GCSE Trigonometric ratios Practice Questions

Free GCSE Trigonometric ratios practice questions with full step-by-step worked solutions. Covers finding a missing side, SOHCAHTOA, exact trigonometric values, special angles. Practise exam-style problems and check your method.

finding a missing sideSOHCAHTOAexact trigonometric valuesspecial anglesrounding to a given accuracyfinding a missing angle
GCSE Foundation70 questionsStep-by-step solutions
Question 1
2 markseasy
A right-angled triangle has an acute angle of 3030^\circ. The hypotenuse has length 1212 cm. Work out the length of the side opposite this angle. Give your answer as an exact value.
Show worked solution

Worked solution

  1. Label the sides of the triangle with respect to the given angle.

    hypotenuse=12 cm,opposite=x cm\text{hypotenuse} = 12\text{ cm}, \quad \text{opposite} = x\text{ cm}

    The hypotenuse is the side opposite the right angle. Looking from the 3030^\circ angle, the opposite side is the one across the triangle from it and the adjacent side is the one beside it. Here you are given the hypotenuse and asked for the opposite.

  2. Choose the ratio that links the side you know to the side you want.

    sinθ=opphyp\sin \theta = \frac{\text{opp}}{\text{hyp}}

    The hypotenuse and the opposite are exactly the two sides in the sine ratio (SOH), so use sin\sin. Neither of the other two ratios uses both of these sides.

  3. State the length of the opposite side.

    x=12×12=6 cmx = 12 \times \frac{1}{2} = 6\text{ cm}

    The opposite side is 66 cm long.

Answer
x=6 cmx = 6\text{ cm}
Question 2
1 markeasy
In a right-angled triangle you know the length of the hypotenuse and the size of one acute angle θ\theta. You want to work out the length of the adjacent side. Which trigonometric ratio links these two sides?
Show worked solution

Worked solution

  1. Write down the two sides the question involves.

    known=hypotenuse,wanted=adjacent\text{known} = \text{hypotenuse}, \quad \text{wanted} = \text{adjacent}

    You have the hypotenuse and you want the adjacent side. The right ratio is the one whose fraction contains BOTH of these and nothing else.

  2. Recall the three trigonometric ratios.

    sinθ=opphyp,cosθ=adjhyp,tanθ=oppadj\sin \theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos \theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan \theta = \frac{\text{opp}}{\text{adj}}

    SOHCAHTOA: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.

  3. State the ratio to use.

    cosθ=adjacenthypotenuse\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}

    The cosine ratio links the hypotenuse and the adjacent side, so it is the one to use.

Answer
cosθ=adjacenthypotenuse\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}
Question 3
2 marksintermediate
In a right-angled triangle, the side opposite the angle θ\theta has length 55 cm and the hypotenuse has length 1313 cm. Which of these calculations gives the size of the angle θ\theta in degrees?
Show worked solution

Worked solution

  1. Label the two known sides with respect to the unknown angle.

    opposite=5 cm,hypotenuse=13 cm\text{opposite} = 5\text{ cm}, \quad \text{hypotenuse} = 13\text{ cm}

    The hypotenuse is the side opposite the right angle. Looking from the unknown angle, the opposite side is the one across the triangle from it and the adjacent side is the one beside it. Here you know the opposite and the hypotenuse.

  2. Choose the ratio that uses the two sides you know.

    sinθ=opphyp\sin \theta = \frac{\text{opp}}{\text{hyp}}

    The opposite and the hypotenuse are the two sides in the sine ratio (SOH), so use sin\sin. Because the ANGLE is the unknown, you will need the inverse, sin1\sin^{-1}.

  3. Write the ratio the two sides give.

    sinθ=513\sin \theta = \frac{5}{13}

    The opposite goes on top and the hypotenuse underneath, so the ratio is 513\frac{5}{13}.

  4. Undo the ratio with the inverse function.

    θ=sin1(513)\theta = \sin^{-1} \left(\frac{5}{13}\right)

    The ratio is known and the angle is not, so apply sin1\sin^{-1} — the function that turns a ratio back into an angle.

  5. Rule out the first wrong inverse.

    cos1(513)θ\cos^{-1} \left(\frac{5}{13}\right) \ne \theta

    cos1\cos^{-1} would be the right function only if the two sides given were the adjacent and the hypotenuse. They are not.

  6. State the correct calculation.

    θ=sin1(513)\theta = \sin^{-1} \left(\frac{5}{13}\right)

    The two sides given are the opposite and the hypotenuse, so the sine ratio applies, and the angle is recovered with sin1\sin^{-1}.

Answer
θ=sin1(513)\theta = \sin^{-1} \left(\frac{5}{13}\right)
Question 4
4 markshard
In triangle XYZXYZ, angle XZYXZY is a right angle. With respect to angle ZYXZYX, which of these statements is correct?
Show worked solution

Worked solution

  1. Find the hypotenuse first.

    XZY=90hyp=YX\angle XZY = 90^\circ \Rightarrow \text{hyp} = YX

    The right angle is at ZZ, so the hypotenuse is the side opposite it, YXYX. The hypotenuse is never the opposite or the adjacent side.

  2. Look at the triangle from the given angle.

    θ=ZYX sits at Y\theta = \angle ZYX \text{ sits at } Y

    The middle letter of ZYXZYX is YY, so you are standing at YY and looking across the triangle. Opposite and adjacent always depend on WHICH angle you look from.

  3. Find the opposite side.

    opp=ZX\text{opp} = ZX

    The side across the triangle from YY — the one that does not touch YY at all — is ZXZX, so that is the opposite side.

  4. Find the adjacent side.

    adj=YZ\text{adj} = YZ

    The remaining side, YZYZ, touches YY and is not the hypotenuse, so it is the adjacent side: it lies between the angle and the right angle.

  5. Check every side has exactly one job.

    {YX,ZX,YZ}=hyp, opp, adj\{YX, ZX, YZ\} = \text{hyp, opp, adj}

    The three sides take the three roles, one each. Any option that gives the hypotenuse a second job as well must be wrong.

  6. Rule out the options that use the hypotenuse.

    YXopp,YXadjYX \ne \text{opp}, \quad YX \ne \text{adj}

    YXYX is the hypotenuse, so three of the options can be discarded immediately for calling it the opposite or the adjacent side.

  7. Rule out the option that swaps the two.

    YZoppYZ \ne \text{opp}

    YZYZ touches the angle at YY, so it cannot be the side OPPOSITE that angle. Swapping opposite and adjacent is the commonest slip here.

  8. Note what happens at the other acute angle.

    at X:opp=YZ,adj=ZX\text{at } X: \quad \text{opp} = YZ, \quad \text{adj} = ZX

    Looked at from XX instead, the two labels swap over. That is why you must always name the angle before you label the sides.

  9. Recall the three trigonometric ratios.

    sinθ=opphyp,cosθ=adjhyp,tanθ=oppadj\sin \theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos \theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan \theta = \frac{\text{opp}}{\text{adj}}

    SOHCAHTOA: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.

  10. State the correct labelling.

    opposite=ZX,adjacent=YZ\text{opposite} = ZX, \quad \text{adjacent} = YZ

    From angle ZYXZYX the opposite side is ZXZX and the adjacent side is YZYZ.

Answer
opposite=ZX,adjacent=YZ\text{opposite} = ZX, \quad \text{adjacent} = YZ
Question 5
5 markschallenging
Which of these has the smallest value?
Show worked solution

Worked solution

  1. Note what the question is asking for.

    compare the five values\text{compare the five values}

    Each option is a trigonometric ratio of an acute angle, so each one is just a number. Work all five out and compare them.

  2. Check the calculator is in degrees mode.

    sin30=0.5\sin 30^\circ = 0.5

    Every angle here is in degrees. If the calculator is in radians, sin30\sin 30 comes out as 0.988-0.988\ldots instead of 0.50.5, so test it on an angle you know before you start.

  3. Work out the cosine of 65 degrees.

    cos65=0.4226\cos 65^\circ = 0.4226\ldots

    A calculator gives cos65=0.4226\cos 65^\circ = 0.4226\ldots.

  4. Work out the sine of 35 degrees.

    sin35=0.5735\sin 35^\circ = 0.5735\ldots

    A calculator gives sin35=0.5735\sin 35^\circ = 0.5735\ldots, which is more than 0.42260.4226\ldots.

  5. Work out the cosine of 40 degrees.

    cos40=0.7660\cos 40^\circ = 0.7660\ldots

    A calculator gives cos40=0.7660\cos 40^\circ = 0.7660\ldots, which is more than 0.42260.4226\ldots.

  6. Work out the sine of 65 degrees.

    sin65=0.9063\sin 65^\circ = 0.9063\ldots

    A calculator gives sin65=0.9063\sin 65^\circ = 0.9063\ldots, which is more than 0.42260.4226\ldots.

  7. Work out the tangent of 65 degrees.

    tan65=2.1445\tan 65^\circ = 2.1445\ldots

    A calculator gives tan65=2.1445\tan 65^\circ = 2.1445\ldots, which is more than 0.42260.4226\ldots.

  8. Compare the five values.

    2.144>0.906>0.766>0.573>0.4222.144 > 0.906 > 0.766 > 0.573 > 0.422

    Putting the five values in order shows that 0.42260.4226\ldots is the smallest, so cos65\cos 65^\circ is the smallest of the five.

  9. Note the range of sine and cosine.

    0<sinθ<1,0<cosθ<10 < \sin \theta < 1, \quad 0 < \cos \theta < 1

    For an acute angle both sine and cosine lie strictly between 00 and 11, so neither can ever be very large.

  10. Note the range of tangent.

    0<tanθ<0 < \tan \theta < \infty

    Tangent is not capped at 11: it passes 11 at 4545^\circ and grows without limit as the angle approaches 9090^\circ. So a tangent can beat any sine or cosine if the angle is big enough.

  11. Note how sine and cosine behave as the angle grows.

    sin increases,cos decreases\sin \text{ increases}, \quad \cos \text{ decreases}

    As an acute angle gets bigger, its sine gets bigger and its cosine gets smaller. That alone settles several of these comparisons without a calculator.

  12. Note the connection between sine and cosine.

    sinθ=cos(90θ)\sin \theta = \cos (90^\circ - \theta)

    The sine of an angle equals the cosine of its complement, so the same value can be written in two different ways. Two options that look different can be equal.

  13. Note where tangent overtakes sine and cosine.

    tan45=1\tan 45^\circ = 1

    Tangent passes 11 at exactly 4545^\circ, so any tangent of an angle above 4545^\circ beats every sine and every cosine, and any tangent below it need not.

  14. Recall the three trigonometric ratios.

    sinθ=opphyp,cosθ=adjhyp,tanθ=oppadj\sin \theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos \theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan \theta = \frac{\text{opp}}{\text{adj}}

    SOHCAHTOA: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.

  15. State the smallest of the five values.

    cos65=0.4226\cos 65^\circ = 0.4226\ldots

    cos65\cos 65^\circ is the smallest, at 0.42260.4226\ldots.

Answer
cos65\cos 65^\circ

Unlock 65 more Trigonometric ratios questions

Create a free account to work through every GCSE Trigonometric ratios question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Trigonometric ratios practice

Related Geometry & Measures topics