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Question 92 of 105 in Unit 9

Unit 9

7 marks · Easy

A straight road runs up a hill and is shown on a diagram below from the side. The road runs from point A at (0, 1) to point B at (5, 7), where the coordinates are given in metres.

(a)

Find the gradient of the road from A to B.

[2 marks]
(b)

A signpost is placed vertically at B and is 3 m tall. Calculate the angle between the signpost at B and the sloping road AB. Give your answer in degrees to 1 decimal place.

[5 marks]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

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(a) Gradient = (7 - 1)/(5 - 0) = 6/5 = 1.2

(b) Angle = arctan(1/gradient) = arctan(1/1.2) = 39.8° (to 1 decimal place) degrees

Common mistakes: Swapping rise and run in gradient; Using the horizontal instead of vertical as the reference for the angle

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Attempts to find the difference in y- and x-coordinates

A1 (a) · 1 mark · Correct calculation of the gradient as 1.2

M2 (b) · 1 mark · Recognises that the angle required is between the road and the vertical (signpost)

M3 (b) · 2 marks · Forms correct right triangle or uses correct trigonometric relation: tan θ = 1/gradient (follow-through)

A2 (b) · 2 marks · Accurately finds θ = arctan(1/1.2) and gives answer as 39.8° (to 1 decimal place) (follow-through)

Worked solution

a) The gradient of AB is given by the change in y divided by the change in x:

b) The signpost at B is vertical. The road makes an angle θ with the horizontal, where

But the question asks for the angle between the road (AB) and the signpost (vertical), so we want the angle between AB and the vertical.

Final answer: Angle = 39.8° to 1 decimal place

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MATHEMATICS A question: Unit 9 | SubjectCover