Right Triangle Word Problems: 20 Solved Examples

Every right triangle word problem reduces to the same four steps: sketch the triangle, label the known sides and angles, choose the relationship that connects the known to the unknown, then solve and check. The wording changes from ladders to ramps to sailing routes, but the mathematics does not. Below are 20 fully solved examples covering Pythagorean problems, trigonometric side problems, angle problems and multi-step problems, each with the given information, the relationship used, the substitution, the arithmetic and a rounded answer with units.

How to set up any right triangle word problem

The four steps are worth doing in order every single time, because most wrong answers come from skipping the first two.

Step 1: sketch the triangle. Draw the right angle first and mark it with a small square. In applied problems the right angle is almost always where a vertical meets a horizontal: a wall meeting the ground, a mast meeting the earth, a building edge meeting a floor. Nothing in the wording needs to say “right angle” for the right angle to be there.

Step 2: label what you know. Write each given number on the side or angle it belongs to, with its unit. Mark the quantity you want with a letter. The single most useful question at this stage is which side is the hypotenuse, because the hypotenuse is always opposite the right angle and is always the longest side. The ladder, the ramp surface, the guy wire, the line of sight and the zip line are hypotenuses. The wall, the shadow, the rise and the run are legs.

Step 3: choose the relationship. If your three quantities are all sides, use the Pythagorean theorem. If an angle is involved, use a trigonometric ratio. The naming of “opposite” and “adjacent” is always relative to the acute angle you are working with, which is why the labels swap when you move to the other acute angle. The ratios themselves are covered in detail in the SOHCAHTOA guide.

Step 4: solve and check. Keep full calculator precision until the final line, then round and attach the unit. Check three things: the hypotenuse came out longest, the two acute angles add to 90°, and the answer is physically sensible. A ladder that reaches 40 m up a two-storey house is a signal that something went wrong.

Figure 1

Which relationship to use

This table turns step 3 into a lookup. Find the row that matches what you have and what you want.

Two shortcuts sit on top of the table. When a problem gives you one acute angle you get the other one free by subtracting from 90°, which often converts an awkward setup into an easy one. And when the numbers are 3-4-5, 5-12-13, 8-15-17 or a multiple of those, the Pythagorean arithmetic collapses to a single recognisable step.

Pythagorean problems

These six problems involve three sides and no angles, so the Pythagorean theorem does all the work. The only decision is whether the unknown is the hypotenuse (add the squares) or a leg (subtract the squares). For a deeper set of these, see the dedicated collection of Pythagorean theorem word problems.

Problem 1: ladder against a wall

The wall is vertical, the ground is horizontal, and the ladder is the hypotenuse because it faces the right angle. The ladder length is known and one leg is known, so the unknown leg comes from subtracting squares.

Figure 2

The picture also tells you the ladder sits at 73.7° to the ground, since tan⁻¹(7.2 ÷ 2.1) = 73.740°. That is close to the 75° usually recommended for a straight ladder, which is the kind of sanity check step 4 is for.

Problem 2: missing hypotenuse

A diagonal brace across a rectangular frame is a hypotenuse spanning two legs you already know. Both legs are given, so the squares get added.

The numbers 1.2, 1.6 and 2.0 are the 3-4-5 triple scaled by 0.4, which is why the answer is exact. Spotting that saves the arithmetic entirely.

Problem 3: missing leg

Here the hypotenuse is the string and one leg is the horizontal distance, so the vertical leg comes from subtracting squares. Note the answer is a height above the hand, not above the ground.

Check that 75 is less than 85. A leg that comes out longer than the hypotenuse means the squares were added when they should have been subtracted.

Problem 4: building diagonal

A diagonal across the face of a building joins two opposite corners of a rectangle, so it is the hypotenuse of a triangle whose legs are the width and the height.

That is the 5-12-13 triple scaled by 9, giving 45-108-117. A cable run or a sightline across a rectangular face always reduces to this same shape.

Problem 5: rectangular field diagonal

Same geometry as the building, different orientation. The field’s two sides are the legs and the diagonal is the hypotenuse.

√17725 simplifies to 5√709, since 17725 = 25 à - 709 and 709 is prime. That exact form is the honest answer, and 133.1 m is a rounding of it made at the last step.

Problem 6: shortest distance across a corner

Walking around two sides of a rectangle covers the sum of the legs; cutting straight across covers the hypotenuse. The saving is the difference, which is why corner-cutting footpaths appear on every rectangular lawn.

The diagonal is 300 m, a 3-4-5 triple scaled by 60. The shortcut saves about 29% of the walking distance.

Trigonometric side problems

In these six problems an angle is given and a side is wanted, so a ratio replaces the theorem. Decide first which side is opposite the given angle, which is adjacent, and which is the hypotenuse, then read the relationship straight off the decision table. Set the calculator to degree mode before starting.

Problem 7: ramp length from height and angle

The dock height is opposite the 15° angle and the ramp surface is the hypotenuse, so the pairing is sine. Because the hypotenuse is the unknown, the known side gets divided by the sine.

Check it backwards: 4.6364 à - sin 15° = 1.2 m, the height you started with. The ramp also needs 4.48 m of clear floor space, from 1.2 ÷ tan 15°.

Problem 8: tree height from its shadow

The shadow lies on the ground next to the angle, so it is adjacent; the tree is opposite. Opposite over adjacent is tangent, and the unknown is on top, so multiply.

The sun’s elevation and the tree height rise together: at 52° the tree is taller than its shadow, and at exactly 45° the two would be equal.

Problem 9: shadow length from the sun’s angle

This is problem 8 run in reverse. The pole is opposite the angle and the shadow is adjacent, so tangent applies again, but this time the unknown is on the bottom and the known side gets divided.

A high sun makes a short shadow. If the elevation dropped to 30° the same pole would throw a shadow of 9.2 ÷ tan 30° = 15.93 m.

Problem 10: tower height from an angle of elevation

The observer’s distance is adjacent to the angle of elevation and the tower is opposite, so this is the tangent multiply again. The answer is the height above eye level, which here is ground level because the angle was measured from the ground.

Figure 3

If the instrument had been on a 1.6 m tripod, the 66.41 m would be the height above the instrument and the tower would be 68.01 m tall. Problems that mention eye height always need that final addition.

Problem 11: guy wire length and anchor distance

The mast height is opposite the 58° angle at the anchor. Sine gives the wire because the wire is the hypotenuse, and tangent gives the ground distance because it is the adjacent leg.

Check with the theorem: 12² + 7.4984² = 144 + 56.23 = 200.23, and √200.23 = 14.15 m. When a trig answer and a Pythagorean answer agree, the setup was right.

Problem 12: trigonometric side problem on an incline

The belt itself is the hypotenuse and the rise is opposite the angle of inclination, so sine applies with the unknown on top.

The horizontal footprint is 8.5 à - cos 22° = 7.88 m. Sine and cosine of the same angle give the two legs of the same triangle, so one calculation rarely comes alone.

Angle problems

When two sides are known and the angle is wanted, use an inverse function: sin⁻¹, cos⁻¹ or tan⁻¹. Pick the one whose ratio is built from the two sides you actually have, and keep the calculator in degree mode so the result is an angle and not a radian measure. The elevation and depression cases below are treated at more length in the guide to angles of elevation and depression.

Problem 13: wheelchair ramp angle

The rise is opposite the ramp angle and the run is adjacent, so the ratio is a tangent and the angle comes from tan⁻¹.

That ratio is 1:11.67, slightly steeper than the 1:12 maximum written into most accessibility codes, since a 1:12 slope is tan⁻¹(1 ÷ 12) = 4.8°. To pass, the same 0.9 m rise would need a run of 0.9 à - 12 = 10.8 m. Angle problems like this one are usually really compliance questions in disguise.

Problem 14: angle of elevation

Both legs are known, so tangent again. The angle of elevation is measured up from the horizontal, and it sits at the observer, not at the top of the building.

The other acute angle, at the roof looking down to the observer, is 90° − 37.4° = 52.6°. The two always sum to 90°, which is a free check on any angle answer.

Problem 15: angle of depression

The angle of depression is measured down from the horizontal at the observer’s eye. It equals the angle of elevation from the boat back up to the observer, because those two angles are alternate angles between parallel horizontals.

Figure 4

The line of sight itself is √(85² + 240²) = 254.61 m. Putting the angle inside the triangle at the cliff top instead of above the horizontal is the most common error here, and it would give 70.5° rather than 19.5°.

Problem 16: trigonometric angle problem with a hypotenuse

This time the two known sides are the rise (opposite) and the slope length (hypotenuse), which is the sine pairing, so the inverse function is sin⁻¹.

Using tan⁻¹(7.2 ÷ 18) here would give 21.8°, which is wrong because 18 m is the slope, not the horizontal run. The horizontal run is √(18² − 7.2²) = 16.50 m, and tan⁻¹(7.2 ÷ 16.50) does return 23.6°.

Multi-step problems

The last four problems need two relationships in sequence. The rule that saves marks is to carry the unrounded intermediate value into the second step, and to round only the final answer.

Problem 17: roof rafter length and pitch

The rise and run are the legs of the rafter triangle, so the theorem gives the rafter and an inverse tangent gives the pitch angle. Both come from the same two numbers.

Figure 5

Builders often quote pitch as a ratio instead of an angle. A rise of 2.4 over a run of 4.5 is a 16-in-30 pitch, or roughly 6.4 in 12 in imperial framing terms. Real rafters are cut longer than 5.10 m to allow for the eaves overhang.

Problem 18: television screen with an aspect ratio

The diagonal alone is not enough, so the aspect ratio supplies the missing relationship. Write the sides as 16k and 9k, put them into the theorem, solve for k, then scale.

Figure 6

Check: √(47.9367² + 26.9644²) = 55.00 in, the advertised diagonal. This is why a 55 in 16:9 screen is narrower than a 55 in 4:3 screen from the 1990s, whose width would have been 44 in.

Problem 19: navigation with two perpendicular legs

Sailing east then north traces the two legs of a right triangle, so the theorem gives the straight-line distance and an inverse tangent gives the direction. The bearing is measured clockwise from north, so the east leg is opposite and the north leg is adjacent.

The return trip is the same 20 km on the reciprocal bearing of 216.9°. Swapping the ratio to tan⁻¹(16 ÷ 12) gives 53.1°, which is the angle measured from the east line instead of from north, so it answers a different question.

Problem 20: mixed multi-step problem with two triangles

A flagpole standing on a roof creates two triangles sharing the same base distance. Find the height to the top of the pole, find the height to the bottom of the pole, and subtract. Neither height alone is the answer.

The building alone is 43.59 m tall and the roofline plus pole reaches 55.95 m. Rounding each of those to 2 significant figures first, as 44 and 56, would give 12 m and lose the precision the question asked for. You can solve either triangle with the calculator to confirm both intermediate heights before subtracting.

Common mistakes

Using the hypotenuse as a leg. In problem 1 the ladder is the hypotenuse, so the working is √(7.5² − 2.1²) = 7.20 m. Adding instead gives √(7.5² + 2.1²) = 7.79 m, which quietly claims a 7.5 m ladder reaches 7.79 m up a wall. Whenever a leg comes out longer than the hypotenuse, the operation was wrong.

Mixing units inside one calculation. A rise given in millimetres and a run given in metres must be converted before dividing. In problem 13 the rise of 900 mm has to become 0.9 m to sit alongside a 10.5 m run. The same trap appears with inches and feet, and with kilometres and metres.

Leaving the calculator in radian mode. In problem 10, tan 38° = 0.7813 and the tower is 66.41 m tall. In radian mode the machine reads 38 as 38 radians, returns 0.3103, and the tower shrinks to 26.38 m. Answers that are wrong by a strange factor with no obvious arithmetic slip are usually this.

Rounding early and reusing the rounded value. In problem 20, rounding tan 36° to 0.73 and tan 43° to 0.93 gives a flagpole of exactly 12.00 m instead of 12.36 m, an error of 36 cm. Carry the full value in the calculator’s memory and round once, at the end.

Answering the wrong quantity. Problem 3 asks for height above the hand, not above the ground. Problem 20 asks for the pole, not the building. Problem 6 asks for the distance saved, not the diagonal. Before writing the final line, reread the sentence beginning with “find” and confirm the number you are about to write is that quantity, with its unit.

Frequently Asked Questions

How do you know whether to use the Pythagorean theorem or trigonometry?

Count the angles in the problem: if the only things mentioned are three side lengths, use the Pythagorean theorem, and if any acute angle is given or wanted, use a trigonometric ratio. Problems 1 to 6 above never mention an angle, so they are pure Pythagoras. Problems 7 to 16 each include an acute angle, so they need sine, cosine or tangent. Multi-step problems such as 17 and 19 use both because they ask for a length and an angle from the same pair of sides.

Do right triangle word problems always state that the angle is 90°?

No, the right angle is usually implied by the physical setup rather than stated. A wall meeting level ground, a mast standing vertically, a plumb line, a shadow cast on flat ground and the corner of a rectangular room or field are all right angles by construction. If nothing in the wording produces a perpendicular pair, the problem is not a right triangle problem and needs the law of sines or the law of cosines instead.

Is the angle of elevation always equal to the angle of depression?

Yes, for the same pair of points the angle of elevation from the lower point equals the angle of depression from the higher point. Both are measured from a horizontal line, and the two horizontals are parallel, so the angles are alternate angles across the line of sight. In problem 15 the observer’s angle of depression of 19.5° is also the angle of elevation a person on the boat would measure looking back up at the cliff top.

How much should you round in a right triangle word problem?

Round only on the final line, and match the precision of the data you were given. Measurements quoted to 2 decimal places support an answer to 2 decimal places; angles measured to the nearest degree rarely justify more than 1 decimal place in a derived angle. Keeping the unrounded value in the calculator between steps is what protects the last digit, and every answer above states its own precision for that reason.

Where can you check the answers to these problems?

Enter any two known parts into the right triangle calculator and it returns the remaining sides, both acute angles, the area and the perimeter, so every problem on this page can be verified in one step. It is a fast way to confirm a multi-step answer such as problem 17 or problem 20, where an early rounding error is otherwise hard to spot.