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Geometry

Geometry Help: Solving Right Triangles Pythagorean Theorem Trigonometry

Review real example questions for Solving Right Triangles Pythagorean Theorem Trigonometry in Geometry.

Question 1

In the right triangle shown, ∠A\angle A∠A is a right angle. The hypotenuse is BC‾\overline{BC}BC. If BC=30BC=30BC=30 and ∠C=22∘\angle C=22^\circ∠C=22∘, what is the length of ABABAB?

  1. 30cos⁡(22∘)30\cos(22^\circ)30cos(22∘)
  2. 30sin⁡(22∘)\dfrac{30}{\sin(22^\circ)}sin(22∘)30​
  3. 30sin⁡(22∘)30\sin(22^\circ)30sin(22∘)
  4. 30cos⁡(22∘)\dfrac{30}{\cos(22^\circ)}cos(22∘)30​
Explanation: This problem requires finding the side opposite to a given angle in a right triangle. We have hypotenuse BC = 30, angle C = 22°, and angle A is the right angle. Since AB is opposite to angle C and BC is the hypotenuse, we use sine: sin(C) = opposite/hypotenuse = AB/BC. Rearranging: AB = BC × sin(C) = 30sin(22°). The answer is 30sin(22°) because sine relates the opposite side to the hypotenuse. A common error would be using cosine (30cos(22°)), which would give the adjacent side AC instead. When solving with trigonometry, always identify which side you need relative to the given angle: opposite uses sine, adjacent uses cosine.

Question 2

In triangle ABC, angle C is a right angle. If sin⁡(A)=35\sin(A) = \frac{3}{5}sin(A)=53​, what is the value of cos⁡(B)+sin⁡(B)\cos(B) + \sin(B)cos(B)+sin(B)?

  1. 75\frac{7}{5}57​
  2. 45\frac{4}{5}54​
  3. 35\frac{3}{5}53​
  4. 85\frac{8}{5}58​
Explanation: Since angles A and B are complementary in a right triangle, cos⁡(B)=sin⁡(A)=35\cos(B) = \sin(A) = \frac{3}{5}cos(B)=sin(A)=53​. To find sin⁡(B)\sin(B)sin(B), we use sin⁡(B)=cos⁡(A)\sin(B) = \cos(A)sin(B)=cos(A). Since sin⁡(A)=35\sin(A) = \frac{3}{5}sin(A)=53​, we can find cos⁡(A)\cos(A)cos(A) using the Pythagorean identity: cos⁡(A)=1−sin⁡2(A)=1−925=1625=45\cos(A) = \sqrt{1 - \sin^2(A)} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}cos(A)=1−sin2(A)​=1−259​​=2516​​=54​. Therefore, sin⁡(B)=45\sin(B) = \frac{4}{5}sin(B)=54​, and cos⁡(B)+sin⁡(B)=35+45=75\cos(B) + \sin(B) = \frac{3}{5} + \frac{4}{5} = \frac{7}{5}cos(B)+sin(B)=53​+54​=57​. Choice B gives only cos⁡(A)\cos(A)cos(A), choice C gives only sin⁡(A)\sin(A)sin(A), and choice D incorrectly adds the squares.

Question 3

Given that cos⁡(α)=513\cos(\alpha) = \frac{5}{13}cos(α)=135​ where α\alphaα is acute, what is the value of sin⁡(90°−α)+cos⁡(90°−α)\sin(90° - \alpha) + \cos(90° - \alpha)sin(90°−α)+cos(90°−α)?

  1. 1813\frac{18}{13}1318​
  2. 1213\frac{12}{13}1312​
  3. 1713\frac{17}{13}1317​
  4. 111
Explanation: When you encounter trigonometric expressions involving complementary angles like 90°−α90° - \alpha90°−α, remember that complementary angle identities are key: sin⁡(90°−α)=cos⁡(α)\sin(90° - \alpha) = \cos(\alpha)sin(90°−α)=cos(α) and cos⁡(90°−α)=sin⁡(α)\cos(90° - \alpha) = \sin(\alpha)cos(90°−α)=sin(α). Since you're given cos⁡(α)=513\cos(\alpha) = \frac{5}{13}cos(α)=135​ with α\alphaα acute, you first need to find sin⁡(α)\sin(\alpha)sin(α). Using the Pythagorean identity sin⁡2(α)+cos⁡2(α)=1\sin^2(\alpha) + \cos^2(\alpha) = 1sin2(α)+cos2(α)=1: sin⁡2(α)=1−cos⁡2(α)=1−(513)2=1−25169=144169\sin^2(\alpha) = 1 - \cos^2(\alpha) = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169}sin2(α)=1−cos2(α)=1−(135​)2=1−16925​=169144​ Since α\alphaα is acute, sin⁡(α)=1213\sin(\alpha) = \frac{12}{13}sin(α)=1312​. Now you can evaluate the expression: sin⁡(90°−α)+cos⁡(90°−α)=cos⁡(α)+sin⁡(α)=513+1213=1713\sin(90° - \alpha) + \cos(90° - \alpha) = \cos(\alpha) + \sin(\alpha) = \frac{5}{13} + \frac{12}{13} = \frac{17}{13}sin(90°−α)+cos(90°−α)=cos(α)+sin(α)=135​+1312​=1317​ Looking at the wrong answers: Choice A (1813\frac{18}{13}1318​) likely comes from incorrectly calculating sin⁡(α)\sin(\alpha)sin(α) or making an arithmetic error. Choice B (1213\frac{12}{13}1312​) represents just sin⁡(α)\sin(\alpha)sin(α) alone, missing the cosine term entirely. Choice D (111) might result from incorrectly thinking that sin⁡(90°−α)+cos⁡(90°−α)\sin(90° - \alpha) + \cos(90° - \alpha)sin(90°−α)+cos(90°−α) equals the Pythagorean identity, but that's sin⁡2(α)+cos⁡2(α)=1\sin^2(\alpha) + \cos^2(\alpha) = 1sin2(α)+cos2(α)=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.

Question 4

In the right triangle △DEF\triangle DEF△DEF shown, ∠E\angle E∠E is a right angle and the hypotenuse is DF‾\overline{DF}DF. The acute angle at DDD is 35∘35^\circ35∘, and DE=9DE=9DE=9. What is the length of DF‾\overline{DF}DF?

  1. 9sin⁡35∘\dfrac{9}{\sin 35^\circ}sin35∘9​
  2. 9cos⁡35∘\dfrac{9}{\cos 35^\circ}cos35∘9​
  3. 9cos⁡35∘9\cos 35^\circ9cos35∘
  4. 9tan⁡35∘9\tan 35^\circ9tan35∘
Explanation: This problem involves finding the hypotenuse of a right triangle using trigonometry. We are given the adjacent side DE = 9 and the angle at D = 35°, and need to find the hypotenuse DF. Since we have the adjacent side and need the hypotenuse, we use the cosine ratio: cos(angle) = adjacent/hypotenuse. Setting up the equation: cos(35°) = 9/DF, which rearranges to DF = 9/cos(35°). This is justified because cosine relates the adjacent side to the hypotenuse in a right triangle. A common mistake is using sine instead of cosine, which would give 9/sin(35°). Before computing, identify which sides you have relative to the given angle to select the correct trigonometric ratio.

Question 5

In right triangle △RST\triangle RST△RST, the right angle is at SSS. One leg is RS=10RS=10RS=10, and ∠R=28∘\angle R = 28^\circ∠R=28∘. What is the length of hypotenuse RTRTRT?

  1. 10cos⁡(28∘)10\cos(28^\circ)10cos(28∘)
  2. 10cos⁡(28∘)\dfrac{10}{\cos(28^\circ)}cos(28∘)10​
  3. 10sin⁡(28∘)10\sin(28^\circ)10sin(28∘)
  4. 10tan⁡(28∘)\dfrac{10}{\tan(28^\circ)}tan(28∘)10​
Explanation: This problem requires finding the hypotenuse when given a leg and an acute angle. We have a right angle at S, leg RS = 10, and angle R = 28°. Since we know an angle and a leg, we use trigonometry. To find hypotenuse RT, we note that RS is adjacent to angle R, so we use cosine: cos(28°) = adjacent/hypotenuse = 10/RT. Solving for RT: RT = 10/cos(28°). This correctly isolates the hypotenuse by dividing the adjacent side by cosine. A common error would be multiplying by cosine (10·cos(28°)), which would give a smaller value. When finding the hypotenuse using trigonometry, you divide the known side by the appropriate trig function.

Question 6

In the right triangle shown, ∠B\angle B∠B is a right angle. The hypotenuse is AC‾\overline{AC}AC. If AB=9AB=9AB=9 and BC=12BC=12BC=12, what is the length of ACACAC?

  1. 151515
  2. 21\sqrt{21}21​
  3. 63\sqrt{63}63​
  4. 225\sqrt{225}225​
Explanation: This problem asks us to find the hypotenuse of a right triangle using the Pythagorean theorem. We're given the two legs: AB = 9 and BC = 12, with angle B being the right angle. Since we have both legs and need the hypotenuse, we apply the Pythagorean theorem: c² = a² + b². Setting up the equation: AC² = AB² + BC² = 9² + 12² = 81 + 144 = 225, so AC = √225 = 15. The answer is 15 because we take the positive square root when finding a length. A common error would be to subtract instead of add (√63 = √(144-81)), which happens when students confuse finding a leg versus finding the hypotenuse. Remember: when finding the hypotenuse, always add the squares of the legs.

Question 7

In right triangle △GHI\triangle GHI△GHI shown, ∠H\angle H∠H is a right angle. The leg GH=4GH=4GH=4 and hypotenuse GI=9GI=9GI=9. What is the measure of ∠G\angle G∠G (nearest degree)?

  1. sin⁡−1 ⁣(49)\sin^{-1}\!\left(\dfrac{4}{9}\right)sin−1(94​)
  2. cos⁡−1 ⁣(49)\cos^{-1}\!\left(\dfrac{4}{9}\right)cos−1(94​)
  3. tan⁡−1 ⁣(49)\tan^{-1}\!\left(\dfrac{4}{9}\right)tan−1(94​)
  4. sin⁡ ⁣(49)\sin\!\left(\dfrac{4}{9}\right)sin(94​)
Explanation: This problem requires finding an angle when given two sides of a right triangle. We have leg GH = 4 (adjacent to angle G) and hypotenuse GI = 9. Since we have the adjacent side and hypotenuse, we use cosine: cos(G) = adjacent/hypotenuse = 4/9. To find angle G, we use the inverse cosine: G = cos⁻¹(4/9). This gives the angle measure in degrees. A common error is using regular cosine cos(4/9) instead of inverse cosine, or choosing the wrong trig function. When finding angles from sides, use inverse trig functions with the correct ratio.

Question 8

In the plane, right triangle △GHI\triangle GHI△GHI is shown. The right angle is explicitly marked at HHH (∠GHI=90∘\angle GHI = 90^\circ∠GHI=90∘). The hypotenuse is GI‾\overline{GI}GI and is labeled 151515. The leg GH‾\overline{GH}GH is labeled 999. The leg HI‾\overline{HI}HI is unlabeled.

Which method should be used to find the length of HI‾\overline{HI}HI?

(Diagram is not drawn to scale. No acute angle measures are given.)

  1. Use the Pythagorean Theorem with 151515 and 999.
  2. Use sin⁡(90∘)=915\sin(90^\circ)=\dfrac{9}{15}sin(90∘)=159​.
  3. Use tan⁡(9∘)=HI15\tan(9^\circ)=\dfrac{HI}{15}tan(9∘)=15HI​.
  4. Use a 303030-606060-909090 triangle relationship.
Explanation: 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.

Question 9

In the plane, right triangle △JKL\triangle JKL△JKL is shown. The right angle is explicitly marked at KKK (∠JKL=90∘\angle JKL = 90^\circ∠JKL=90∘). The hypotenuse is JL‾\overline{JL}JL (explicitly identified) and is labeled 101010. The acute angle at JJJ is marked and labeled 35∘35^\circ35∘. The leg JK‾\overline{JK}JK is unlabeled.

What is the length of JK‾\overline{JK}JK?

(Diagram is not drawn to scale. No other angles are marked.)

  1. 10sin⁡(35∘)10\sin(35^\circ)10sin(35∘)
  2. 10cos⁡(35∘)10\cos(35^\circ)10cos(35∘)
  3. 10sin⁡(35∘)\dfrac{10}{\sin(35^\circ)}sin(35∘)10​
  4. 10tan⁡(35∘)10\tan(35^\circ)10tan(35∘)
Explanation: Solving right triangles involves using the Pythagorean theorem or trigonometry to find unknown sides or angles. In this problem, we are given the hypotenuse JL‾=10\overline{JL} = 10JL=10 and the acute angle at J=35∘J = 35^\circJ=35∘. Since we have the hypotenuse and an angle, and need the adjacent leg JK‾\overline{JK}JK, trigonometric ratios apply. The setup is cos⁡(35∘)=JK10\cos(35^\circ) = \frac{\text{JK}}{10}cos(35∘)=10JK​. This gives JK=10cos⁡(35∘)\text{JK} = 10 \cos(35^\circ)JK=10cos(35∘), matching choice B. A common misconception is using sine for the adjacent side, as in choice A, confusing opposite and adjacent. To transfer this strategy, always choose whether to use the Pythagorean theorem or trig ratios based on the given information before computing.

Question 10

In the right triangle shown, ∠B\angle B∠B is a right angle. The hypotenuse is AC‾\overline{AC}AC. If AB=11AB=11AB=11 and BC=4BC=4BC=4, which expression represents the correct setup to find ACACAC?

  1. AC=112+42AC=\sqrt{11^2+4^2}AC=112+42​
  2. AC=112−42AC=\sqrt{11^2-4^2}AC=112−42​
  3. AC=(11+4)2AC=\sqrt{(11+4)^2}AC=(11+4)2​
  4. AC=114AC=\dfrac{11}{4}AC=411​
Explanation: This problem asks which expression correctly sets up the Pythagorean theorem to find the hypotenuse. We have legs AB = 11 and BC = 4, with angle B being the right angle, and need to find hypotenuse AC. The Pythagorean theorem states that the hypotenuse squared equals the sum of the legs squared: c² = a² + b². Therefore, AC² = AB² + BC² = 11² + 4², which gives us AC = √(11² + 4²). The correct setup is AC = √(11² + 4²) because we add the squares of the legs when finding the hypotenuse. Option B incorrectly subtracts (used for finding a leg), while option C incorrectly adds before squaring. Before computing, always set up the correct equation based on what you're finding.