Right Triangle Trigonometry Practice Problems With Answers

Here are 30 right triangle trigonometry practice problems split into three levels: 10 easy, 10 medium and 10 hard. Every problem has a fully worked solution further down the page, with the formula chosen, the substitution shown, the arithmetic carried at full precision and the rounding stated. Two of the problems are deliberately impossible or inconsistent, and for those the correct answer is to explain why.

Work through the question lists first. The solutions live in a single section at the end, so you can attempt a whole level before checking anything.

One-screen refresher

This page is for practice, not for learning the ratios from zero. If any of the next three lines is unfamiliar, read the full SOHCAHTOA lesson at how to choose sine, cosine or tangent and come back.

Figure 1

Decision table

Name what you know and what you want, then read the ratio straight off this table.

The last three rows are the inverse trig cases, covered in more depth in the guide to finding a missing angle from two sides.

Exact values worth memorising

Several problems below ask for an exact answer rather than a decimal. These nine values are all you need.

Level 1: easy (problems 1 to 10)

These are single-step problems. Each one needs one ratio and one calculator keystroke sequence.

Figure 2

1. In the triangle in Figure 2, the right angle is at R, side QR = 6, side PR = 8 and side PQ = 10. Taking angle P as the reference angle, name the opposite side, the adjacent side and the hypotenuse.
2. A right triangle has an acute angle of 40°. You know the hypotenuse and want the side adjacent to the 40° angle. Which ratio do you use, and what equation do you write? Do not compute a number.
3. A right triangle has an acute angle of 35° and a hypotenuse of 12 cm. Find the side opposite the 35° angle, to 2 decimal places.
4. A right triangle has an acute angle of 62° and a hypotenuse of 20 m. Find the side adjacent to the 62° angle, to 2 decimal places.
5. A right triangle has an acute angle of 28°, and the leg adjacent to it measures 15 in. Find the leg opposite the 28° angle, to 2 decimal places.
6. In a right triangle, the leg opposite a 41° angle is 9 units. Find the hypotenuse, to 2 decimal places.
7. A right triangle has a hypotenuse of 10 and a leg of 7 opposite angle θ. Find θ, to 2 decimal places.
8. A right triangle has a hypotenuse of 14 and a leg of 9 adjacent to angle θ. Find θ, to 2 decimal places.
9. A right triangle has legs of 5 and 12, with the 5 opposite angle θ. Find θ, to 2 decimal places.
10. A right triangle has two 45° angles and each leg measures 7. Find the hypotenuse in exact form, then as a decimal to 2 decimal places.

Level 2: medium (problems 11 to 20)

These add exact values, unknowns in the denominator, short word problems, and one set of data that cannot describe a real triangle.

Figure 3

11. A right triangle has a 60° angle and a hypotenuse of 14. Find both legs in exact form, then to 2 decimal places.
12. In the triangle in Figure 3, the leg opposite the 30° angle measures 9. Find the other leg and the hypotenuse in exact form, and state which special triangle this is.
13. In a right triangle, the leg adjacent to a 33° angle is 18 cm. Find the hypotenuse, to 2 decimal places.
14. In a right triangle, the leg opposite a 52° angle is 25 mm. Find the leg adjacent to the 52° angle, to 2 decimal places.
15. A 13 ft ladder leans against a wall and makes an angle of 68° with level ground. How far up the wall does it reach? Round to 2 decimal places.
16. A wheelchair ramp rises 1.2 m over a horizontal run of 14 m. Find the angle the ramp makes with the ground, to 2 decimal places.
17. From a point 50 m from the base of a tower on level ground, the angle of elevation to the top is 32°. Find the height of the tower, to 2 decimal places.
18. A right triangle has legs of 8 and 15. Find the hypotenuse and both acute angles, with angles to 2 decimal places.
19. A student writes: “In right triangle ABC the hypotenuse is 8 cm and the side opposite angle θ is 10 cm. Find θ.” Find θ or explain why it cannot exist.
20. A right triangle has a 45° angle, and the leg adjacent to that angle measures 5√2. Find the hypotenuse exactly, without a calculator.

Level 3: hard (problems 21 to 30)

These need two relationships, a diagram you draw yourself, or a careful check of whether the given data is consistent.

Figure 4

21. Standing on level ground, you measure the angle of elevation to the top of a flagpole as 34°. You walk 20 m directly toward the pole and measure the elevation again as 52°. Find the height of the pole, to 2 decimal places.
22. A right triangle has legs of 9 and 12. Find the length of the altitude drawn from the right angle to the hypotenuse, using a trig ratio rather than an area formula. Give the exact value.
23. From the top of a 85 m cliff, the angle of depression to a boat is 14°. Find the horizontal distance from the base of the cliff to the boat, to 2 decimal places.
24. A worksheet states: “Right triangle PQR has hypotenuse 20, a leg of 12, and the angle opposite the leg of 12 measures 40°.” Decide whether this data is consistent, and if not, say which value is wrong and what it should be.
25. A right triangle has a 30° angle, and the leg adjacent to it measures 12. Find the other leg and the hypotenuse in exact radical form, then verify with the Pythagorean theorem.
26. The two legs of a right triangle are in the ratio 2 : 5. Find both acute angles, to 2 decimal places, and explain why no actual side lengths are needed.
27. A kite is flying on 120 m of taut string. The string makes an angle of 39° with the horizontal, and the hand holding it is 1.5 m above the ground. Find the height of the kite above the ground, to 2 decimal places.
28. A mountain road climbs at a constant 7° grade. A car drives 2.4 km along the road surface. Find the horizontal distance covered and the altitude gained, both to 3 decimal places in kilometres.
29. An isosceles triangle has a base of 16 cm and two equal sides of 17 cm. Find its height and its apex angle, with the angle to 2 decimal places.
30. In right triangle ABC the right angle is at C, angle A = 47°, and side a (opposite A) = 9.4. Solve the triangle completely: find angle B, side b and side c. Round lengths to 2 decimal places.

Answers and worked solutions

Every calculation below was carried at full precision and rounded only at the last step. Where an answer is exact, the radical form is given first and the decimal second.

Easy solutions

If you had taken angle Q instead, the 6 and the 8 would swap jobs and the 10 would stay put. The hypotenuse is the only label that does not depend on your choice of reference angle.

The two sides in play are the adjacent and the hypotenuse, and CAH is the only part of SOHCAHTOA that pairs those two. Naming the pair of sides first is faster than testing all three ratios.

The unknown started under the division bar. Cross-multiplying first and dividing second keeps you from reaching for 9 à - sin 41°, which is the single most common slip on this type. The answer is larger than 9, which it must be, because the hypotenuse is always the longest side.

Note that 5, 12 and 13 form a Pythagorean triple, so the hypotenuse here is exactly 13 and the other acute angle is 90° − 22.62° = 67.38°.

You can reach the same place with cosine: cos 45° = 7 / c gives c = 7 / (√2/2) = 14/√2 = 7√2. Recognising the special triangle just saves the step.

Medium solutions

Check: 7² + (7√3)² = 49 + 147 = 196 = 14². The data was a 30-60-90 triangle in disguise, since the third angle had to be 30°.

Verify with tangent: tan 60° = 15.588457… / 9 = 1.732050…, which is √3 as it should be. The scale factor here is 9, applied to the base ratio 1 : √3 : 2.

Since 52° is larger than 45°, the side opposite it must be the longer leg, and 19.53 is indeed shorter than 25. That one-line sanity check catches a flipped fraction immediately.

Rise over run is the opposite over the adjacent, so tangent is the ratio, never sine. Sine would need the length of the sloping surface, which was not given.

Figure 5

The angle of elevation is always measured from the horizontal, which is why the horizontal distance is the adjacent side. More setups of this kind are worked through in the article on elevation and depression angles.

The two angles sum to exactly 90°, and 17 is longer than both legs, so the solution is self-consistent. The 8-15-17 side lengths are a Pythagorean triple, which is why the hypotenuse came out whole.

A calculator confirms this by refusing the operation: sin⁻¹(1.25) returns a domain error rather than a number. The underlying geometry is the point to state in your answer, not just the error message. The hypotenuse faces the 90° angle, the largest angle in the triangle, and the largest angle always faces the longest side.

The shortcut is to recognise the 45-45-90 ratio 1 : 1 : √2. A leg of 5√2 gives a hypotenuse of 5√2 à - √2 = 5 à - 2 = 10, with no trig at all.

Hard solutions

Two relationships were needed because neither distance was given directly. Factoring h out of the subtraction avoids solving a system. Substituting back: 28.519468…/tan 34° = 42.2818… and 28.519468…/tan 52° = 22.2818…, and those differ by exactly 20 m.

The altitude splits the original triangle into two smaller triangles similar to it, so the same acute angle B appears in both. That similarity is the second relationship, and it is unpacked in the article on the altitude to the hypotenuse. The area shortcut agrees: (9 Ã - 12) / 15 = 7.2.

Figure 6

The common error is to use 14° as an angle inside the triangle at the cliff top. The angle of depression is measured downward from the horizontal, so the angle inside the triangle at the top is 90° − 14° = 76°. Using the equal alternate angle at the boat, as above, sidesteps the confusion.

Unlike Problem 19, this triangle is not impossible: 12 is a perfectly legal leg for a hypotenuse of 20. The three numbers simply cannot all be true at once, and a good answer names the two ways to repair the data rather than picking one silently.

Figure 7

Pythagorean check: (4√3)² + 12² = 48 + 144 = 192, and (8√3)² = 64 à - 3 = 192. Note that 12 is the long leg here, so the hypotenuse is 8√3 and not 24. Reading “adjacent to 30°” as “short leg” is the trap.

No actual lengths are needed because every ratio is a quotient of two sides. Legs of 2 and 5, of 20 and 50, or of 2.4 and 6 all give the same 0.4 and therefore the same angles. Similar right triangles share their angles exactly.

The trig gives the vertical rise from the hand, not from the ground. Adding the 1.5 m at the very end is the second step that a one-ratio answer misses.

The 2.4 km is the hypotenuse because it is measured along the slope. Check: 2.382110…² + 0.292486…² = 5.76 = 2.4².

Figure 4 shows the right half. The 8-15-17 triple appears again, which is why the height came out whole. Doubling the half-angle at the very end, rather than doubling a rounded 28.07°, keeps the result accurate.

All four checks pass: the angles sum to 47 + 43 + 90 = 180, c is longer than both legs, √(9.4² + 8.765641…²) = 12.852878…, and the larger angle A faces the longer leg a. You can paste the same two starting values into the right triangle calculator to confirm the whole solution at once.

How to check your answer

Four checks catch almost every error in this kind of problem, and all four take seconds.

1. The hypotenuse is the longest side. If a leg came out longer than the hypotenuse, you divided when you should have multiplied, or you named the sides from the wrong angle. Problem 19 is this check applied to the given data instead of the answer.
2. The two acute angles sum to 90°. After finding one angle, the other is 90° minus it. If you computed both independently and they do not add to 90°, one of them is wrong. Problem 18 and Problem 26 both pass this test.
3. Substitute back into the original relationship. Put your answer into the ratio you started with and see whether it reproduces the given number. In Problem 30, √(9.4² + 8.765641²) returned 12.852878, matching the hypotenuse found from sine.
4. The bigger angle faces the bigger side. A 52° angle must face a longer side than a 38° angle in the same triangle. This one-glance check caught the flipped fraction risk in Problem 14.

For a fifth check, or when you want every remaining value at once, enter what you know into the triangle solver on the homepage and compare all six measurements. It is faster than repeating the arithmetic, and it will show the angles and sides you did not compute.

Common mistakes

Using the rounded value in the next step. If you round 14 · sin 60° to 12.12 and then square it, you get 146.89 instead of 147. Keep full precision on your calculator until the final line, then round once and say what precision you used.

Calculator in radians. sin 30° is 0.5; sin(30 radians) is −0.988. Every answer on a radian-mode page is wrong, usually by an amount that looks plausible enough to miss.

Multiplying when the unknown is in the denominator. sin 41° = 9 / c does not rearrange to c = 9 · sin 41°. Cross-multiply to c · sin 41° = 9, then divide. Problems 6, 13 and 14 all test this.

Naming opposite and adjacent from the triangle instead of the angle. The same side is opposite one acute angle and adjacent to the other. Write the reference angle down before labelling anything.

Reaching for sine when only the two legs are known. Sine and cosine both involve the hypotenuse. Two legs means tangent, as in Problems 9, 16 and 26.

Confusing a depression angle with an interior angle. Angles of elevation and depression are measured from the horizontal, not from the vertical or from the slope.

Assuming “adjacent to 30°” means “the short leg”. In Problem 25 the side adjacent to the 30° angle was the long leg, and treating 12 as the short leg would have produced a hypotenuse of 24 instead of 8√3.

Frequently Asked Questions

How many right triangle trigonometry practice problems are on this page?

There are 30, split evenly into 10 easy, 10 medium and 10 hard, and each one has a full worked solution showing the formula, the substitution and the rounding. Two of them, Problems 19 and 24, have deliberately faulty data, and the expected answer is an explanation rather than a number.

Do I need a calculator for all of these?

No. Problems 10, 11, 12, 20 and 25 are built on the exact values of 30°, 45° and 60° and on the 1 : 1 : √2 and 1 : √3 : 2 side ratios, so they can be finished in exact radical form by hand. The rest need a calculator in degree mode.

How do I know whether to use sine, cosine or tangent?

Name the two sides involved in the problem, the one you know and the one you want, then pick the ratio that pairs exactly those two: opposite and hypotenuse is sine, adjacent and hypotenuse is cosine, opposite and adjacent is tangent. The decision table near the top of this page lists all nine cases, including the three inverse cases.

Why does my answer differ from the answer key in the last decimal place?

Almost always because a value was rounded partway through. Rounding 9 / sin 41° to 13.7 and then using 13.7 in a later step shifts the final digits. Carry the full display value through every intermediate step and round only the number you write down.

Can a right triangle have a leg longer than its hypotenuse?

No. The hypotenuse faces the 90° angle, which is the largest angle in any right triangle, and the largest angle always faces the longest side. That is exactly why Problem 19 has no solution: a hypotenuse of 8 cannot contain a leg of 10, and sin θ = 1.25 is outside the range of the sine function.