Mathematics · May / June 2022 · Paper 11 · Question 2

sin

x°

sin 25

°

=

o.e

21

13

sin

x°=

0.6827

(awrt)

sin (0.6827)

−

1

=

...

(43.05°)

180

°−

sin (0.6827)

−

1

=

...

(AC < AB so

∠

ABC

<∠

ACB

so) required angle is

So x = awrt 136.95

Notes:

Condone the omission of the ° symbol. Mark (a) and (b) as one

(a)

M1: A correct statement of the sine rule with sides and angles in the correct position.

Implied by sin x°= awrt 0.68

A1: sin x ° = awrt 0.6827

ISW for instance if they go on to find the value of x in (a)

(b)

M1: Applies inverse sine to their value found in part (a) to find the angle in degrees (using degree mode)

correct to nearest degree. For a correct (a) awrt 43° is sufficient

Implied by a correct answer for their value of sin x, either for the acute or the obtuse angle..

M1: Attempts to find the correct angle, 180

°−

arcsin''0.6827''

°. Award even if other angles are given

No reasoning need be given, but may see diagram drawn, size of angles, or comparison of sides used.

A1: awrt 136.95 but it is A0 if two angles are given

..................................................................................................................................................................

There are alternative methods so look at the candidates work carefully.

E.g.

Uses the cosine rule once to find AC

2

2

13

21

2

2 21

cos25

=

+

y

−×

×

y

y

2

−

38.1..

y

+

272

=

0

y

=

9.533...

2

2

2

13

+

9.533

−

21

cos

136.95

...and then a second time to find x

x

=

x

=

°

2 13 9.533

×

×

In this solution both M's in part (b) are scored together.

To score marks in (a) (even though they haven't used the sine rule) they would need to simply find sin (their

136.95) for M1, and A1 for an accurate answer.

Writes 180 as 6 5 or 80 as 4 5