Mathematics · Oct / Nov 2021 · Paper 11 · Question 7

2

2

2

2

OB

=

0.6

+

1.4

−×

2 0.6 1.4cos2

×

⇒

OB

=

... or

OB

=

...

OB =

1.738

sin

AOB

sin 2

=

⇒

AOB

=

0.822

1.4

''1.738''

or e.g.

sin

ABO

sin 2

=

⇒

ABO

=

0.319...

⇒

AOB

= π −

2

−

0.319...

0.6

''1.738''

θ =

2

×

AOB

=

2 0.822

×

=

1.64*

Attempts 0.6×α with

α

= 2

π −

1.64 or

α

=

π −

1.64

or

Attempts 2

0.6

0.6 1.64

×π×

−

×

(

)

0.6

×

2

π −

1.64

+

2.8

=

5.6

m

1

0.6

2

α

= 2

π −

1.64 or

α

=

π −

1.64

Attempts

2 ×

×α with

or

1

2

2

0.6

0.6

1.64

Attempts

π×

−

×

×

2

Attempts 0.6 1.4sin 2

×

1

2

(

)

Full method

2 ×

0.6

×

2

π −

1.64

+

0.6 1.4sin 2

×

=

1.6

m2

In part (a), candidates may not be careful with the use of θ. E.g. in their working, their θ may be

angle AOB which they correctly double to get 1.64 – condone this poor notation and give credit

if the intention is clear.

Example:

2

2

2

OB

0.6

1.4

2 0.6 1.4cos2

OB

1.73756...

=

+

−×

×

⇒

=

2

2

2

0.6

+

1.738

−

1.4

cos

θ

=

⇒

θ

=

0.822

2 0.6 1.738

×

×

2 0.822

1.64

θ =

×

=

Is acceptable for full marks in (a)

(a)

M1 Attempts the cosine rule to get OB or OB2 seen in part (a) only

A1 OB = awrt 1.74 or truncated as 1.7 or e.g. 1.73 (may be implied)

dM1

Attempts the sine rule with the sides and angles in the correct positions in an attempt to

find AOB or 1/2 AOB. This must be a full attempt including using inverse sin to find

the angle.

This may be achieved by attempting the sine rule to find angle ABO first and then using

the angle sum. This requires use of the sine rule with the sides and angles in the correct

positions in an attempt to find ABO followed by

π −

2 "0.319..."

−

or e.g.

2

π −

4

−×

2 "0.319..."

Alternatively uses the cosine rule to find angle AOB

θ

θ

2

2

2

E.g.

1.4

=

0.6

+

1.738

−×

2 0.6 1.738cos

×

⇒

=

...

2

2

Depends on the first method mark.

A1*

Fully correct work leading to the given answer with no obvious rounding errors.

E.g. if they obtain angle AOB = 0.827 following correct a method then state 2×0.827 =

1.64 this mark should be withheld.

Alternatives working backwards:

θ

BC

0.6

θ =

1.64

⇒

=

0.82,

=

⇒

BC

=

1.4

so

θ =

1.64

(

)

2

sin 0.82

sin

π −

2

−

0.82

θ

1.4

OC

Or

θ =

1.64

⇒

=

0.82,

=

⇒

OC

=

0.6

so

θ =

1.64

(

)

2

sin 0.82

sin

π −

2

−

0.82

θ and uses angle sum of triangle and sine rule

M1: Finds 2

A1: Correct sine rule statement

dM1: Rearranges for BC or OC. Depends on first method mark.

A1: Fully correct work to obtain BC = 1.4 or OC = 0.6 with minimal conclusion e.g.

tick, QED etc.

(b)

M1

Attempts 0.6×α with an allowable α

For an allowable angle accept 2

π −

1.64

(awrt 4.64) or

π −

1.64

(awrt 1.50)

An alternative is to find the circumference and subtract the minor arc AC

For reference the correct value is 2.78… which may imply the method (3.7699… −

0.984)

(

)

A1

0.6

×

2

π −

1.64

+

2.8

=

awrt 5.6

m. Condone lack of units

(c) In general the marks in part (c) are M1: Attempting the major sector, M2 attempting

the kite area or half of it e.g. area of triangle AOB but not e.g. area of triangle OCA (they

would need to find the area of triangle ABC as well, or half of both of these), dM3: A

complete and correct method for the total area, A1: awrt 1.6

M1

This mark is for an attempt at the sector area OCXA:

1

2 ×

0.6

2

×α with

α

= 2

π −

1.64 or

α

=

π −

1.64

E.g. Attempts

An alternative is to find the area of the circle and subtract the area of the minor sector

For reference the correct value is 0.835… which may imply the method (1.130… −

0.2952)

M1

This mark is for an attempt at the kite ABCO (or half of it):

which may be part of 1

Examples: Attempts 0.6 1.4sin 2

×

2 ×

0.6 1.4sin 2

×

1

1

2

2

(

Attempts OCA + ABC e.g.

×

0.6 sin1.64

+

×

1.4 sin 2

π

−

4 1.64

−

2

2

θ

which may be part of 1

θ

Attempts "1.74" 0.6sin 2

×

×

"1.74" 0.6sin

×

2

2

ddM1 This mark is for a complete and correct attempt at the total area and depends on

both previous method marks:

1

2

(

)

e.g.

2 ×

0.6

×

2

π −

1.64

+

0.6 1.4sin 2

×

o.e

A1

awrt 1.6 m2 Condone a lack of units

Alternative which doesn’t follow the above but is equivalent:

1

1

2

2

(

)

2

(

)

Area =

π×

0.6

−

×

0.6

1.64

−

sin1.64

+

×

1.4 sin 2

π

−

4 1.64

−

2

2

Award M1 for the attempt at the major segment and M1 for the attempt at triangle ABC (or half

of it) then ddM1A1 as above.

Some lengths and angles for reference:

OB2 = 3.019…

OB = 1.7375…

AC2 = 0.7697…

AC = 0.8773…

Angle ABO = 0.322

Angle COA = 4.64…

Solutions where candidates change to degrees

NB 2 radians is 114.591559…°

(a)

2

2

2

2

OB

=

0.6

+

1.4

−×

2 0.6 1.4cos114.59...

×

⇒

OB

=

... or

OB

=

...

OB =

1.738

sin

AOB

sin114.59...

0.822

=

⇒

AOB

=

1.4

''1.738''

or e.g.

sin

ABO

sin114.59...

=

⇒

ABO

=

18.3

⇒

AOB

=

180 114.59... 18.3

−

−

0.6

''1.738''

θ =

2

×

AOB

=

2 47.1

×

=

94.2

° =

1.64*

(b)

α × ×π×

Attempts

2

0.6

with

α

= 360

−

awrt94

°

360

or

awrt94

°

2

×π×

0.6

−

× ×π×

2

0.6

Attempts

360

360

−

awrt94

× ×π×

2

0.6

+

2.8

=

5.6

m

360

(c)

α ×π×

0.6

2

α

= 360

−

awrt94

°

Attempts

with

360

or

awrt94

°

2

2

π×

0.6

−

×π×

0.6

Attempts

360

Attempts e.g. 0.6 1.4sin114.59...

×

360

−

awrt94

° ×π×

0.6

2

+

0.6 1.4sin114.59...

×

=

1.6

Full method e.g.

m2

360

Questio

n