios in right triangles, remember that $\frac{\sin(A)}{\cos(A)} = \tan(A)$. This identity is crucial for solving problems efficiently. Given that $\tan(A) = \frac{4}{3}$, you can visualize this as a right triangle where the opposite side to angle A is 4 and the adjacent side is 3. Using the Pythagorean theorem, the hypotenuse equals $\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$. Since angles A and B are complementary in a right triangle (they sum to 90°), angle B's opposite side is 3 and its adjacent side is 4. Therefore: - $\sin(B) = \frac{3}{5}$ (opposite over hypotenuse) - $\cos(B) = \frac{4}{5}$ (adjacent over hypotenuse) So $\sin(B) \cdot \cos(B) = \frac{3}{5} \cdot \frac{4}{5} = \frac{12}{25}$, which is answer C. The wrong answers represent common calculation errors. Answer A ($\frac{24}{25}$) comes from incorrectly multiplying $\frac{3}{5} \cdot \frac{4}{5}$ as $\frac{3 \cdot 4 \cdot 2}{25}$. Answer B ($\frac{7}{25}$) results from adding the numerators instead of multiplying: $\frac{3+4}{25}$. Answer D ($\frac{16}{25}$) occurs when students confuse which angle they're working with and calculate $\cos^2(B)$ instead. Strategy tip: Always draw the triangle and label all three sides when given a tangent ratio. This visual approach prevents confusion about which angle you're analyzing and makes complementary angle relationships clear.
8. A right triangle $\triangle HIJ$ is shown with the right angle at $I$. The hypotenuse is $HJ=25$, and one leg is $IJ=20$. What is the length of leg $HI$?
   Answer: $\sqrt{225}$ This problem requires finding a leg when given the hypotenuse and the other leg. We have a right angle at I, hypotenuse HJ = 25, and leg IJ = 20. Using the Pythagorean theorem: HJ² = HI² + IJ², we can solve for the unknown leg. Rearranging: HI² = HJ² - IJ² = 25² - 20² = 625 - 400 = 225, so HI = √225 = 15. The answer is √225 or 15 (both represent the same value). This is another 3-4-5 triangle scaled by 5 (15-20-25). A common error would be computing √1025 by adding the squares instead of subtracting. When finding a leg, always subtract the known leg squared from the hypotenuse squared.
9. In the right triangle shown, $\angle C$ is a right angle. The hypotenuse is $\overline{AB}$. If $AC=7$ and $\angle A=58^\circ$, what is the length of $AB$?
   Answer: $\dfrac{7}{\cos(58^\circ)}$ This problem asks us to find the hypotenuse when given a leg and an acute angle. We have AC = 7 (adjacent to angle A = 58°) and need to find hypotenuse AB, with angle C being the right angle. Since we know the adjacent side to angle A and need the hypotenuse, we use cosine: cos(A) = adjacent/hypotenuse = AC/AB. Solving for AB: AB = AC/cos(A) = 7/cos(58°). The answer is 7/cos(58°) because we divide the adjacent side by cosine to get the hypotenuse. A common mistake is multiplying instead of dividing (7cos(58°)), which would give a shorter length than the leg itself. Remember: the hypotenuse is always the longest side, so when finding it, you'll divide by a trig ratio less than 1.
10. Given that $\cos(\alpha) = \frac{5}{13}$ where $\alpha$ is acute, what is the value of $\sin(90° - \alpha) + \cos(90° - \alpha)$?
    Answer: $\frac{17}{13}$ When you encounter trigonometric expressions involving complementary angles like $90° - \alpha$, remember that complementary angle identities are key: $\sin(90° - \alpha) = \cos(\alpha)$ and $\cos(90° - \alpha) = \sin(\alpha)$. Since you're given $\cos(\alpha) = \frac{5}{13}$ with $\alpha$ acute, you first need to find $\sin(\alpha)$. Using the Pythagorean identity $\sin^2(\alpha) + \cos^2(\alpha) = 1$: $\sin^2(\alpha) = 1 - \cos^2(\alpha) = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169}$ Since $\alpha$ is acute, $\sin(\alpha) = \frac{12}{13}$. Now you can evaluate the expression: $\sin(90° - \alpha) + \cos(90° - \alpha) = \cos(\alpha) + \sin(\alpha) = \frac{5}{13} + \frac{12}{13} = \frac{17}{13}$ Looking at the wrong answers: Choice A ($\frac{18}{13}$) likely comes from incorrectly calculating $\sin(\alpha)$ or making an arithmetic error. Choice B ($\frac{12}{13}$) represents just $\sin(\alpha)$ alone, missing the cosine term entirely. Choice D ($1$) might result from incorrectly thinking that $\sin(90° - \alpha) + \cos(90° - \alpha)$ equals the Pythagorean identity, but that's $\sin^2(\alpha) + \cos^2(\alpha) = 1$, not the sum of the functions themselves. The correct answer is C. Study tip: Always memorize complementary angle identities and practice finding missing trigonometric values using the Pythagorean identity—these appear frequently together on geometry exams.
11. In right triangle $\triangle WXY$ shown, $\angle X$ is a right angle. The hypotenuse is $WY=18$, and $\angle W=25^\circ$. Which expression represents the correct setup to find the length of leg $XY$?
    Answer: $\sin(25^\circ)=\dfrac{XY}{18}$ This problem asks for the correct trigonometric setup to find a leg in a right triangle. We have hypotenuse WY = 18, angle W = 25°, and need leg XY (opposite to angle W). Since we need the opposite side and have the hypotenuse, we use sine: sin(angle) = opposite/hypotenuse. The correct setup is sin(25°) = XY/18, which matches option B. This can be solved by multiplying both sides by 18. A common error is confusing sine with cosine or setting up the ratio incorrectly. Always identify whether the unknown side is opposite or adjacent to the given angle before choosing the trig function.
12. In the right triangle $\triangle ABC$ shown, $\angle C$ is a right angle. The hypotenuse is $\overline{AB}$. If $AC=6$ and $BC=8$, what is the length of $\overline{AB}$?
    Answer: $10$ This problem requires solving for the hypotenuse of a right triangle using the Pythagorean theorem. We are given the two legs: AC = 6 and BC = 8, and need to find the hypotenuse AB. Since we have both legs of a right triangle, the Pythagorean theorem applies: a² + b² = c², where c is the hypotenuse. Setting up the equation: 6² + 8² = AB², which gives us 36 + 64 = 100, so AB = √100 = 10. The answer is justified because 10² = 100 = 36 + 64. A common error would be adding the legs directly (6 + 8 = 14) instead of using the Pythagorean theorem. When solving right triangles, first identify what information you have to choose between the Pythagorean theorem and trigonometric ratios.
13. In the diagram, $\triangle PQR$ is a right triangle with the right angle explicitly marked at $Q$. The hypotenuse is $\overline{PR}$. The side $PQ$ is labeled $12$, and the acute angle at $P$ is labeled $35^\circ$. What is the length of $\overline{PR}$? (Diagram is not drawn to scale; no other angles are marked.)
    Answer: $\dfrac{12}{\cos 35^\circ}$ Solving right triangles involves finding unknown sides or angles using the Pythagorean theorem or trigonometric ratios. In this problem, one leg is 12, and an adjacent acute angle is 35°. Cosine applies because it relates the adjacent side to the hypotenuse for the given angle. The equation is cos(35°) = 12 / PR, so PR = 12 / cos(35°). This matches choice B, as it correctly isolates the hypotenuse. A common misconception is using sine instead, leading to 12 / sin(35°) in choice C, which would be for the opposite side. To transfer this strategy, always choose between the Pythagorean theorem or trigonometry based on whether sides or angles are provided before computing.
14. In the right triangle $\triangle MNO$ shown, $\angle N$ is a right angle and the hypotenuse is $\overline{MO}$. If $MN=7$ and $NO=24$, what is the length of $\overline{MO}$?
    Answer: $\sqrt{7^2+24^2}$ This problem requires finding the hypotenuse of a right triangle using the Pythagorean theorem. We are given the two legs: MN = 7 and NO = 24, and need to find the hypotenuse MO. Since we have both legs, the Pythagorean theorem applies: a² + b² = c². Setting up the equation: 7² + 24² = MO², which gives us MO² = 49 + 576 = 625, so MO = √(7² + 24²) = √625 = 25. This is justified because the hypotenuse squared equals the sum of the squares of the legs. A common error is subtracting the squares (√(24² - 7²)) as if finding a leg instead of the hypotenuse. When finding the hypotenuse, always add the squares of the two legs.
15. In the right triangle shown, $\angle A$ is a right angle. The hypotenuse is $\overline{BC}$. If $BC=20$ and $\angle B=35^\circ$, what is the length of $AB$?
    Answer: $20\cos(35^\circ)$ This problem involves finding a leg adjacent to a given angle in a right triangle using trigonometry. We have hypotenuse BC = 20, angle B = 35°, and angle A is the right angle. Since AB is adjacent to angle B and BC is the hypotenuse, we use cosine: cos(B) = adjacent/hypotenuse. Rearranging: AB = BC × cos(B) = 20cos(35°). The answer is 20cos(35°) because cosine relates the adjacent side to the hypotenuse. Students might incorrectly use sine (20sin(35°)) by confusing which side is opposite versus adjacent to the angle. When solving right triangles with trigonometry, always identify the relationship between the known angle and the side you're finding.
16. In the figure, triangle PQR is a right triangle with right angle at Q. If $\sin(P) = \frac{\sqrt{7}}{4}$, what is $\cos(R) - \sin(R)$?
    Answer: $\frac{\sqrt{7} - 3}{4}$ Since P and R are complementary angles in right triangle PQR, $\cos(R) = \sin(P) = \frac{\sqrt{7}}{4}$. To find $\sin(R) = \cos(P)$, we use $\cos^2(P) + \sin^2(P) = 1$: $\cos^2(P) = 1 - \frac{7}{16} = \frac{9}{16}$, so $\cos(P) = \frac{3}{4}$ (positive since P is acute). Therefore, $\sin(R) = \frac{3}{4}$, and $\cos(R) - \sin(R) = \frac{\sqrt{7}}{4} - \frac{3}{4} = \frac{\sqrt{7} - 3}{4}$. Choice B reverses the subtraction, while choices C and D incorrectly add the terms.
17. In the right triangle $\triangle PQR$ shown, $\angle Q$ is a right angle and the hypotenuse is $\overline{PR}$. The acute angle at $P$ is $22^\circ$, and $PQ=12$. Which expression represents the correct setup to find the length of $\overline{QR}$?
    Answer: $\tan 22^\circ=\dfrac{QR}{12}$ This problem asks for the correct setup to find the opposite side using trigonometry. We are given the adjacent side PQ = 12 and the angle at P = 22°, and need to find the opposite side QR. Since we have the adjacent side and need the opposite side, we use the tangent ratio: tan(angle) = opposite/adjacent. The correct setup is: tan(22°) = QR/12, which can be rearranged to find QR. This is justified because tangent relates the opposite side to the adjacent side. A common mistake is using sine or cosine, which would require the hypotenuse. When setting up trigonometric equations, identify which sides you have and need relative to the given angle.
18. In the right triangle $\triangle JKL$ shown, $\angle K$ is a right angle and the hypotenuse is $\overline{JL}$. The acute angle at $J$ is $58^\circ$, and $JL=20$. What is the length of $\overline{JK}$?
    Answer: $20\cos 58^\circ$ This problem involves finding the adjacent side of a right triangle using trigonometry. We are given the hypotenuse JL = 20 and the angle at J = 58°, and need to find the adjacent side JK. Since we have the hypotenuse and need the adjacent side, we use the cosine ratio: cos(angle) = adjacent/hypotenuse. Setting up the equation: cos(58°) = JK/20, which gives us JK = 20·cos(58°). This is justified because cosine relates the adjacent side to the hypotenuse. A common mistake is using sine (20·sin(58°)) which would give the opposite side instead. Before solving, identify which side you need relative to the given angle to choose the correct trigonometric function.
19. In the plane, right triangle $\triangle GHI$ is shown. The right angle is explicitly marked at $H$ ($\angle GHI = 90^\circ$). The hypotenuse is $\overline{GI}$ and is labeled $15$. The leg $\overline{GH}$ is labeled $9$. The leg $\overline{HI}$ is unlabeled. Which method should be used to find the length of $\overline{HI}$? (Diagram is not drawn to scale. No acute angle measures are given.)
    Answer: Use the Pythagorean Theorem with $15$ and $9$. Solving right triangles involves using the Pythagorean theorem or trigonometry to find unknown sides or angles. In this problem, we are given the hypotenuse GI = 15 and one leg GH = 9. Since we have the hypotenuse and one leg, and need the other leg HI, the Pythagorean theorem applies. The equation is HI² = 15² - 9². This setup correctly finds HI, matching the method in choice A. A common misconception is assuming a special triangle like 30-60-90 without angle information, as in choice D. To transfer this strategy, always choose whether to use the Pythagorean theorem or trig ratios based on the given information before computing.
20. In right triangle $\triangle JKL$, the right angle is at $K$. The hypotenuse is $JL=15$, and $\angle J = 50^\circ$. What is the length of leg $JK$?
    Answer: $15\cos(50^\circ)$ This problem requires finding a leg adjacent to a given angle in a right triangle. We have a right angle at K, hypotenuse JL = 15, and angle J = 50°. Since we know an angle and the hypotenuse, trigonometry is the appropriate method. To find leg JK (adjacent to angle J), we use cosine: cos(50°) = adjacent/hypotenuse = JK/15. Solving for JK: JK = 15·cos(50°). This correctly uses cosine to relate the adjacent side to the hypotenuse. A common mistake would be using sine (15·sin(50°)), which would give the opposite leg KL instead. Always identify which leg you're finding relative to the given angle before choosing the trig function.

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