Question 1 of 24
Triangle ABC is right-angled at B. Which side is the hypotenuse?
Show a hint
The hypotenuse lies opposite the 90° angle.
Cambridge O Level · Mathematics 4024
Use Pythagoras' theorem and the sine, cosine and tangent ratios to solve fully described two-dimensional right-triangle problems. This original practice pack targets Cambridge O Level Mathematics 4024 sections 6.1 and 6.2, including perpendicular shortest distance and angles of elevation and depression. Coordinate distance is used only as a text-based application of Pythagoras; non-right-angled trigonometry, three-dimensional problems and standalone bearings techniques are excluded.
2025 / 2026 / 2027 · Academic review not recorded · Published 2026-09-08
AI-assisted practice — not independently academically reviewed. Answers may contain errors; check important results against your course materials.
Guided tutorial
19 lesson sections
Explanation
A right-angled triangle contains one 90° angle. The side opposite that angle is the hypotenuse; it is always the longest side. The other two sides are perpendicular legs. In triangle ABC with angle B = 90°, AC is the hypotenuse, while AB and BC are the legs. Naming the right angle first prevents the most common labelling error.
When one acute angle is selected, the two legs receive relative names. The leg touching the selected angle is adjacent, unless it is the hypotenuse; the leg across from the angle is opposite. These labels can change when the selected angle changes, but the hypotenuse never changes. Write the labels beside the stated lengths before selecting Pythagoras or a trigonometric ratio.
Worked example
Triangle PQR is right-angled at Q. The perpendicular legs are PQ = 7 cm and QR = 24 cm, so PR is the hypotenuse. Pythagoras gives PR² = PQ² + QR² = 7² + 24² = 49 + 576 = 625.
Taking the positive square root gives PR = 25 cm. A length uses the positive root, and 25 is longer than both 7 and 24, as a hypotenuse must be. The reverse check 25² - 24² = 625 - 576 = 49 = 7² confirms the calculation.
Worked example
Triangle ABC is right-angled at B. Its hypotenuse AC is 13 m and leg AB is 5 m. Let BC = x m. Because the unknown is a leg, subtract the known leg's square from the hypotenuse's square: x² = 13² - 5² = 169 - 25 = 144.
Therefore x = 12 and BC = 12 m. Adding instead of subtracting would produce a leg longer than the stated hypotenuse, which is impossible. The check 5² + 12² = 25 + 144 = 169 = 13² verifies both the arithmetic and the placement of the sides.
Explanation
If the squared length is not a perfect square, an exact answer can be left as a square root, such as √74 cm. Give a decimal only when requested or when a practical context needs one. Keep the full calculator value during later steps and round only the final result; early rounding can move a final angle or length outside the required accuracy.
To test three positive side lengths, first identify the longest as c. Compare c² with the sum a² + b² for the other two sides. Equality confirms a right angle opposite c; inequality means the three lengths are not the sides of a right-angled triangle. This check does not identify every property of a non-right triangle, so do not extend it into non-right-angled trigonometry.
Worked example
Point A has coordinates (-3, 2) and point B has coordinates (5, 8). The horizontal change is 5 - (-3) = 8 units and the vertical change is 8 - 2 = 6 units. These changes are perpendicular legs of a right triangle whose hypotenuse is the direct distance AB.
Thus AB = √(8² + 6²) = √100 = 10 units. Coordinate signs affect the changes, but the squared changes are non-negative. The direct distance must be shorter than travelling 8 units horizontally and then 6 units vertically, and 10 < 14 is a useful reasonableness check.
Worked example
A builder marks three points so that two sides from one point measure 9 m and 40 m, while the segment joining their other ends measures 41 m. The longest side is 41 m. Compare 41² = 1681 with 9² + 40² = 81 + 1600 = 1681.
The values are equal, so the angle between the 9 m and 40 m sides is 90°. This is a measurement check based on Pythagoras. It does not prove that an entire four-sided structure is a rectangle; it confirms only the described corner.
Explanation
For an acute angle θ in a right-angled triangle, sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse and tan θ = opposite/adjacent. A compact memory aid is SOH-CAH-TOA, but the words matter more than the initials. Label the sides relative to θ before writing a ratio.
Choose the ratio containing the known side and the required side. If the opposite side and hypotenuse are involved, use sine; if adjacent and hypotenuse are involved, use cosine; if opposite and adjacent are involved, use tangent. Calculator angle mode must be degrees for this syllabus context.
Worked example
Triangle DEF is right-angled at E. The hypotenuse DF is 18 cm, angle D is 32°, and the required side EF is opposite angle D. Write sin 32° = EF/18, then multiply by 18: EF = 18 sin 32°.
The calculator value is 9.5385..., so EF = 9.54 cm to three significant figures. The result is shorter than the 18 cm hypotenuse, as expected. Retaining the unrounded value 18 sin 32° until the final rounding avoids accumulated error.
Worked example
Triangle JKL is right-angled at K. Its hypotenuse JL is 20 m and angle J is 57°. Side JK touches angle J and is not the hypotenuse, so it is adjacent. Write cos 57° = JK/20 and rearrange to JK = 20 cos 57°.
This gives JK = 10.8927..., so JK = 10.9 m to three significant figures. Since cos 57° is between 0 and 1, multiplying 20 by it must give a value below 20. That quick check would expose an accidental division by cosine.
Worked example
Triangle MNO is right-angled at N. Relative to angle M = 41°, leg NO = 8 m is opposite and leg MN = x m is adjacent. Therefore tan 41° = 8/x. Multiplying by x and dividing by tan 41° gives x = 8/tan 41°.
The value is 9.2029..., so MN = 9.20 m to three significant figures. Because 41° is below 45°, the opposite leg should be shorter than the adjacent leg; 8 < 9.20 agrees with that expectation. This angle-size check is useful but does not replace the calculation.
Explanation
When two sides are known and an acute angle is required, first form the appropriate side ratio and then use the matching inverse function. For example, if opposite/adjacent = 3/4, then θ = tan⁻¹(3/4). The superscript -1 here means inverse function, not reciprocal: tan⁻¹ is not 1/tan.
Use degree mode and normally give an angle answer to one decimal place, following the syllabus instruction. A right triangle's two non-right angles add to 90°, so an acute answer must lie strictly between 0° and 90°. Keep the side ratio unrounded when entering the inverse function.
Worked example
Triangle RST is right-angled at S. Relative to angle R, the opposite leg ST is 7 cm and the adjacent leg RS is 24 cm. Therefore tan R = 7/24, so R = tan⁻¹(7/24). The complete side description makes the ratio unambiguous without a picture.
A degree-mode calculator gives R = 16.2602..., so R = 16.3° to one decimal place. The opposite leg is much shorter than the adjacent leg, so a relatively small acute angle is sensible. The other acute angle would be 90° - 16.3° = 73.7° after rounding.
Worked example
Triangle UVW is right-angled at V. The hypotenuse UW is 17 m and the leg UV adjacent to angle U is 8 m. Hence cos U = 8/17 and U = cos⁻¹(8/17). The remaining leg is not needed to find this angle.
The calculator gives U = 61.9275..., so U = 61.9° to one decimal place. The adjacent leg is less than half the hypotenuse, so the cosine is below 0.5 and an angle a little greater than 60° is reasonable. This estimate helps detect the wrong inverse function.
Explanation
In an applied problem, identify a horizontal line and a vertical or explicitly perpendicular line. Their intersection supplies the right angle. State what every side represents and keep all lengths in compatible units. The shortest distance from a point to a straight line is measured along the perpendicular segment, not along a sloping route.
An angle of elevation is measured upward from an observer's horizontal line; an angle of depression is measured downward from a horizontal line at the observer. Horizontal lines at different heights are parallel, so the matching acute angle inside the right triangle can be identified. Eye height or instrument height must be added or subtracted when the question distinguishes it from ground level.
Worked example
A vertical pole is 9 m tall. A straight guy wire runs from its top to a ground anchor 12 m horizontally from the foot of the pole. The vertical pole and horizontal ground distance are perpendicular legs, and the wire is the hypotenuse.
The wire length is √(9² + 12²) = √225 = 15 m. This model assumes level ground and a straight taut wire, both stated by the description. The answer exceeds each leg but is less than their sum, so it satisfies two useful length checks.
Worked example
From point A on level ground, the horizontal distance to the foot B of a vertical tower BT is 45 m. The angle of elevation from A to the top T is 32°. Relative to the 32° angle, BT is opposite and AB is adjacent, so tan 32° = BT/45.
Therefore BT = 45 tan 32° = 28.1191..., giving a tower height of 28.1 m to three significant figures. No observer height is added because the line of sight starts at ground point A as explicitly stated. The units remain metres throughout.
Worked example
An observer's eyes are 1.6 m above level ground. The observer stands 30 m horizontally from the foot of a vertical tree and measures an angle of elevation of 38° to its top. The vertical rise from eye level to the top is 30 tan 38° = 23.4386... m.
The total tree height is 23.4386... + 1.6 = 25.0386..., so it is 25.0 m to one decimal place. Adding the eye height is essential because tangent found only the rise above the observer's horizontal line, not the height above the ground.
Worked example
A point C is at the top of a vertical 40 m cliff directly above its foot F. A boat B is at sea level, and FB is horizontal. The angle of depression from C to B is 27°, measured from the horizontal through C. Parallel horizontals make the angle of elevation from B to C also 27°.
In right triangle BFC, the vertical side CF = 40 m is opposite 27° and the horizontal distance FB = x is adjacent. Thus tan 27° = 40/x, so x = 40/tan 27° = 78.5044.... The boat is 78.5 m from the cliff's foot to one decimal place.
Worked example
A vertical mast CT and the straight level ground line through A, C and B lie in one vertical plane. A support cable AT is 25 m long, and its ground anchor A is 7 m horizontally from the mast's foot C. In right triangle ACT, CT = √(25² - 7²) = √576 = 24 m.
An observer at B is 18 m horizontally from C on the same ground line. In right triangle BCT, the angle of elevation θ from B to the mast top T satisfies tan θ = 24/18. Therefore θ = 53.1301...°, or 53.1° to one decimal place. The shared height CT links two right triangles that remain in the stated single vertical plane.
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Question 1 of 24
The hypotenuse lies opposite the 90° angle.
Quick recall
Card 1 of 14
Front
Review the essentials
Use Pythagoras when a right angle is known and the information concerns side lengths only. Mark the hypotenuse first. To find it, add the squares of the legs; to find a leg, subtract the square of the known leg from the square of the hypotenuse. Take the positive square root for a geometric length.
Before accepting an answer, ask whether the hypotenuse is the longest side, whether the unit is a length rather than a squared unit, and whether the precision instruction has been followed. For coordinate points, state both coordinate changes. For a right-angle test, square the longest side and compare it with the sum of the other two squares.
Label opposite, adjacent and hypotenuse relative to the named acute angle. Select the one ratio containing both the known side and the target. When finding a side, rearrange the ratio algebraically; when finding an angle, apply the matching inverse function to a ratio of two sides.
Check degree mode before evaluating. Leg lengths must be positive and shorter than the hypotenuse. Acute-angle answers must lie between 0° and 90° and should normally be written to one decimal place. Preserve unrounded calculator values through a multi-step solution and attach the requested unit only to lengths, not to ratios.
For every problem: identify the right angle, name the target, label the sides from the relevant acute angle, choose Pythagoras or one trigonometric ratio, substitute unrounded values, and check units and size. Elevation and depression begin at a horizontal line. Include eye or instrument height only when the description requires it.
This pack covers the mapped two-dimensional content of sections 6.1 and 6.2. Coordinate distance is included only as an explicit Pythagoras application. Bearings may appear as supporting direction information in a 6.2 problem, but this pack does not teach bearings as a standalone technique. It does not teach sine rule, cosine rule, area using 1/2 ab sin C, ambiguous cases, three-dimensional trigonometry, or angles between a line and a plane.
Every example and question supplies coordinates or a complete text description of the right triangle; no answer depends on a missing diagram. Consequently, this text-only pack practises calculation and interpretation but does not claim to cover skills that require constructing, measuring or reading an unstated graphical figure.
Authorship: original ai assisted.
Original practice, not an official examination paper. Readnary is not affiliated with the awarding body. Prepared with AI assistance.