Date | May 2019 | Marks available | 3 | Reference code | 19M.2.SL.TZ2.S_3 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 2 |
Command term | Find | Question number | S_3 | Adapted from | N/A |
Question
The following diagram shows the quadrilateral ABCD.
AB = 6.73 cm, BC = 4.83 cm, BĈD = 78.2° and CD = 3.80 cm.
Find BD.
The area of triangle ABD is 18.5 cm2. Find the possible values of θ.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
choosing cosine rule (M1)
eg c2=a2+b2−2abcosC
correct substitution into RHS (A1)
eg 4.832+3.802−2×4.83×3.80×cos78.2 , 30.2622,
4.832+3.802−2(4.83)(3.80)×cos1.36
5.50111
5.50 (cm) A1 N2
[3 marks]
correct substitution for area of triangle ABD (A1)
eg 12×6.73×5.50111sinθ
correct equation A1
eg 12×6.73×5.50111sinθ=18.5 , sinθ=0.999393
88.0023, 91.9976, 1.53593, 1.60566
θ = 88.0 (degrees) or 1.54 (radians)
θ = 92.0 (degrees) or 1.61 (radians) A1A1 N2
[4 marks]