Date | November 2018 | Marks available | 5 | Reference code | 18N.2.SL.TZ0.T_5 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Calculate | Question number | T_5 | Adapted from | N/A |
Question
A flat horizontal area, ABC, is such that AB = 100 m , BC = 50 m and angle AĈB = 43.7° as shown in the diagram.
Show that the size of angle BÂC is 20.2°, correct to 3 significant figures.
Calculate the area of triangle ABC.
Find the length of AC.
A vertical pole, TB, is constructed at point B and has height 25 m.
Calculate the angle of elevation of T from, M, the midpoint of the side AC.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)(A1)
Note: Award (M1) for substitution into sine rule formula, (A1) for correct substitution.
BAC = 20.2087… = 20.2° (A1)(AG)
Note: Award (A1) only if both the correct unrounded and rounded answers are seen.
[3 marks]
units are required in part (b)
(A1)(M1)(A1)
Note: Award (A1) for 116.1 or unrounded value or 116 seen, (M1) for substitution into area of triangle formula, (A1) for correct substitution.
= 2250 m2 (2245.06… m2) (A1)(G3)
Note: The answer is 2250 m2; the units are required. Use of 20.2087… gives 2245.23….
[4 marks]
(M1)(A1)(ft)
Note: Award (M1) for substitution into sine rule formula, (A1)(ft) for their correct substitution. Follow through from their 116.1.
AC = 130 (m) (129.982… (m)) (A1)(ft)(G2)
Note: Use of 20.2087… gives 129.992….
OR
AC2 = 1002 + 502 −2(100)(50) cos (116.1) (M1)(A1)(ft)
Note: Award (M1) for substitution into cosine rule formula, (A1)(ft) for their correct substitution. Follow through from their 116.1.
AC = 130 (m) (129.997… (m)) (A1)(ft)(G2)
Note: Award (M1) for substitution into cosine rule formula, (A1)(ft) for their correct substitution.
[3 marks]
BM2 = 1002 + 652 − 2(100)(65) cos (20.2°) (M1)(A1)(ft)
OR
BM2 = 502 + 652 − 2(50)(65) cos (43.7°) (M1)(A1)(ft)
Note: Award (M1) for substitution into cosine rule formula, (A1)(ft) for correct substitution, including half their AC.
BM = 45.0 (44.9954… OR 45.0079…) (A1)(ft)
Note: Use of 20.2052… gives 45. Award (G2) for 45.0 seen without working.
(M1)
Note: Award (M1) for correct substitution into tangent formula.
TB = 29.1° (29.0546…°) (A1)(ft)(G4)
Note: Follow through within part (d) provided their BM is seen. Use of 44.9954 gives 29.0570… and use of 45.0079… gives 29.0503…. Follow through from their AC in part (c).
[5 marks]