Solve right triangles in applied problems using trigonometric ratios and the Pythagorean Theorem
Several sessions: ruler-and-protractor experiments or coordinate work, a proof discussed as a class, modeled examples, independent practice and a written argument.
math-2017:G.SRT.8 Solve problems involving right triangles.
math-2017:G.SRT.8a Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems if one of the two acute angles and a side length is given. (G, M2)
math-2017:G.SRT.8b Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. (A2, M3)
Provisional targets. Clause-by-clause educational review is still pending.
📘Section 1 · LearnLearn the idea
From an instrument 1.6 m above level ground and 50 m from the base of a tower, the angle of elevation to the top is 32°. The horizontal leg is 50 m, adjacent to the 32° angle; the vertical leg is 50 tan 32° ≈ 31.24 m; adding the instrument height gives a tower height of about 32.8 m. The short green posts mark the instrument and tower bases on the ground line. The drawing is to scale at 6 pixels per meter.
Learning goal: Solve problems involving right triangles: use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems, including when one of the two acute angles and a side length is given, and when two sides are given.
Scope note. Part b of G.SRT.8 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full as part of the Geometry course, as extension content for students on that path.
What "solve a right triangle" means. A right triangle has three sides and two acute angles, five unknowns in all. Solving it means finding every one of them from the given information. Two pieces of information beyond the right angle are enough: either two sides, or one side and one acute angle. The tools are the Pythagorean Theorem, a² + b² = c² for legs a, b and hypotenuse c; the angle sum, so the two acute angles add to 90°; and the three trigonometric ratios, sine = opposite/hypotenuse, cosine = adjacent/hypotenuse, tangent = opposite/adjacent, with the inverse ratios on a calculator to recover an angle from a ratio. Throughout, a calculator in degree mode gives decimals, and the rule is to round only at the end and to state the unit.
Case 1: one acute angle and a side are given. Choose the ratio that connects the known side to the wanted side through the known angle. A 6 m ladder leans against a vertical wall making a 70° angle with level ground. How high up the wall does it reach, and how far is its foot from the wall? The ladder is the hypotenuse; the height is opposite the 70° angle; the distance from the wall is adjacent. Height = 6 sin 70° ≈ 6 × 0.9397 ≈ 5.64 m. Distance = 6 cos 70° ≈ 6 × 0.3420 ≈ 2.05 m. Check with the Pythagorean Theorem: 5.638² + 2.052² ≈ 31.79 + 4.21 = 36.0 = 6². The other acute angle, between the ladder and the wall, is 90° − 70° = 20°. All five parts are now known, so the triangle is solved.
Case 1, a surveying-style measurement. From a point 50 m from the base of a tower on level ground, the angle of elevation to the top of the tower, measured from an instrument 1.6 m above the ground, is 32°. The angle of elevation is the angle between the horizontal and the line of sight upward. The horizontal distance 50 m is adjacent to the 32° angle and the height above the instrument is opposite, so use tangent: height above instrument = 50 tan 32° ≈ 50 × 0.6249 ≈ 31.24 m. The tower's height is that plus the instrument height, 31.24 + 1.6 ≈ 32.84 m, about 32.8 m. Skipping the instrument height is a common error; the right triangle's horizontal leg is at eye level, not at the ground.
Case 2: two sides are given. A ramp rises 0.75 m over a horizontal run of 9 m. Find the length of the ramp and the angle it makes with the ground. The length is the hypotenuse: √(0.75² + 9²) = √(0.5625 + 81) = √81.5625 ≈ 9.03 m. The angle has opposite 0.75 and adjacent 9, so tan θ = 0.75/9 = 1/12, and θ = tan⁻¹(1/12) ≈ 4.76°. The other acute angle is 90° − 4.76° ≈ 85.24°. The inverse tangent key answers "which angle has this tangent?"; on a calculator it is often labeled tan⁻¹ or atan and must be used in degree mode.
Which ratio to use. Label the triangle from the point of view of the known or wanted angle: hypotenuse, opposite, adjacent. If the hypotenuse is involved with the opposite side, use sine; with the adjacent side, use cosine; if the two legs are involved, use tangent. When the unknown is in the denominator, as in "the guy wire makes 58° with the ground and reaches a point 12 m up the pole; how long is the wire?", write sin 58° = 12/wire and solve: wire = 12/sin 58° ≈ 12/0.8480 ≈ 14.15 m. The foot of the wire is 12/tan 58° ≈ 12/1.6003 ≈ 7.50 m from the pole. Check: 14.15² ≈ 200.2 and 12² + 7.498² ≈ 144 + 56.2 = 200.2.
A worked model, fully solved. A right triangle has hypotenuse 20 cm and one acute angle 38°. The leg opposite 38° is 20 sin 38° ≈ 20 × 0.6157 ≈ 12.31 cm; the leg adjacent to 38° is 20 cos 38° ≈ 20 × 0.7880 ≈ 15.76 cm; the other acute angle is 52°. Pythagorean check: 12.313² + 15.760² ≈ 151.6 + 248.4 = 400.0 = 20². The check catches arithmetic slips, but it cannot catch a swapped sine and cosine: legs 15.76 opposite and 12.31 adjacent satisfy the same equation. What catches the swap is the labels: 38° is less than 45°, so the leg opposite it must be the shorter leg, 12.31 cm.
Precision and rounding. Keep at least four decimal places in the ratios and round the final answer to a precision that matches the data: a distance given as 50 m deserves an answer to the nearest tenth of a meter, not to the nearest millimeter. Rounding sin 70° to 0.94 before multiplying gives 5.64 m, the same to two decimals here, but rounding to 0.9 gives 5.4 m, which is wrong at the level of precision the problem carries.
The misconception. Students often apply a trigonometric ratio to a triangle that has no right angle, or to the wrong angle of a right triangle. The ratios in this lesson are defined for an acute angle in a right triangle; the angle of elevation problem works because the tower is vertical and the ground horizontal, which creates the right angle. A triangle formed by two hills and a valley floor has no right angle and needs the tools of a later lesson.
✏️Section 2 · ModelWork through the examples
Worked, angle and hypotenuse given: A kite string 40 m long makes an angle of 55° with the horizontal, assumed straight. Height of the kite above the hand: 40 sin 55° ≈ 40 × 0.8192 ≈ 32.77 m. Horizontal distance: 40 cos 55° ≈ 40 × 0.5736 ≈ 22.94 m. Check: 32.766² + 22.943² ≈ 1073.6 + 526.4 = 1600 = 40². The other angle, between the string and the vertical, is 35°.
Worked, two legs given: A right triangle has legs 5 and 12. Hypotenuse: √(25 + 144) = 13. The angle opposite the leg 5: tan θ = 5/12, so θ = tan⁻¹(5/12) ≈ 22.62°; the other acute angle is 90° − 22.62° = 67.38°. Check with sine: sin 22.62° ≈ 0.3846 = 5/13. Solved: sides 5, 12, 13; angles 22.62°, 67.38°, 90°.
Worked, angle and adjacent leg, with an instrument height: A tree casts no usable shadow, so a surveyor stands 18 m from its base and measures the angle of elevation to the top as 41° from an instrument 1.5 m above the ground. Height above the instrument: 18 tan 41° ≈ 18 × 0.8693 ≈ 15.65 m. Tree height ≈ 15.65 + 1.5 = 17.15 m, about 17.1 m. Leaving out the 1.5 m would understate the height by the instrument height.
Worked, unknown in the denominator: A plane flies 2000 m along a straight path inclined 3° below the horizontal. Altitude lost: 2000 sin 3° ≈ 2000 × 0.05234 ≈ 104.7 m. Horizontal ground distance: 2000 cos 3° ≈ 1997.3 m. To lose 2000 m of altitude at that angle, the path length would be 2000/sin 3° ≈ 38,215 m, about 38.2 km, because the altitude is opposite the 3° angle and the path is the hypotenuse.
🧩Section 3 · DoPractice, demonstrate, and explain
Recording task: For the ladder problem (6 m ladder at 70° to level ground), draw the right triangle to scale at 1 cm per meter, label the hypotenuse 6, the angle 70°, and mark which leg is opposite and which is adjacent. Record the computed height 6 sin 70° ≈ 5.64 m and distance 6 cos 70° ≈ 2.05 m, then measure the drawing and record the measured values beside them. Record the Pythagorean check 5.64² + 2.05² ≈ 36.0 and the other acute angle 20°. Then repeat the record for the ramp problem (rise 0.75 m, run 9 m), recording the ramp length 9.03 m and angle 4.76° from tan⁻¹(1/12).
Independent performance: Solve four applied right-triangle problems completely, each giving all three sides and both acute angles: a guy wire making 58° with the ground attached 12 m up a pole; a 20 cm hypotenuse with a 38° angle; a tower whose angle of elevation is 32° from an instrument 1.6 m above the ground at a horizontal distance of 50 m; and a rectangular field 120 m by 50 m crossed by a diagonal path. For each, state which ratio you chose and why (which side is opposite, adjacent or the hypotenuse relative to the known angle), give each answer with its unit and a sensible rounding, and check every triangle with the Pythagorean Theorem. Success criteria: each ratio choice is justified by the labels; instrument height is handled explicitly; no ratio is rounded before the final step; each Pythagorean check agrees to the stated precision; and the diagonal path problem uses the inverse tangent to report the angle the path makes with the long side, about 22.6°.
✅Section 4 · Check8-question lesson practice test
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Problem 1
A 6 m ladder leans against a vertical wall at 70° to level ground. Find how high it reaches, how far its foot is from the wall, and the other acute angle, and check with the Pythagorean Theorem.
Show worked solution
The ladder is the hypotenuse. Height (opposite 70°) = 6 sin 70° ≈ 6 × 0.9397 ≈ 5.64 m. Foot distance (adjacent) = 6 cos 70° ≈ 6 × 0.3420 ≈ 2.05 m. The other acute angle is 90° − 70° = 20°. Check: 5.638² + 2.052² ≈ 31.79 + 4.21 = 36.0 = 6², so the arithmetic is consistent. The check cannot show which leg is which (swapped legs pass it too); the picture does: at a steep 70° the height must be the longer leg, and 5.64 m > 2.05 m.
Problem 2
From an instrument 1.6 m above level ground, 50 m from the base of a tower, the angle of elevation to the top is 32°. Find the height of the tower.
Show worked solution
The horizontal distance 50 m is adjacent to the 32° angle and the height above the instrument is opposite, so tan 32° = h/50 and h = 50 tan 32° ≈ 50 × 0.6249 ≈ 31.24 m. The tower is 31.24 + 1.6 ≈ 32.84 m, about 32.8 m tall. The instrument height must be added because the right triangle’s horizontal leg is at instrument level, not at the ground.
Problem 3
A ramp rises 0.75 m over a horizontal run of 9 m. Find its length and the angle it makes with the ground.
Show worked solution
Length = √(0.75² + 9²) = √81.5625 ≈ 9.03 m by the Pythagorean Theorem. The angle has opposite 0.75 and adjacent 9, so tan θ = 0.75/9 = 1/12 and θ = tan⁻¹(1/12) ≈ 4.76°. Check: sin 4.76° ≈ 0.0830 and 0.75/9.031 ≈ 0.0830, so the angle and the sides agree.
Problem 4
A guy wire makes an angle of 58° with level ground and is attached to a pole 12 m above the ground. Find the length of the wire and the distance from its foot to the pole.
Show worked solution
The 12 m height is opposite the 58° angle; the wire is the hypotenuse, so sin 58° = 12/wire and wire = 12/sin 58° ≈ 12/0.8480 ≈ 14.15 m. The foot distance is adjacent: tan 58° = 12/d, so d = 12/tan 58° ≈ 12/1.6003 ≈ 7.50 m. Check: 12² + 7.498² ≈ 144 + 56.2 = 200.2 and 14.150² ≈ 200.2.
Problem 5
A right triangle has hypotenuse 20 cm and one acute angle 38°. Solve the triangle and explain what the Pythagorean check would reveal if sine and cosine had been swapped.
Show worked solution
Opposite leg = 20 sin 38° ≈ 12.31 cm; adjacent leg = 20 cos 38° ≈ 15.76 cm; other acute angle = 52°. Check: 12.313² + 15.760² ≈ 151.6 + 248.4 = 400 = 20². Swapping sine and cosine gives the same two numbers in the other positions, so the Pythagorean check alone would still pass; what catches the swap is the meaning of the labels, the leg opposite 38° must be the shorter one, since 38° < 45°. Use the check together with the labels.
Problem 6
A rectangular field is 120 m by 50 m. Find the length of the diagonal path and the angle it makes with the long side.
Show worked solution
Diagonal = √(120² + 50²) = √(14400 + 2500) = √16900 = 130 m. The angle with the long side has opposite 50 and adjacent 120, so tan θ = 50/120 = 5/12 and θ = tan⁻¹(5/12) ≈ 22.62°. The angle with the short side is 90° − 22.62° ≈ 67.38°. Both sides given, so the Pythagorean Theorem gives the third side and an inverse ratio gives the angle.
Problem 7
A student computes the height reached by the 6 m ladder at 70° as 6 cos 70° ≈ 2.05 m. Identify the error and explain how the size of the answer could have flagged it.
Show worked solution
The height is opposite the 70° angle, so it needs sine, not cosine: 6 sin 70° ≈ 5.64 m. A ladder at a steep 70° angle should reach most of its length up the wall, and 2.05 m is barely a third of 6 m, which fits a shallow angle, not a steep one. Comparing the answer with the picture is a fast check on which ratio was used.
Problem 8
A plane flies 2000 m along a straight path inclined 3° below the horizontal. How much altitude does it lose, and how far along the path would it have to fly at that angle to lose 2000 m of altitude?
Show worked solution
Altitude lost is opposite the 3° angle with the path as hypotenuse: 2000 sin 3° ≈ 2000 × 0.05234 ≈ 104.7 m. To lose 2000 m, the path length L satisfies sin 3° = 2000/L, so L = 2000/sin 3° ≈ 38,215 m, about 38.2 km. The unknown was in the denominator, so the equation was solved for L rather than multiplied out.