IB Revision Bank
About

← back to Mathematics topic 3

21N.2.SL.TZ0.3

pestleMathematicsAASLPaper 221N· sl-3-2-2d-and-3d-trig-sine-rule-cosine-rule-areasource ↗

Consider a triangle ABC, where AC=12, CB=7 and BA^C=25°.

Find the smallest possible perimeter of triangle ABC.

Markscheme / solution

EITHER

attempt to use cosine rule               (M1)

122+AB2-2×12×cos25°×AB=72  OR  AB2-21.7513AB+95=0                 (A1)

at least one correct value for AB                 (A1)

AB=6.05068  OR  AB=15.7007

using their smaller value for AB to find minimum perimeter               (M1)

12+7+6.05068


OR

attempt to use sine rule               (M1)

sinB12=sin25°7  OR  sinB=0.724488  OR  B^=133.573°  OR  B^=46.4263°                 (A1)

at least one correct value for C                 (A1)

C^=21.4263°  OR  C^=108.573°

using their acute value for C^ to find minimum perimeter               (M1)

12+7+122+72-2×12×7cos21.4263°  OR  12+7+7sin21.4263°sin25°


THEN

25.0506

minimum perimeter =25.1.                       A1

 

[5 marks]

Examiners’ report
[N/A]