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Geometry Help: Solving Right Triangles Pythagorean Theorem Trigonometry

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

Question 1 / 10

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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.)

All questions

Question 1

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 $9$ . (correct answer)
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.

Question 2

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.)

$10\sin(35^\circ)$
$10\cos(35^\circ)$ (correct answer)
$\dfrac{10}{\sin(35^\circ)}$
$10\tan(35^\circ)$

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.

Question 3

In the diagram, $\triangle RST$ is a right triangle with the right angle explicitly marked at $S$ . The hypotenuse is $\overline{RT}$ . The legs are labeled $RS=5$ and $ST=12$ . Which method should be used to find the length of $\overline{RT}$ ?

(Diagram is not drawn to scale; no acute angles are labeled.)

Use the Pythagorean Theorem because two side lengths are known. (correct answer)
Use sine because an acute angle and the opposite side are known.
Use cosine because an acute angle and the adjacent side are known.
Use tangent because an acute angle and two legs are known.

Explanation: Solving right triangles involves finding unknown sides or angles using the Pythagorean theorem or trigonometric ratios. In this problem, both legs are given as 5 and 12. The Pythagorean theorem applies because both legs are known, and we need the hypotenuse. The equation is RT = √(5² + 12²). This justifies using the theorem over trigonometry, matching choice A. A common misconception is opting for sine when no angles are given, as in choice B. To transfer this strategy, always choose between the Pythagorean theorem or trigonometry based on whether sides or angles are provided before computing.

Question 4

In the right triangle $\triangle VWX$ shown, $\angle W$ is a right angle and the hypotenuse is $\overline{VX}$ . The acute angle at $V$ is $41^\circ$ , and $WX=10$ . What is the length of $\overline{VX}$ ?

$\dfrac{10}{\cos 41^\circ}$
$\dfrac{10}{\sin 41^\circ}$ (correct answer)
$10\sin 41^\circ$
$10\cos 41^\circ$

Explanation: This problem involves finding the hypotenuse using trigonometry. We are given the opposite side WX = 10 and the angle at V = 41°, and need to find the hypotenuse VX. Since we have the opposite side and need the hypotenuse, we use the sine ratio: sin(angle) = opposite/hypotenuse. Setting up the equation: sin(41°) = 10/VX, which rearranges to VX = 10/sin(41°). This is justified because sine relates the opposite side to the hypotenuse in a right triangle. A common mistake is using cosine (10/cos(41°)) which would be used if WX were adjacent to angle V. Before solving, identify the position of the known side relative to the given angle.

Question 5

In right triangle $\triangle DEF$ shown, $\angle E$ is a right angle. The hypotenuse is $DF=13$ and one leg is $DE=5$ . What is the length of leg $EF$ ?

$8$
$\sqrt{194}$
$\sqrt{144}$ (correct answer)
$18$

Explanation: This problem involves finding a missing leg when given the hypotenuse and one leg of a right triangle. We have hypotenuse DF = 13 and leg DE = 5, and need to find leg EF. The Pythagorean theorem applies in the form a² + b² = c², where c is the hypotenuse. Rearranging to find the missing leg: EF² = DF² - DE² = 13² - 5² = 169 - 25 = 144. Therefore, EF = √144 = 12. A common mistake is adding instead of subtracting when finding a leg (13² + 5² = 194). Remember: when finding a leg, subtract the known leg squared from the hypotenuse squared.

Question 6

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.)

$\sin^{-1}\!\left(\dfrac{8}{17}\right)$ (correct answer)
$\cos^{-1}\!\left(\dfrac{8}{17}\right)$
$\tan^{-1}\!\left(\dfrac{17}{8}\right)$
$\sin\!\left(\dfrac{8}{17}\right)$

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.

Question 7

A right triangle $\triangle MNO$ is shown with the right angle at $N$ . What is the length of hypotenuse $MO$ ?

$21$
$\sqrt{63}$
$\sqrt{225}$
$15$ (correct answer)

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.

Question 8

In the coordinate plane, triangle $\triangle ABC$ is shown. $\angle B$ is a right angle (marked with a square). The segment $\overline{AB}$ is horizontal from $A(-1,2)$ to $B(5,2)$ , and $\overline{BC}$ is vertical from $B(5,2)$ to $C(5,10)$ . The hypotenuse is $\overline{AC}$ . The diagram is not drawn to scale. No other angles or lengths are marked.

What is the length of $\overline{AC}$ ?

$6+8=14$
$\sqrt{6^2+8^2}$ (correct answer)
$\sqrt{8^2-6^2}$
$\sqrt{6^2-8^2}$

Explanation: Solving right triangles involves finding unknown sides or angles using the Pythagorean theorem or trigonometric ratios. In this problem, the coordinates provide the lengths of the legs AB = 6 and BC = 8 in right triangle ABC with right angle at B. The Pythagorean theorem applies because both legs are known, and we need the hypotenuse AC. The correct equation is AC = √(6² + 8²). This setup is justified as it directly relates the squares of the legs to the square of the hypotenuse in a right triangle. A common misconception is subtracting the squares, as in choice C, which might confuse finding a leg with finding the hypotenuse. To transfer this strategy, always choose the method—Pythagorean for sides or trig for angles—before computing.

Question 9

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.)

$12\cos 35^\circ$
$\dfrac{12}{\cos 35^\circ}$ (correct answer)
$\dfrac{12}{\sin 35^\circ}$
$12\tan 35^\circ$

Explanation: 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.

Question 10

In right triangle $\triangle ABC$ shown, $\angle C$ is a right angle. The legs are $AC=6$ and $BC=8$ . What is the length of hypotenuse $AB$ ?

$10$ (correct answer)
$14$
$\sqrt{28}$
$\sqrt{100}$

Explanation: 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. 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, always identify what's given and what method applies before computing.