Edexcel GCSE Maths (1MA1), Higher tier · Geometry and measures
Practise Right-angled trigonometry. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy2 marks
The diagram shows a right-angled triangle. Diagram NOT accurately drawn
[object Object]
Work out the value of \(x\).[2]
Show the answer and mark scheme
Answer: \(x = 5\)
M1 for \(\tan 45^\circ = \frac{x}{5}\)
A1 for 5 cao
Worked solution: \(\tan 45^\circ = \frac{x}{5}\), so \(x = 5 \times \tan 45^\circ = 5 \times 1 = 5\).
Question 2Medium3 marks
A ladder 5 m long leans against a vertical wall. The ground is horizontal and the foot of the ladder is 1.2 m from the wall. The ladder is safe to use when the angle between the ladder and the ground is between 70° and 80°.
Is the ladder safe to use? You must show your working.[3]
Show the answer and mark scheme
Answer: Yes: \(\cos\theta = \frac{1.2}{5}\), so \(\theta = 76.1^\circ\), which is between 70° and 80°.
M1 for \(\cos\theta = \frac{1.2}{5}\) (or finding the height \(\sqrt{5^2 - 1.2^2}\) and using sin or tan)
A1 for 76.1... (76.11°)
C1 for 'yes' with 76.1° compared with 70° and 80°
Worked solution: \(\cos\theta = \frac{1.2}{5} = 0.24\), so \(\theta = \cos^{-1}(0.24) = 76.11\ldots^\circ\). 76.1° is between 70° and 80°, so the ladder is safe.
Question 3Hard4 marks
The diagram shows a cuboid ABCDEFGH. AB = 5 cm, BC = 7 cm and CG = 6 cm. Diagram NOT accurately drawn
[object Object]
Work out the size of the angle between the diagonal AG and the base ABCD. Give your answer correct to 1 decimal place.[4]
Show the answer and mark scheme
Answer: 34.9°
P1 for identifying the angle GAC, with AC the projection of AG onto the base
Worked solution: C is directly below G, so the projection of AG onto the base is AC and the angle is GAC (triangle ACG has a right angle at C). \(AC = \sqrt{5^2 + 7^2} = \sqrt{74}\). \(\tan GAC = \frac{6}{\sqrt{74}}\), so angle \(GAC = 34.895... = 34.9\)° (1 d.p.).