Basic Trigonometry: WJEC GCSE Numeracy Intermediate June 2022 Unit 2 Q8

Numeracy Intermediate June 2022 Unit 2 Q8

WJECIntermediateCurrent spec5 marksBasic Trigonometry

8. A motorcyclist leans into a corner on a motorcycle.
The angle of lean is the angle between the vertical and the motorcycle.

When the motorcycle is upright, the centre of the handlebars is 110 cm above the ground.
The diagrams below illustrate a front view of a motorbike as its rider goes into a corner.

The motorcycle is vertical to begin with. It then leans 30 cm horizontally into the corner.
The motorcycle then leans a further 30 cm into the corner, with a total horizontal lean distance of 60 cm.

Three diagrams. Before the corner: a vertical bar 110 cm long. Lean of 30 cm into the corner: the 110 cm bar leans so its top is 30 cm horizontally from the vertical dashed line, angle of lean marked between the vertical and the bar. Lean of 60 cm into the corner: the same with a horizontal distance of 60 cm

Diagrams not drawn to scale

Show that the angle of lean more than doubles as the motorcycle leans from 30 cm horizontally to 60 cm horizontally.
You must show all your working. [5]

Mark scheme

Scheme	Marks	Comments
(Let x be the initial angle of lean) (Let y be the final angle of lean)
sin x = 30/110 sin y = 60/110	M1 M1	Allow M marks for same variable is used for both angles of lean an appropriate statement of the sine rule, e.g. 30/sin x = 110/sin 90 or sin y/60 = sin 90/110
(x =) sin^-1(30/110) or (x =) sin^-1 0.2727... OR (y =) sin^-1(60/110) or (y =) sin^-1 0.5454...	M1	Also implies appropriate previous M1
15.8266...(°) AND 33.0557....(°) (and statement or calculation to show 33.0557....(°) > 2 × 15.8266...(°) )	A2	Accept rounded or truncated angles for A2 or A1 A1 for 15.8266...(°) or 33.0557....(°)

Alternative

8. Alternative method 1

Scheme	Marks	Comments
(To find initial angle of lean) Sin x = 30/110	M1	Allow for an appropriate statement of the sine rule, 30/sin x = 110/sin 90 or sin x/30 = sin 90/110
(x =) sin^-1(30/110) or (x =) sin^-1 0.2727... (x =) 15.8266...(°)	M1 A1	Also implies previous M1 Accept rounded or truncated angles
(To find horizontal lean if angle of lean was doubled) sin ((2 × 15.8266...(°)) =horizontal lean/110 or (Horizontal lean =) 110 × Sin (2 × 15.8266...(°))	M1	FT rounded or truncated double ‘their derived 15.8266...(°)’
57.725 (cm) (and statement that < 60 cm)	A1	FT answer must be < 60 (cm)

Alternative

8. Alternative method 2

Scheme	Marks	Comments
(To find final angle of lean) Sin y = 60/110	M1	Allow for an appropriate statement of the sine rule, 60/sin y = 110/sin 90 or sin y/60 = sin 90/110
(y =) sin^-1(60/110) or (y =) sin^-10.5454… (y =) 33.0557...(°)	M1 A1	Also implies previous M1 Accept rounded or truncated angles
(To find horizontal lean if angle of lean was halved) sin (1/2 × 33.0557...(°)) = horizontal lean/110 or (Horizontal lean =) 110 × Sin(1/2 × 33.0557...(°))	M1	FT rounded or truncated ½ ‘their derived 33.0557...(°)
31.29…(cm) (and statement that > 30 cm)	A1	FT answer must be > 30 (cm)