Date | May 2015 | Marks available | 2 | Reference code | 15M.1.sl.TZ2.8 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 8 | Adapted from | N/A |
Question
The diagram shows a triangle ABC. The size of angle CˆAB is 55∘ and the length of AM is 10 m, where M is the midpoint of AB. Triangle CMB is isosceles with CM=MB.
Write down the length of MB.
Find the size of angle CˆMB.
Find the length of CB.
Markscheme
10 m (A1)(C1)
AˆMC=70∘ORAˆCM=55∘ (A1)
CˆMB=110∘ (A1) (C2)
CB2=102+102−2×10×10×cos110∘ (M1)(A1)(ft)
Notes: Award (M1) for substitution into the cosine rule formula, (A1)(ft) for correct substitution. Follow through from their answer to part (b).
OR
CBsin110∘=10sin35∘ (M1)(A1)(ft)
Notes: Award (M1) for substitution into the sine rule formula, (A1)(ft) for correct substitution. Follow through from their answer to part (b).
OR
AˆCB=90∘ (A1)
sin55∘=CB55ORcos35∘=CB20 (M1)
Note: Award (A1) for some indication that AˆCB=90∘, (M1) for correct trigonometric equation.
OR
Perpendicular MN is drawn from M to CB. (A1)
12CB10=cos35∘ (M1)
Note: Award (A1) for some indication of the perpendicular bisector of BC, (M1) for correct trigonometric equation.
CB=16.4 (m)(16.3830… (m)) (A1)(ft)(C3)
Notes: Where a candidate uses CˆMB=90∘ and finds CB=14.1 (m) award, at most, (M1)(A1)(A0).
Where a candidate uses CˆMB=60∘ and finds CB=10 (m) award, at most, (M1)(A1)(A0).