Pearson Edexcel International GCSE (9–1) · MATHEMATICS A · Unit
Unit 10
108 practice questions
Topics covered
Practice questions
21–30 of 41 · Easy
Question 21
1 mark · Easy
Answer and see the solution →What is the value of the gradient at a stationary point of a curve?
Question 22
1 mark · Easy
Answer and see the solution →Write down the formula for the area of a triangle when two sides are of length and and the included angle is .
Question 23
2 marks · Easy
Answer and see the solution →Calculate the value of .
Question 24
2 marks · Easy
Answer and see the solution →In triangle XYZ, sides XY = 9 cm, XZ = 7 cm, and YZ = 12 cm. Angle YXZ is opposite side YZ. Can you use the sine rule to find angle YXZ? Explain your answer.
Question 25
3 marks · Easy
Answer and see the solution →The triangle below has sides PQ = 5.0 cm and QR = 6.0 cm, and angle PQR = 60°.
Triangle PQR (not to scale) (a)[1 mark]Write down the lengths of the two sides and the angle needed to calculate the area of triangle PQR using trigonometry.
(b)[2 marks]Calculate the area of triangle PQR, giving your answer to 2 decimal places.
Question 26
3 marks · Easy
Answer and see the solution →The table below gives values for the function . Complete the missing values for and .
Values for \( y = \sin \theta \) (degrees) 0 90 180 270 360 Question 27
1 mark · Easy
Answer and see the solution →State whether the sine rule or cosine rule would be used to find the third side of a triangle when you know two sides and the included angle.
Question 28
2 marks · Easy
Answer and see the solution →Calculate the area of a triangle with sides 9.0 cm and 4.5 cm and included angle 45°. Give your answer to 2 decimal places.
Question 29
2 marks · Easy
Answer and see the solution →A coin is tossed twice. Calculate the probability that both tosses show heads. Give your answer as a decimal.
Question 30
2 marks · Easy
Answer and see the solution →A bag contains 1 red, 2 blue and 3 green counters. You take one at random, replace it, and then take a second counter at random. Explain why a tree diagram can be used to represent all possible outcomes for this problem.