Trigonometry
Geometry and measures
- Crossover
- Calculator allowed
- 8 questions
- 29 marks
Questions marked Higher are aimed at the Higher tier.
These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.
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A right-angled triangle has opposite side $3$ cm and adjacent side $4$ cm. Find the angle to $1$ decimal place.
Right-angled triangle with opposite 3 cm and adjacent 4 cm 2 marks
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The hypotenuse is $10$ cm and one angle is $30°$. Find the side opposite the $30°$ angle.
2 marks
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The hypotenuse is $12$ cm and the angle is $40°$. Find the opposite side to $3$ significant figures.
3 marks
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The hypotenuse is $20$ cm and the angle is $35°$. Find the adjacent side to $3$ significant figures.
3 marks
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A $4$ m ladder makes an angle of $65°$ with the ground. How high up the wall does it reach, to $3$ significant figures?
4 marks
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A ramp is $5$ m long and rises $0.8$ m. Find the angle the ramp makes with the horizontal, to $1$ decimal place, and the horizontal distance, to $3$ significant figures.
4 marks
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A ship sails $20$ km north and then east. From the start, the bearing of the ship is $025°$. Find how far east it sailed, and its distance from the start, both to $3$ significant figures. Higher
5 marks
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In a right-angled triangle the angle is $42°$ and the opposite side is $11$ cm. Find the hypotenuse and the adjacent side, each to $3$ significant figures. Use Pythagoras to check the rounded lengths. Higher
6 marks
Mark scheme
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Answer. $36.9°$
- $\tan \theta = \dfrac{3}{4}$
- $\theta = \tan^{-1}(0.75) = 36.9°$
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Answer. $5$ cm
- $\sin 30° = \dfrac{1}{2}$
- Opposite $= 10 \times \dfrac{1}{2} = 5$ cm
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Answer. $7.71$ cm
- Opposite $= 12 \sin 40°$
- $7.713\ldots = 7.71$ cm to $3$ s.f.
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Answer. $16.4$ cm
- Adjacent $= 20 \cos 35°$
- $16.383\ldots = 16.4$ cm to $3$ s.f.
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Answer. $3.63$ m
- Height $= 4 \sin 65°$
- $3.625\ldots = 3.63$ m to $3$ s.f.
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Answer. $9.2°$ and $4.94$ m
- $\sin \theta = \dfrac{0.8}{5} = 0.16$, so $\theta = 9.2°$ to $1$ d.p.
- Horizontal distance $= \sqrt{5^{2} - 0.8^{2}} = \sqrt{24.36} = 4.94$ m to $3$ s.f.
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Answer. East $9.33$ km, distance $22.1$ km
- East $= 20 \tan 25° = 9.326\ldots = 9.33$ km.
- Distance $= \dfrac{20}{\cos 25°} = 22.067\ldots = 22.1$ km.
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Answer. Hypotenuse $16.4$ cm, adjacent $12.2$ cm
- Hypotenuse $= \dfrac{11}{\sin 42°} = 16.4$ cm to $3$ s.f.
- Adjacent $= \dfrac{11}{\tan 42°} = 12.2$ cm to $3$ s.f.
- $\sqrt{11^{2} + 12.2^{2}} \approx 16.4$, which agrees with the hypotenuse at this accuracy.