Solving Right Triangles: Pythagorean Theorem, Trigonometry
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Geometry › Solving Right Triangles: Pythagorean Theorem, Trigonometry
In the right triangle $\triangle DEF$ shown, $\angle E$ is a right angle and the hypotenuse is $\overline{DF}$. The acute angle at $D$ is $35^\circ$, and $DE=9$. What is the length of $\overline{DF}$?
$9\tan 35^\circ$
$\dfrac{9}{\cos 35^\circ}$
$9\cos 35^\circ$
$\dfrac{9}{\sin 35^\circ}$
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.
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 $\overline{QR}$?
$\sin 22^\circ=\dfrac{QR}{12}$
$\cos 22^\circ=\dfrac{QR}{12}$
$\tan 22^\circ=\dfrac{12}{QR}$
$\tan 22^\circ=\dfrac{QR}{12}$
Explanation
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.
A right triangle $\triangle MNO$ is shown with the right angle at $N$. What is the length of hypotenuse $MO$?
$\sqrt{225}$
$15$
$\sqrt{63}$
$21$
Explanation
This problem asks us to find the hypotenuse of a right triangle given both legs. We have a right angle at N, with legs MN = 9 and NO = 12. The Pythagorean theorem applies directly since we know both legs: MO² = MN² + NO². Setting up the calculation: MO² = 9² + 12² = 81 + 144 = 225, so MO = √225 = 15. The answer is 15 because this is a 3-4-5 right triangle scaled by 3 (9-12-15). A common error would be adding the legs directly (9 + 12 = 21) instead of using the Pythagorean theorem. Recognizing special right triangle ratios like 3-4-5 can help verify your answer quickly.
In the right triangle shown, $\angle A$ is a right angle. The hypotenuse is $\overline{BC}$. If $BC=30$ and $\angle C=22^\circ$, what is the length of $AB$?
$30\sin(22^\circ)$
$\dfrac{30}{\sin(22^\circ)}$
$\dfrac{30}{\cos(22^\circ)}$
$30\cos(22^\circ)$
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.
In the right triangle $\triangle STU$ shown, $\angle T$ is a right angle and the hypotenuse is $\overline{SU}$. If $SU=15$ and $TU=9$, what is the length of $\overline{ST}$?
$\sqrt{15^2-9^2}$
$\sqrt{15^2+9^2}$
$15-9$
$\sqrt{(15-9)^2}$
Explanation
This problem requires finding a leg of a right triangle when given the hypotenuse and the other leg. We are given the hypotenuse SU = 15 and one leg TU = 9, and need to find the other leg ST. Using the Pythagorean theorem: a² + b² = c², we solve for the unknown leg. Setting up the equation: ST² + 9² = 15², which gives us ST² = 225 - 81 = 144, so ST = √(15² - 9²) = √144 = 12. This is justified because we subtract the square of the known leg from the square of the hypotenuse. A common error is adding the squares (√(15² + 9²)) as if finding a hypotenuse. When finding a missing leg, subtract squares rather than add them.
A right triangle $\triangle JKL$ is shown in the plane. $\angle K$ is marked as a right angle. The hypotenuse is $\overline{JL}$ and is labeled $25$. The leg $\overline{KL}$ is labeled $7$. The diagram is not drawn to scale. No other angles or lengths are marked.
What is the length of $\overline{JK}$?
$25-7$
$\sqrt{25^2-7^2}$
$\sqrt{7^2-25^2}$
$\sqrt{25^2+7^2}$
Explanation
Solving right triangles involves finding unknown sides or angles using the Pythagorean theorem or trigonometric ratios. In this problem, the hypotenuse JL = 25 and one leg KL = 7 are given in right triangle JKL with right angle at K. The Pythagorean theorem applies because we need the other leg JK, with the hypotenuse and one leg known. The correct equation is JK = √(25² - 7²). This is justified as rearranging the theorem isolates the unknown leg by subtracting squares. A distractor like choice B adds squares, which finds the hypotenuse instead of a leg. To transfer this strategy, always choose the method—Pythagorean for sides or trig for angles—before computing.
Right triangle $\triangle STU$ is shown in the plane. The right angle is explicitly marked at $T$ ($\angle STU = 90^\circ$). The hypotenuse is $\overline{SU}$ and is labeled $17$. The leg $\overline{TU}$ is labeled $8$. The acute angle at $S$ is unlabeled.
What is the measure of $\angle S$?
(Diagram is not drawn to scale. No other angles are marked.)
$\cos^{-1}!\left(\dfrac{8}{17}\right)$
$\sin!\left(\dfrac{8}{17}\right)$
$\tan^{-1}!\left(\dfrac{17}{8}\right)$
$\sin^{-1}!\left(\dfrac{8}{17}\right)$
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 SU = 17 and the opposite leg to angle S, TU = 8. Since we have the opposite side and hypotenuse, and need the angle at S, inverse trigonometric ratios apply. The setup is angle S = sin⁻¹(8/17). This correctly finds the angle, matching choice A. A common misconception is using inverse cosine instead, as in choice B, which would apply to the adjacent side. To transfer this strategy, always choose whether to use the Pythagorean theorem or trig ratios based on the given information before computing.
A right triangle $\triangle KLM$ is shown in the plane. The right angle is explicitly marked at $L$ ($\angle KLM = 90^\circ$). The legs are $\overline{KL}=6$ and $\overline{LM}=8$. The slanted side $\overline{KM}$ is explicitly identified as the hypotenuse.
Which claim about the triangle is NOT justified?
(Diagram is not drawn to scale. No angle measures other than the right angle are marked.)
$\angle K=60^\circ$.
$KL^2+LM^2=KM^2$.
$KM=10$.
$\sin(\angle K)=\dfrac{8}{10}$.
Explanation
Solving right triangles involves using the Pythagorean theorem or trigonometry to find unknown sides or angles. In this problem, we are given the two legs KL = 6 and LM = 8. To evaluate the claims, the Pythagorean theorem and trig ratios apply to find the hypotenuse and angles. The equation for the hypotenuse is KM² = 6² + 8², giving KM = 10. Then, sin(∠K) = 8/10, so ∠K ≈ 53.13°, not 60°, making claim D unjustified. A common misconception is assuming a 60° angle based on side ratios without calculating, 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.
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.)
Use the Pythagorean Theorem with $15$ and $9$.
Use $\sin(90^\circ)=\dfrac{9}{15}$.
Use $\tan(9^\circ)=\dfrac{HI}{15}$.
Use a $30$-$60$-$90$ 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.
In the plane, right triangle $\triangle JKL$ is shown. The right angle is explicitly marked at $K$ ($\angle JKL = 90^\circ$). The hypotenuse is $\overline{JL}$ (explicitly identified) and is labeled $10$. The acute angle at $J$ is marked and labeled $35^\circ$. The leg $\overline{JK}$ is unlabeled.
What is the length of $\overline{JK}$?
(Diagram is not drawn to scale. No other angles are marked.)
$\dfrac{10}{\sin(35^\circ)}$
$10\tan(35^\circ)$
$10\cos(35^\circ)$
$10\sin(35^\circ)$
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 $ \overline{JL} = 10 $ and the acute angle at $ J = 35^\circ $. Since we have the hypotenuse and an angle, and need the adjacent leg $ \overline{JK} $, trigonometric ratios apply. The setup is $ \cos(35^\circ) = \frac{\text{JK}}{10} $. This gives $ \text{JK} = 10 \cos(35^\circ) $, 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.