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Matemáticas GCSE

SOHCAHTOA Practice Questions with Worked Answers

Twelve right-angled triangle questions, missing sides and angles, each labelling opposite, adjacent and hypotenuse first and showing the calculator keystrokes.

La respuesta corta

Twelve right-angled triangle questions: six find a missing side, three find a missing angle, two are worded problems with a ladder and an angle of elevation, and one asks for exact surd answers. Every solution labels opposite, adjacent and hypotenuse first, then picks sin, cos or tan.

Los ejercicios

  1. Ejercicio 1Básico2 puntos

    A right-angled triangle has an angle of 30° and a hypotenuse of 12 cm. Find the side opposite the 30° angle.

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    6 cm

    1. From the 30° angle: 12 cm is the hypotenuse, and the unknown side faces the angle, so it is the opposite.
    2. Opposite and hypotenuse means sin: sin 30 = x / 12.
    3. x = 12 × sin 30
    4. Keys: 1 2 × sin 3 0 =. sin 30 = 0.5 exactly, so x = 6 cm.
  2. Ejercicio 2Básico2 puntos

    A right-angled triangle has an angle of 40° and a hypotenuse of 9 cm. Find the side adjacent to the 40° angle, correct to 1 decimal place.

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    6.9 cm

    1. From the 40° angle: 9 cm is the hypotenuse, and the unknown lies between the angle and the right angle, so it is the adjacent.
    2. Adjacent and hypotenuse means cos: cos 40 = x / 9.
    3. x = 9 × cos 40
    4. Keys: 9 × cos 4 0 =, giving 6.8944, so x = 6.9 cm to 1 d.p.
  3. Ejercicio 3Básico2 puntos

    A right-angled triangle has an angle of 35° and the side adjacent to it is 10 cm. Find the opposite side, correct to 1 decimal place.

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    7.0 cm

    1. From the 35° angle: 10 cm is the adjacent and the unknown is the opposite. The hypotenuse is not involved at all.
    2. Opposite and adjacent means tan: tan 35 = x / 10.
    3. x = 10 × tan 35 = 7.0021
    4. x = 7.0 cm to 1 d.p. The trailing zero is part of the answer at 1 d.p.
  4. Ejercicio 4Estándar3 puntos

    In a right-angled triangle the side opposite angle θ is 5 cm and the side adjacent to it is 12 cm. Find θ, correct to 1 decimal place.

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    22.6°

    1. The two known sides are the opposite (5 cm) and the adjacent (12 cm), so the ratio is tan.
    2. tan θ = 5 / 12 = 0.41666...
    3. θ = tan⁻¹(5/12). Keys: shift tan ( 5 ÷ 1 2 ) =.
    4. θ = 22.6199..., so θ = 22.6° to 1 d.p.
  5. Ejercicio 5Estándar3 puntos

    In a right-angled triangle the hypotenuse is 14 cm and the side opposite angle θ is 6 cm. Find θ, correct to 1 decimal place.

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    25.4°

    1. 14 cm is the hypotenuse and 6 cm faces the angle, so the ratio is sin.
    2. sin θ = 6 / 14 = 0.428571...
    3. θ = sin⁻¹(6/14). Keys: shift sin ( 6 ÷ 1 4 ) =.
    4. θ = 25.3769..., so θ = 25.4° to 1 d.p.
  6. Ejercicio 6Estándar3 puntos

    In a right-angled triangle the hypotenuse is 17 cm and the side adjacent to angle θ is 8 cm. Find θ, correct to 1 decimal place.

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    61.9°

    1. 8 cm lies between the angle and the right angle, so it is the adjacent; 17 cm is the hypotenuse. The ratio is cos.
    2. cos θ = 8 / 17 = 0.470588...
    3. θ = cos⁻¹(8/17). Keys: shift cos ( 8 ÷ 1 7 ) =.
    4. θ = 61.9275..., so θ = 61.9° to 1 d.p. It is more than 45°, which fits: the adjacent is much shorter than the hypotenuse.
  7. Ejercicio 7Estándar3 puntos

    A right-angled triangle has an angle of 52°, and the side opposite it is 9 cm. Find the hypotenuse, correct to 1 decimal place.

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    11.4 cm

    1. From the 52° angle: 9 cm is the opposite and the unknown is the hypotenuse, so the ratio is sin.
    2. sin 52 = 9 / h - the unknown is underneath.
    3. Rearranging: h = 9 ÷ sin 52.
    4. Keys: 9 ÷ sin 5 2 =, giving 11.4212, so h = 11.4 cm to 1 d.p.
    5. Multiplying instead would give 7.1 cm, a hypotenuse shorter than the opposite side, which is impossible.
  8. Ejercicio 8Estándar3 puntos

    A right-angled triangle has an angle of 63°, and the side opposite it is 15 cm. Find the adjacent side, correct to 1 decimal place.

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    7.6 cm

    1. From the 63° angle: 15 cm is the opposite, the unknown is the adjacent, so the ratio is tan.
    2. tan 63 = 15 / x, with the unknown underneath.
    3. x = 15 ÷ tan 63
    4. Keys: 1 5 ÷ tan 6 3 =, giving 7.6429, so x = 7.6 cm to 1 d.p.
    5. 63° is a large angle, so the side facing it should be much the longer of the two - and 15 cm against 7.6 cm is.
  9. Ejercicio 9Estándar3 puntos

    A right-angled triangle has an angle of 25°, and the side adjacent to it is 20 m. Find the hypotenuse, correct to 1 decimal place.

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    22.1 m

    1. From the 25° angle: 20 m is the adjacent and the unknown is the hypotenuse, so the ratio is cos.
    2. cos 25 = 20 / h, so h = 20 ÷ cos 25.
    3. Keys: 2 0 ÷ cos 2 5 =, giving 22.0676.
    4. h = 22.1 m to 1 d.p. - only a little longer than the adjacent, which fits a small angle of 25°.
  10. Ejercicio 10Avanzado5 puntos

    A 6 m ladder leans against a vertical wall, making an angle of 72° with the ground. How far up the wall does it reach, and how far is its foot from the wall? Give both to 1 decimal place.

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    5.7 m up the wall, with the foot 1.9 m from the wall

    1. From the 72° angle at the ground: the ladder (6 m) is the hypotenuse, the wall is the opposite, the ground is the adjacent.
    2. Up the wall: sin 72 = h / 6, so h = 6 × sin 72 = 5.7063, that is 5.7 m.
    3. Out from the wall: cos 72 = d / 6, so d = 6 × cos 72 = 1.8541, that is 1.9 m.
    4. Check with Pythagoras on the unrounded figures: 5.7063² + 1.8541² = 32.56 + 3.44 = 36 = 6².
  11. Ejercicio 11Avanzado4 puntos

    From a point 45 m from the base of a tower, the angle of elevation of the top of the tower is 38°. Find the height of the tower, correct to 1 decimal place.

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    35.2 m

    1. The 45 m runs along the ground away from the angle, so it is the adjacent. The tower height faces the angle, so it is the opposite.
    2. Opposite and adjacent means tan: tan 38 = h / 45.
    3. h = 45 × tan 38 = 35.1579
    4. The tower is 35.2 m tall to 1 d.p., taking the angle as measured from ground level.
  12. Ejercicio 12Avanzado4 puntos

    A right-angled triangle has a hypotenuse of 8 cm and an angle of 60°. Find the other two sides, giving exact answers.

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    Opposite = 4√3 cm, adjacent = 4 cm

    1. From the 60° angle: 8 cm is the hypotenuse, and the two unknowns are the opposite and the adjacent.
    2. Opposite: sin 60 = √3/2, so the opposite = 8 × √3/2 = 4√3 cm.
    3. Adjacent: cos 60 = 1/2, so the adjacent = 8 × 1/2 = 4 cm.
    4. Check with Pythagoras: (4√3)² + 4² = 48 + 16 = 64 = 8².
    5. As a decimal 4√3 is 6.93 cm, but a question asking for an exact answer wants the surd.

Dónde se falla

  • Labelling opposite and adjacent from the right angle instead of from the angle in the question. Only the hypotenuse is fixed by the right angle; the other two labels move with the angle.
  • Picking sin because the word hypotenuse appears, without checking whether the second known side is the opposite or the adjacent.
  • Multiplying when the unknown is in the denominator: writing h = 9 × sin 52 = 7.1 rather than h = 9 ÷ sin 52 = 11.4, and not noticing the hypotenuse came out shorter than a shorter side.
  • Leaving the calculator in radians, which turns tan 35 into 0.474 instead of 0.700 and sin 30 into -0.988 instead of 0.5.
  • Rounding the ratio before taking the inverse: using 0.43 rather than 6/14 shifts the angle from 25.4° to 25.5°.

Label from the angle, not from the right angle

The hypotenuse never moves: it is opposite the right angle, always. The other two labels do move, and they move with the angle named in the question. The side facing that angle is the opposite; the remaining side, the one between the angle and the right angle, is the adjacent.

Swap to the other acute angle in the same triangle and opposite and adjacent trade places. That is why the labelling has to be redone for every question rather than pencilled on once at the start, and why every solution here begins by naming the three sides from the angle actually given.

When the unknown ends up underneath

Three of these twelve put the unknown in the denominator. sin 52 = 9 / h does not rearrange to h = 9 × sin 52; it rearranges to h = 9 ÷ sin 52. The two give 7.1 cm and 11.4 cm, and only one of them can be a hypotenuse when the opposite side is 9 cm.

The sense check is quicker than the algebra. The hypotenuse is the longest side, so an answer for it that is shorter than a side you were given is wrong. Likewise, opposite a large angle sits a long side, and opposite a small angle a short one, so a 63° angle facing a side of 15 cm should leave an adjacent much shorter than 15.

The keys, and the one setting that ruins everything

Check the display says DEG before starting. A calculator left in radians gives sin 30 as -0.988 instead of 0.5 and tan 35 as 0.474 instead of 0.700 - numbers that are not obviously absurd, so the mistake survives all the way to the answer line. This is the single most common cause of a whole page of trigonometry being wrong for one reason.

For a side, the order is: type the known length, then × or ÷, then the trig key, then the angle. For an angle, use the inverse: shift and sin, cos or tan, then the ratio in brackets. Bracketing the ratio matters, because sin⁻¹ 6 ÷ 14 and sin⁻¹ (6 ÷ 14) are not the same instruction on most calculators.

  • Finding a side with the unknown on top: known length, ×, trig key, angle - as in 9 × cos 40
  • Finding a side with the unknown underneath: known length, ÷, trig key, angle - as in 9 ÷ sin 52
  • Finding an angle: shift, then sin/cos/tan, then the ratio in brackets
  • Round only at the end, never the intermediate ratio

Preguntas frecuentes

What does SOHCAHTOA actually stand for?

Sin = Opposite over Hypotenuse, Cos = Adjacent over Hypotenuse, Tan = Opposite over Adjacent. It is a memory aid for three ratios, not a method. The work is deciding which two sides you have, and that decision comes before the mnemonic is any use.

How do I choose between sin, cos and tan?

By the two sides involved in the question - the one you have and the one you want. Opposite with hypotenuse is sin, adjacent with hypotenuse is cos, opposite with adjacent is tan. Label all three sides first and the choice makes itself; guessing at the ratio first is what produces wrong answers with correct-looking working.

When do I use the sin⁻¹, cos⁻¹ and tan⁻¹ buttons?

Only when the answer you want is an angle. If you have two sides and need the angle between or opposite them, form the ratio and apply the inverse. If you have an angle and one side and need another side, no inverse is involved - the ordinary sin, cos or tan key does the job.

How many decimal places should the answer have?

Whatever the question specifies, and 1 decimal place or 3 significant figures if it does not. The important part is where the rounding happens: keep the full calculator value all the way through and round only the final line. Rounding an intermediate step is what turns 25.4° into 25.5°.

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