A vertical tower casts a shadow of length 48 m on horizontal ground. The angle of elevation of the sun is 58°. Work out the height of the tower. Give your answer correct to 1 decimal place.

Draw the right-angled triangle made by the tower and its shadow.: 58° = (h)/(48) Choose the ratio that links the side you know to the side you want.: = oppadj Label the sides of the triangle with respect to the given angle.: adjacent = 48 m, opposite = h m Substitute the lengths you have into the ratio.: 58° = (h)/(48) Rearrange to make the unknown side the subject.: h = 48 × 58° Work the calculation out on a calculator.: 58° = 1.60033 h = 48 × 1.60033 = 76.81605 Round the answer to the accuracy the question asks for.: h = 76.8160 = 76.8 m Check the rounding decision.: 76.816 digit 1 round down Check the answer is shorter than the hypotenuse.: 76.8 < 90.57 Check the answer by putting it back into the ratio.: (76.8)/(48) = 1.6000 1.6003 = 58° Recall the three trigonometric ratios.: = opphyp, = adjhyp, = oppadj Note the commonest mistake in this type of question.: , Note the other common mistake.: h h 58° Check the units of the answer.: 76.8 m State the height of the tower.: h = 48 × 58° = 76.8160 = 76.8 m Answer: h = 76.8 m

An isosceles triangle has a base of length 26 cm and both of its base angles are 67°. Work out the perpendicular height of the triangle. Give your answer correct to 2 decimal places.

Split the isosceles triangle into two right-angled triangles.: (26)/(2) = 13 Choose the ratio that links the side you know to the side you want.: = oppadj Label the sides of the triangle with respect to the given angle.: adjacent = 13 cm, opposite = h cm Substitute the lengths you have into the ratio.: 67° = (h)/(13) Rearrange to make the unknown side the subject.: h = 13 × 67° Work the calculation out on a calculator.: 67° = 2.35585 h = 13 × 2.35585 = 30.62608 Round the answer to the accuracy the question asks for.: h = 30.62608 = 30.63 cm Check the rounding decision.: 30.6260 digit 6 round up Check the answer is shorter than the hypotenuse.: 30.63 < 33.27 Check the answer by putting it back into the ratio.: (30.63)/(13) = 2.3561 2.3558 = 67° Recall the three trigonometric ratios.: = opphyp, = adjhyp, = oppadj Note the commonest mistake in this type of question.: , Note the other common mistake.: h h 67° Check the units of the answer.: 30.63 cm State the perpendicular height.: h = 13 × 67° = 30.6260 = 30.63 cm Answer: h = 30.63 cm

Hard GCSE Trigonometric ratios Questions

Challenging, exam-style GCSE Trigonometric ratios questions with worked solutions. Stretch yourself on the hardest finding a missing side, SOHCAHTOA, multi-step problem, area of a triangle problems.

finding a missing sideSOHCAHTOAmulti-step problemarea of a triangleperimeterfinding a missing angle

GCSE Foundation34 questionsStep-by-step solutions

Question 1

5 markschallenging

Which of these has the smallest value?

Show worked solution

Worked solution

Note what the question is asking for.
$\text{compare the five values}$
Each option is a trigonometric ratio of an acute angle, so each one is just a number. Work all five out and compare them.
Check the calculator is in degrees mode.
$\sin 30^\circ = 0.5$
Every angle here is in degrees. If the calculator is in radians, $\sin 30$ comes out as $-0.988\ldots$ instead of $0.5$ , so test it on an angle you know before you start.
Work out the cosine of 65 degrees.
$\cos 65^\circ = 0.4226\ldots$
A calculator gives $\cos 65^\circ = 0.4226\ldots$ .
Work out the sine of 35 degrees.
$\sin 35^\circ = 0.5735\ldots$
A calculator gives $\sin 35^\circ = 0.5735\ldots$ , which is more than $0.4226\ldots$ .
Work out the cosine of 40 degrees.
$\cos 40^\circ = 0.7660\ldots$
A calculator gives $\cos 40^\circ = 0.7660\ldots$ , which is more than $0.4226\ldots$ .
Work out the sine of 65 degrees.
$\sin 65^\circ = 0.9063\ldots$
A calculator gives $\sin 65^\circ = 0.9063\ldots$ , which is more than $0.4226\ldots$ .
Work out the tangent of 65 degrees.
$\tan 65^\circ = 2.1445\ldots$
A calculator gives $\tan 65^\circ = 2.1445\ldots$ , which is more than $0.4226\ldots$ .
Compare the five values.
$2.144 > 0.906 > 0.766 > 0.573 > 0.422$
Putting the five values in order shows that $0.4226\ldots$ is the smallest, so $\cos 65^\circ$ is the smallest of the five.
Note the range of sine and cosine.
$0 < \sin \theta < 1, \quad 0 < \cos \theta < 1$
For an acute angle both sine and cosine lie strictly between $0$ and $1$ , so neither can ever be very large.
Note the range of tangent.
$0 < \tan \theta < \infty$
Tangent is not capped at $1$ : it passes $1$ at $45^\circ$ and grows without limit as the angle approaches $90^\circ$ . So a tangent can beat any sine or cosine if the angle is big enough.
Note how sine and cosine behave as the angle grows.
$\sin \text{ increases}, \quad \cos \text{ decreases}$
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.
Note the connection between sine and cosine.
$\sin \theta = \cos (90^\circ - \theta)$
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.
Note where tangent overtakes sine and cosine.
$\tan 45^\circ = 1$
Tangent passes $1$ at exactly $45^\circ$ , so any tangent of an angle above $45^\circ$ beats every sine and every cosine, and any tangent below it need not.
Recall the three trigonometric ratios.
$\sin \theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos \theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan \theta = \frac{\text{opp}}{\text{adj}}$
SOHCAHTOA: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.
State the smallest of the five values.
$\cos 65^\circ = 0.4226\ldots$
$\cos 65^\circ$ is the smallest, at $0.4226\ldots$ .

Answer

\cos 65^\circ

Question 2

6 markschallenging

A right-angled triangle has an acute angle of

27^\circ

, and its side opposite this angle has length

9

cm. The side adjacent to this angle has length

x

cm. Which of these equations is correct?

Show worked solution

Worked solution

Label the sides of the triangle with respect to the given angle.
$\text{opposite} = 9\text{ cm}, \quad \text{adjacent} = x\text{ cm}$
The hypotenuse is the side opposite the right angle. Looking from the $27^\circ$ 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 opposite and asked for the adjacent.
Choose the ratio that links the side you know to the side you want.
$\tan \theta = \frac{\text{opp}}{\text{adj}}$
The opposite and the adjacent are exactly the two sides in the tangent ratio (TOA), so use $\tan$ . Neither of the other two ratios uses both of these sides.
Substitute the lengths you have into the ratio.
$\tan 27^\circ = \frac{9}{x}$
The opposite goes on top and the adjacent goes underneath. The unknown side is written as $x$ .
Check the fraction is the right way up.
$\tan 27^\circ = \frac{\text{opposite}}{\text{adjacent}}$
TOA puts the opposite side on top, so $9$ belongs above the line and $x$ below it.
Rule out the first wrong equation.
$\sin 27^\circ \ne \frac{9}{x}$
This option keeps the correct fraction but changes the ratio in front of it. $\sin$ of the same angle is a different number, so the equation is false.
Rule out the second wrong equation.
$\cos 27^\circ \ne \frac{9}{x}$
The same objection: $\cos 27^\circ$ is not equal to $\tan 27^\circ$ , so this equation cannot hold.
Rule out the upside-down equation.
$\tan 27^\circ \ne \frac{x}{9}$
Turning the fraction over gives the reciprocal of the correct ratio, which is a different number.
Rule out the rearranged equation with the wrong ratio.
$x \ne 9 \times \sin 27^\circ$
This one is rearranged correctly, but it uses $\sin$ where $\tan$ belongs, so it gives the wrong length.
Rearrange to make the unknown side the subject.
$x = \frac{9}{\tan 27^\circ}$
$x$ is underneath, so multiply both sides by $x$ and then divide by $\tan 27^\circ$ . The unknown is the denominator, so this is a DIVISION.
Check the value the correct equation gives.
$x = 17.6634\ldots$
The correct equation gives $x = 17.6634\ldots$ cm, which is a sensible length for this triangle. The other equations give lengths that do not fit.
Estimate the answer before trusting the calculator.
$27^\circ < 45^\circ \Rightarrow \text{opp} < \text{adj}$
Because $27^\circ$ is less than $45^\circ$ , the side opposite it must be shorter than the side next to it. The answer must fit that.
Recall the three trigonometric ratios.
$\sin \theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos \theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan \theta = \frac{\text{opp}}{\text{adj}}$
SOHCAHTOA: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.
Check the calculator is in degrees mode.
$\sin 30^\circ = 0.5$
Every angle here is in degrees. If the calculator is in radians, $\sin 30$ comes out as $-0.988\ldots$ instead of $0.5$ , so test it on an angle you know before you start.
Note the size of the other acute angle.
$90^\circ - 27^\circ = 63^\circ$
The three angles add to $180^\circ$ and one of them is the right angle, so the two acute angles add to $90^\circ$ . The other acute angle is $63^\circ$ , and what is "opposite" for one is "adjacent" for the other.
State the correct equation.
$\tan 27^\circ = \frac{9}{x}$
The tangent ratio links the opposite side and the adjacent side, and the opposite side goes on top.

Answer

\tan 27^\circ = \frac{9}{x}

Question 3

5 markschallenging

In a right-angled triangle, the side adjacent to the angle

\theta

has length

8

cm and the hypotenuse has length

17

cm. Which of these calculations gives the size of the angle

\theta

in degrees?

Show worked solution

Worked solution

Label the two known sides with respect to the unknown angle.
$\text{adjacent} = 8\text{ cm}, \quad \text{hypotenuse} = 17\text{ cm}$
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 adjacent and the hypotenuse.
Choose the ratio that uses the two sides you know.
$\cos \theta = \frac{\text{adj}}{\text{hyp}}$
The adjacent and the hypotenuse are the two sides in the cosine ratio (CAH), so use $\cos$ . Because the ANGLE is the unknown, you will need the inverse, $\cos^{-1}$ .
Write the ratio the two sides give.
$\cos \theta = \frac{8}{17}$
The adjacent goes on top and the hypotenuse underneath, so the ratio is $\frac{8}{17}$ .
Undo the ratio with the inverse function.
$\theta = \cos^{-1} \left(\frac{8}{17}\right)$
The ratio is known and the angle is not, so apply $\cos^{-1}$ — the function that turns a ratio back into an angle.
Rule out the first wrong inverse.
$\sin^{-1} \left(\frac{8}{17}\right) \ne \theta$
$\sin^{-1}$ would be the right function only if the two sides given were the opposite and the hypotenuse. They are not.
Rule out the second wrong inverse.
$\tan^{-1} \left(\frac{8}{17}\right) \ne \theta$
The same objection: $\tan$ uses a pair of sides the question does not give.
Rule out the upside-down ratio.
$\cos^{-1} \left(\frac{17}{8}\right) \ne \theta$
$\frac{17}{8}$ is the reciprocal of the correct ratio, so it describes a different triangle altogether.
Rule out the option that leaves out the inverse.
$\cos \left(\frac{8}{17}\right) \ne \theta$
This applies the ratio instead of undoing it. It treats $\frac{8}{17}$ as an ANGLE and returns a ratio, which is not what the question asks for.
Work the answer out to see it is sensible.
$\theta = \cos^{-1} \left(\frac{8}{17}\right) = 61.9\ldots^\circ$
The correct calculation gives about $61.9^\circ$ , which is an acute angle, as it must be.
Check the ratio is a sensible one.
$0 < \frac{8}{17} < 1$
For $\cos$ the ratio must lie between $0$ and $1$ , because the adjacent side is shorter than the hypotenuse. A ratio bigger than $1$ would have no inverse and would mean the lengths were wrong.
Recall the three trigonometric ratios.
$\sin \theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos \theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan \theta = \frac{\text{opp}}{\text{adj}}$
SOHCAHTOA: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.
Check the triangle with Pythagoras.
$15.00^2 + 8.00^2 \approx 17.00^2$
The three sides come to about $15.00$ , $8.00$ and $17.00$ cm, and those lengths satisfy Pythagoras' theorem, so the triangle the trigonometry has produced is a genuine right-angled triangle.
Check the calculator is in degrees mode.
$\sin 30^\circ = 0.5$
Every angle here is in degrees. If the calculator is in radians, $\sin 30$ comes out as $-0.988\ldots$ instead of $0.5$ , so test it on an angle you know before you start.
Remember that the hypotenuse is the longest side.
$\text{hyp} > \text{opp}, \quad \text{hyp} > \text{adj}$
The hypotenuse sits opposite the right angle and is always the longest side, so any answer that makes another side longer than it must be wrong.
State the correct calculation.
$\theta = \cos^{-1} \left(\frac{8}{17}\right)$
The two sides given are the adjacent and the hypotenuse, so the cosine ratio applies, and the angle is recovered with $\cos^{-1}$ .

Answer

\theta = \cos^{-1} \left(\frac{8}{17}\right)

Question 4

5 markschallenging

A vertical tower casts a shadow of length

48

m on horizontal ground. The angle of elevation of the sun is

58^\circ

. Work out the height of the tower. Give your answer correct to

1

decimal place.

Show worked solution

Worked solution

Draw the right-angled triangle made by the tower and its shadow.
$\tan 58^\circ = \frac{h}{48}$
The shadow lies along the ground, the tower is vertical, and the ray of sunlight from the top of the tower to the tip of the shadow makes the angle of elevation of the sun with the ground.
Choose the ratio that links the side you know to the side you want.
$\tan \theta = \frac{\text{opp}}{\text{adj}}$
The adjacent and the opposite are exactly the two sides in the tangent ratio (TOA), so use $\tan$ . Neither of the other two ratios uses both of these sides.
Label the sides of the triangle with respect to the given angle.
$\text{adjacent} = 48\text{ m}, \quad \text{opposite} = h\text{ m}$
The hypotenuse is the side opposite the right angle. Looking from the $58^\circ$ 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 adjacent and asked for the opposite.
Substitute the lengths you have into the ratio.
$\tan 58^\circ = \frac{h}{48}$
The opposite goes on top and the adjacent goes underneath. The unknown side is written as $h$ .
Rearrange to make the unknown side the subject.
$h = 48 \times \tan 58^\circ$
$h$ is on top of the fraction, so multiply both sides by $48$ . The unknown is the numerator, so this is a MULTIPLICATION.
Work the calculation out on a calculator.
$\tan 58^\circ = 1.60033\ldots \Rightarrow h = 48 \times 1.60033\ldots = 76.81605\ldots$
A calculator gives $\tan 58^\circ = 1.60033\ldots$ , and the calculation then gives $h = 76.81605\ldots$ . Keep all the digits until the very last step — rounding early loses accuracy.
Round the answer to the accuracy the question asks for.
$h = 76.8160\ldots = 76.8\text{ m}$
The question asks for $1$ decimal place. The value is $76.8160\ldots$ , and the digit after the $1$ th decimal place decides the rounding, giving $76.8$ .
Check the rounding decision.
$76.816\ldots \Rightarrow \text{digit } 1 \Rightarrow \text{round down}$
The digit after the last one you keep is $1$ . It is less than 5, so the last digit stays as it is, which turns $76.8\ldots$ into $76.8$ .
Check the answer is shorter than the hypotenuse.
$76.8 < 90.57$
The opposite side is not the hypotenuse, so it must be shorter than it. $76.8$ is less than $90.57$ , so the answer is the right size.
Check the answer by putting it back into the ratio.
$\frac{76.8}{48} = 1.6000\ldots \approx 1.6003\ldots = \tan 58^\circ$
Dividing the two sides gives $1.6000\ldots$ , which matches $\tan 58^\circ = 1.6003\ldots$ , so the answer fits the triangle.
Recall the three trigonometric ratios.
$\sin \theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos \theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan \theta = \frac{\text{opp}}{\text{adj}}$
SOHCAHTOA: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.
Note the commonest mistake in this type of question.
$\sin \ne \tan, \quad \cos \ne \tan$
Reaching for $\sin$ or $\cos$ here would use a side the question never gives you. Pick the ratio from the two sides that appear, not from habit.
Note the other common mistake.
$h \ne \frac{h}{\tan 58^\circ}$
The unknown is on TOP of the fraction here, so you multiply by the ratio. Dividing by it would make the side far too big.
Check the units of the answer.
$76.8\text{ m}$
Every length in the question is in m, so the height of the tower is in m too.
State the height of the tower.
$h = 48 \times \tan 58^\circ = 76.8160\ldots = 76.8\text{ m}$
The tower is $76.8$ m high.

Answer

h = 76.8\text{ m}

Question 5

6 markschallenging

An isosceles triangle has a base of length

26

cm and both of its base angles are

67^\circ

. Work out the perpendicular height of the triangle. Give your answer correct to

2

decimal places.

Show worked solution

Worked solution

Split the isosceles triangle into two right-angled triangles.
$\frac{26}{2} = 13$
The perpendicular height of an isosceles triangle drops onto the MIDPOINT of the base, so it cuts the triangle into two identical right-angled triangles, each with a base of $13$ cm.
Choose the ratio that links the side you know to the side you want.
$\tan \theta = \frac{\text{opp}}{\text{adj}}$
The adjacent and the opposite are exactly the two sides in the tangent ratio (TOA), so use $\tan$ . Neither of the other two ratios uses both of these sides.
Label the sides of the triangle with respect to the given angle.
$\text{adjacent} = 13\text{ cm}, \quad \text{opposite} = h\text{ cm}$
The hypotenuse is the side opposite the right angle. Looking from the $67^\circ$ 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 adjacent and asked for the opposite.
Substitute the lengths you have into the ratio.
$\tan 67^\circ = \frac{h}{13}$
The opposite goes on top and the adjacent goes underneath. The unknown side is written as $h$ .
Rearrange to make the unknown side the subject.
$h = 13 \times \tan 67^\circ$
$h$ is on top of the fraction, so multiply both sides by $13$ . The unknown is the numerator, so this is a MULTIPLICATION.
Work the calculation out on a calculator.
$\tan 67^\circ = 2.35585\ldots \Rightarrow h = 13 \times 2.35585\ldots = 30.62608\ldots$
A calculator gives $\tan 67^\circ = 2.35585\ldots$ , and the calculation then gives $h = 30.62608\ldots$ . Keep all the digits until the very last step — rounding early loses accuracy.
Round the answer to the accuracy the question asks for.
$h = 30.62608\ldots = 30.63\text{ cm}$
The question asks for $2$ decimal places. The value is $30.62608\ldots$ , and the digit after the $2$ th decimal place decides the rounding, giving $30.63$ .
Check the rounding decision.
$30.6260\ldots \Rightarrow \text{digit } 6 \Rightarrow \text{round up}$
The digit after the last one you keep is $6$ . It is 5 or more, so the last digit goes up, which turns $30.62\ldots$ into $30.63$ .
Check the answer is shorter than the hypotenuse.
$30.63 < 33.27$
The opposite side is not the hypotenuse, so it must be shorter than it. $30.63$ is less than $33.27$ , so the answer is the right size.
Check the answer by putting it back into the ratio.
$\frac{30.63}{13} = 2.3561\ldots \approx 2.3558\ldots = \tan 67^\circ$
Dividing the two sides gives $2.3561\ldots$ , which matches $\tan 67^\circ = 2.3558\ldots$ , so the answer fits the triangle.
Recall the three trigonometric ratios.
$\sin \theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos \theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan \theta = \frac{\text{opp}}{\text{adj}}$
SOHCAHTOA: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.
Note the commonest mistake in this type of question.
$\sin \ne \tan, \quad \cos \ne \tan$
Reaching for $\sin$ or $\cos$ here would use a side the question never gives you. Pick the ratio from the two sides that appear, not from habit.
Note the other common mistake.
$h \ne \frac{h}{\tan 67^\circ}$
The unknown is on TOP of the fraction here, so you multiply by the ratio. Dividing by it would make the side far too big.
Check the units of the answer.
$30.63\text{ cm}$
Every length in the question is in cm, so the height is in cm too.
State the perpendicular height.
$h = 13 \times \tan 67^\circ = 30.6260\ldots = 30.63\text{ cm}$
The perpendicular height is $30.63$ cm.

Answer

h = 30.63\text{ cm}

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