Date | May 2015 | Marks available | 3 | Reference code | 15M.2.sl.TZ2.1 |
Level | SL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 1 | Adapted from | N/A |
Question
The following diagram shows triangle ABC.
BC=10 cm,AˆBC=80∘andBˆAC=35∘.
Find AC.
Find the area of triangle ABC.
Markscheme
evidence of choosing sine rule (M1)
egACsin(AˆBC)=BCsin(BˆAC)
correct substitution (A1)
egACsin80∘=10sin35∘
AC=17.1695
AC=17.2 (cm) A1 N2
[3 marks]
AˆCB=65∘(seen anywhere) (A1)
correct substitution (A1)
eg12×10×17.1695×sin65∘
area=77.8047
area=77.8 (cm2) A1 N2
[3 marks]
Total [6 marks]
Examiners report
Most candidates found this question straightforward and accessible.
Most recognized the need for the sine rule in part (a) to solve the problem. Occasionally, the setup had an incorrect match of angle and side. Some used radians instead of degrees, thus losing a mark.
Part (b) was also well done by most of the candidates. Some right triangle trigonometry correct approaches were seen to find the area. A few candidates used the cosine rule or right angled trigonometry, which were less efficient methods and often wasted valuable time.