20 medium SAT Right triangles & trigonometry questions

Medium is where most scores are actually won and lost. These questions are not tricky for the sake of it, but every one of them has a wrong answer built to catch a specific shortcut.

Every question below is a real item from the SAT Climb bank, tagged medium by the same difficulty model the app uses to build your practice. Pick an answer before you open the explanation.

Math · Geometry and Trigonometry~1 per testMedium tier

None of these repeat the 25 hard Right triangles & trigonometry questions, so the two pages together are 45 different questions on this one skill.

Question 1Medium

In right triangle ABC, angle C is the right angle. The length of side AB is 17 and the length of side AC is 8. What is the length of side BC?

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Why B is right

In right triangle ABC with right angle at C, AB is the hypotenuse with length 17, and AC is one leg with length 8. By the Pythagorean theorem, $AC^{2}$ + $BC^{2}$ = $AB^{2}$ , so $8^{2}$ + $BC^{2}$ = $17^{2}$ . This gives 64 + $BC^{2}$ = 289, so $BC^{2}$ = 225, and BC = 15.

Why the others are wrong

AThis results from incorrectly assuming the triangle follows a 8-9-17 pattern, which is not a valid Pythagorean triple.
CThis results from incorrectly treating AC as the hypotenuse and AB as a leg, computing √(17² + 8²) instead of √(17² - 8²).
DThis results from incorrectly adding the given lengths (8 + 17) instead of applying the Pythagorean theorem.

Question 2Medium

In right triangle XYZ, angle Y is the right angle. The length of side XZ is 25 and the length of side YZ is 24. What is the value of cos X?

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Why C is right

Using the Pythagorean theorem, $XY^{2}$ = $25^{2} - 24^{2} = 625 - 576 = 49$ , so XY = 7. The cosine of angle X is the ratio of the adjacent side to the hypotenuse, which is XY/XZ = 7/25.

Why the others are wrong

AThis results from using the calculated leg divided by the given leg instead of by the hypotenuse.
BThis is sin X, not cos X, using the opposite side instead of the adjacent side.
DThis results from inverting the correct ratio.

Question 3Medium

Triangle LMN is similar to triangle QRS. If $\displaystyle \sin (L) = \frac{11}{61}$ , what is $\displaystyle \frac{\sin (Q)}{\cos (Q)}$ ?

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Why A is right

Since the triangles are similar, sin(Q) = $sin(L) = 11/61$ . With opposite side 11 and hypotenuse 61, the adjacent side is 60. Therefore $cos(Q) = 60/61$ , and sin(Q)/ $\displaystyle cos(Q) = \frac{11}{61}/\frac{60}{61} = 11/60$ .

Why the others are wrong

BThis inverts the correct ratio, computing cos(Q)/sin(Q) instead of sin(Q)/cos(Q).
CThis incorrectly uses the hypotenuse in place of the adjacent side.
DThis returns sin(Q) without dividing by cos(Q).

Question 4Medium

In right triangle PQR, angle Q is a right angle. The measure of angle P is 60 degrees, and the length of side PQ is 6. What is the length of side QR?

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Why C is right

In a 30-60-90 triangle, the sides are in the ratio 1 : $\displaystyle \sqrt{3}$ : 2. Since angle P is 60 degrees and angle Q is 90 degrees, angle R must be 30 degrees. The side PQ is adjacent to the 60-degree angle and opposite the 30-degree angle, making it the shorter leg. The side QR is opposite the 60-degree angle, so it equals the shorter leg times $\displaystyle \sqrt{3}$ , which is $\displaystyle 6\sqrt{3}$ .

Why the others are wrong

AThis results from dividing the given side by 2, which would be correct if PQ were the hypotenuse, but PQ is the shorter leg.
BThis assumes QR equals PQ, failing to apply the 30-60-90 triangle ratio where the longer leg is √3 times the shorter leg.
DThis results from multiplying the shorter leg by 2 to get the hypotenuse, but the question asks for QR, the longer leg, not the hypotenuse PR.

Question 5Medium

In right triangle ABC, angle C is the right angle. The length of side AB is 13 and the length of side BC is 5. What is the length of side AC?

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Why B is right

The Pythagorean theorem states that in a right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse. Here AB is the hypotenuse with length 13, and BC is one leg with length 5. Therefore, $AC^{2}$ + $5^{2} = 13^{2}$ , which gives $AC^{2}$ = 169 - 25 = 144, so AC = 12.

Why the others are wrong

AThis results from incorrectly applying the difference 13 - 5 instead of using the Pythagorean theorem.
CThis results from incorrectly treating BC as the hypotenuse and computing √(13² + 5²) = √194.
DThis results from incorrectly adding the given lengths 13 + 5 = 18.

Question 6Medium

In right triangle ABC, angle C is the right angle. If BC = 5 and AC = 12, what is the length of AB?

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Why B is right

The Pythagorean theorem states that in a right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse. Since BC = 5 and AC = 12 are the legs, $AB^{2}$ = $5^{2} + 12^{2} = 25 + 144 = 169$ , so AB = 13.

Why the others are wrong

AThis is the difference 12 - 5, not recognizing that the Pythagorean theorem requires squaring the side lengths.
CThis results from incorrectly using a different Pythagorean triple (8-15-17) instead of computing from the given sides.
DThis is the square root of 12² + 5² computed as √(12 + 5)² = √119, incorrectly applying the square root to the sum before squaring.

Question 7Medium

In right triangle DEF, angle E is the right angle. If DE = 15 and DF = 17, what is the value of sin D?

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Why A is right

First, use the Pythagorean theorem to find EF. Since $DE^{2}$ + $EF^{2}$ = $DF^{2}$ , we have $15^{2}$ + $EF^{2}$ = $17^{2}$ , so $EF^{2}$ = 289 - 225 = 64, and EF = 8. For angle D, the opposite side is EF = 8 and the hypotenuse is DF = 17. Therefore, sin $D = 8/17$ .

Why the others are wrong

BThis results from using the adjacent side instead of the opposite side: sin D should use EF, not DE.
CThis results from dividing the opposite side by the adjacent side, which gives tan D, not sin D.
DThis inverts the correct ratio and misapplies the sine definition.

Question 8Medium

Triangle HIJ is similar to triangle NOP. If $\displaystyle \sin (H) = \frac{20}{29}$ , what is $\displaystyle \frac{\sin (N)}{\cos (N)}$ ?

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Why C is right

Since the triangles are similar, sin(N) = $sin(H) = 20/29$ . With opposite side 20 and hypotenuse 29, the adjacent side is 21. Therefore $cos(N) = 21/29$ , and sin(N)/ $\displaystyle cos(N) = \frac{20}{29}/\frac{21}{29} = 20/21$ .

Why the others are wrong

AThis inverts the correct ratio, computing cos(N)/sin(N) instead of sin(N)/cos(N).
BThis incorrectly uses the hypotenuse in place of the adjacent side.
DThis returns sin(N) without dividing by cos(N).

Question 9Medium

Triangle PQR is a 30-60-90 triangle where angle Q is the right angle and angle P measures 30 degrees. If the length of side QR is $\displaystyle 4\sqrt{3}$ , what is the length of side PQ?

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Why B is right

In a 30-60-90 triangle, the sides are in the ratio 1: $\displaystyle \sqrt{3}$ :2. Since angle P is 30 degrees, angle R is 60 degrees. Side QR is opposite the 60-degree angle, so it has length $\displaystyle x\sqrt{3}$ where x is the shortest side. Therefore, $\displaystyle x\sqrt{3}$ = $\displaystyle 4\sqrt{3}$ , so $x = 4$ . Side PQ is opposite the 60-degree angle from vertex R, making it the longest leg, with length $3x = 12$ .

Why the others are wrong

AThis is the length of the shortest side PR, not the leg PQ.
CThis results from doubling the shortest side instead of tripling it.
DThis assumes PQ equals QR, which is incorrect in a 30-60-90 triangle.

Question 10Medium

In right triangle ABC, angle C is the right angle. The length of side AB is 17, and the length of side BC is 15. What is the length of side AC?

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Why B is right

The Pythagorean theorem states that in a right triangle, the sum of the squares of the legs equals the square of the hypotenuse. Here AB is the hypotenuse (17) and BC is one leg (15). Therefore, $AC^{2}$ + $15^{2} = 17^{2}$ , so $AC^{2}$ = 289 - 225 = 64, and AC = 8.

Why the others are wrong

AThis results from incorrectly subtracting the lengths directly instead of using the Pythagorean theorem.
CThis incorrectly identifies BC as the answer, confusing which side length was being asked for.
DThis results from incorrectly adding the squares instead of subtracting them in the Pythagorean theorem.

Question 11Medium

In right triangle ABC, angle C is the right angle. If AB = 13 and BC = 5, what is the length of AC?

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Why B is right

The Pythagorean theorem states that in a right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse. Here AB = 13 is the hypotenuse and BC = 5 is one leg. Therefore, $AC^{2}$ + $5^{2} = 13^{2}$ , so $AC^{2}$ = 169 - 25 = 144, and AC = 12.

Why the others are wrong

AThis results from incorrectly subtracting the leg from the hypotenuse without squaring: 13 - 5 = 8.
CThis results from incorrectly adding the given values: 13 + 5 = 18.
DThis results from misapplying the relationship between the sides.

Question 12Medium

In right triangle DEF, angle E is a right angle. The sum of the measures of angle D and angle F is 90 degrees. If cos $\displaystyle D = \frac{5}{13}$ , what is the value of sin F?

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Why A is right

In a right triangle, two acute angles are complementary, meaning their measures sum to 90 degrees. The sine of any acute angle equals the cosine of its complement. Since angle D and angle F are complementary angles in right triangle DEF, sin F = cos $D = 5/13$ .

Why the others are wrong

BThis results from finding sin D instead of sin F. Since cos D = 5/13, the adjacent side is 5 and hypotenuse is 13, making the opposite side 12, so sin D = 12/13, not sin F.
CThis results from incorrectly inverting the ratio, taking the reciprocal of cos D instead of recognizing the cofunction identity.
DThis results from inverting sin D = 12/13, combining the error of finding sin D instead of sin F with taking a reciprocal.

Question 13Medium

Triangle XYZ is similar to triangle BCD. If $\displaystyle \sin (X) = \frac{12}{13}$ , what is $\displaystyle \frac{\sin (B)}{\cos (B)}$ ?

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Why C is right

Since the triangles are similar, sin(B) = $sin(X) = 12/13$ . With opposite side 12 and hypotenuse 13, the adjacent side is 5. Therefore $cos(B) = 5/13$ , and sin(B)/ $\displaystyle cos(B) = \frac{12}{13}/\frac{5}{13} = 12/5$ .

Why the others are wrong

AThis inverts the correct ratio, computing cos(B)/sin(B) instead of sin(B)/cos(B).
BThis incorrectly uses the hypotenuse in place of the adjacent side.
DThis returns sin(B) without dividing by cos(B).

Question 14Medium

In right triangle DEF, angle E is the right angle. If angle D measures 60° and the length of side DE is 7, what is the length of side DF?

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Why C is right

In a 30-60-90 triangle, the sides are in the ratio 1: $\displaystyle \sqrt{3}$ :2. Since angle D is 60°, angle F is 30°. The side opposite the 30° angle (DE) has length 7, so the hypotenuse DF has length 2 × 7 = 14.

Why the others are wrong

AThis results from correctly identifying the 30-60-90 ratio but incorrectly treating DE as opposite the 60° angle and computing the longer leg instead of the hypotenuse.
BThis results from incorrectly applying the 45-45-90 triangle ratio (1:1:√2) instead of the 30-60-90 ratio.
DThis results from incorrectly computing the side opposite the 60° angle (the longer leg) by multiplying 7 by 2√3, confusing the ratio relationships.

Question 15Medium

Triangle JKL is a right triangle where angle K is the right angle. The length of side JL is 25, and the length of side JK is 7. What is the length of side KL?

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Why B is right

Using the Pythagorean theorem with JL as the hypotenuse (25) and JK as one leg (7): $7^{2}$ + $KL^{2}$ = $25^{2}$ , so $KL^{2}$ = 625 - 49 = 576, and KL = 24.

Why the others are wrong

AThis results from subtracting the side lengths directly rather than using the Pythagorean theorem correctly.
CThis incorrectly identifies the hypotenuse as the answer instead of computing the missing leg.
DThis results from incorrectly adding 7² and 25² instead of subtracting.

Question 16Medium

In right triangle ABC, angle C is a right angle. The length of side AB is 17, and the length of side BC is 15. What is the length of side AC?

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Why A is right

The Pythagorean theorem states that in a right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse. Here AB is the hypotenuse with length 17, and BC is one leg with length 15. Therefore, $AC^{2}$ + $15^{2} = 17^{2}$ , which gives $AC^{2}$ + 225 = 289, so $AC^{2}$ = 64 and AC = 8.

Why the others are wrong

BThis is the length of BC, one of the given legs, not the unknown leg AC.
CThis results from incorrectly subtracting the leg lengths (17 - 15 = 2) and then taking the square root of 2 × 16, failing to apply the Pythagorean theorem correctly.
DThis results from incorrectly adding the squares: 17² + 15² instead of 17² - 15².

Question 17Medium

In right triangle DEF, angle E is the right angle. If DE = 8 and DF = 17, what is the value of sin F?

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Why A is right

In right triangle DEF with right angle at E, DF is the hypotenuse with length 17, and DE is a leg with length 8. The sine of angle F is the ratio of the length of the side opposite angle F to the length of the hypotenuse. The side opposite angle F is DE, so sin $F = 8/17$ .

Why the others are wrong

BThis uses the ratio of DE to EF (which is 15 by the 8-15-17 triple), treating the adjacent leg as the denominator rather than the hypotenuse.
CThis is cos F, not sin F. It uses the ratio of the adjacent side EF to the hypotenuse instead of the opposite side.
DThis inverts the correct ratio, using hypotenuse/opposite instead of opposite/hypotenuse.

Question 18Medium

Triangle EFG is similar to triangle JKL. If $\displaystyle \sin (E) = \frac{15}{17}$ , what is $\displaystyle \frac{\sin (J)}{\cos (J)}$ ?

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Why B is right

Since the triangles are similar, sin(J) = $sin(E) = 15/17$ . With opposite side 15 and hypotenuse 17, the adjacent side is 8. Therefore $cos(J) = 8/17$ , and sin(J)/ $\displaystyle cos(J) = \frac{15}{17}/\frac{8}{17} = 15/8$ .

Why the others are wrong

AThis inverts the correct ratio, computing cos(J)/sin(J) instead of sin(J)/cos(J).
CThis returns sin(J) without dividing by cos(J).
DThis incorrectly uses the hypotenuse in place of the adjacent side.

Question 19Medium

In right triangle MNO, angle N is the right angle. The measure of angle M is 45 degrees. If MN = 6, what is the length of MO?

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Why B is right

Since angle M is 45 degrees and angle N is 90 degrees, angle O must also be 45 degrees, making this a 45-45-90 triangle. In a 45-45-90 triangle, the sides are in the ratio 1 : 1 : $\displaystyle \sqrt{2}$ . Since MN = 6 is one leg, the hypotenuse MO = $\displaystyle 6\sqrt{2}$ .

Why the others are wrong

AThis incorrectly assumes the hypotenuse equals the leg, missing the √2 ratio in a 45-45-90 triangle.
CThis incorrectly applies the 30-60-90 triangle ratio (√3) instead of the 45-45-90 ratio (√2).
DThis incorrectly uses the 30-60-90 hypotenuse ratio (2 times the short side) instead of the 45-45-90 ratio.

Question 20Medium

In right triangle PQR, angle Q is the right angle. If sin $\displaystyle P = \frac{5}{13}$ and the length of side PR is 39, what is the length of side PQ?

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Why B is right

Since sin P = opposite/hypotenuse = QR/PR, we have QR/39 = 5/13, giving QR = 15. Using the Pythagorean theorem with hypotenuse PR = 39 and leg QR = 15, we get $PQ^{2}$ + $15^{2} = 39^{2}$ , so $PQ^{2}$ = 1521 - 225 = 1296, and PQ = 36.

Why the others are wrong

AThis is the length of QR (the side opposite angle P), not PQ, resulting from confusing which leg is being asked for.
CThis results from incorrectly computing √(39² + 15²) instead of √(39² - 15²).
DThis results from incorrectly using cos P = 5/13 (instead of sin P) to find the adjacent side, leading to PQ = 5 and then miscalculating the other leg.

What to do after medium

Medium is the tier that decides most scores. If these are landing, the hard set is where the remaining points are.

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Questions are written by SAT Climb and drawn from its own item bank. SAT® is a registered trademark of College Board, which is not affiliated with and does not endorse SAT Climb.

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What to do after medium

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