User interface language: English | Español

Date May 2017 Marks available 3 Reference code 17M.2.SL.TZ2.T_4
Level Standard Level Paper Paper 2 Time zone Time zone 2
Command term Calculate Question number T_4 Adapted from N/A

Question

The quadrilateral ABCD represents a park, where AB = 120  m , AD = 95  m and DC = 100  m . Angle DAB is 70° and angle DCB is 110°. This information is shown in the following diagram.

M17/5/MATSD/SP2/ENG/TZ2/04

A straight path through the park joins the points B and D.

A new path, CE, is to be built such that E is the point on BD closest to C.

The section of the park represented by triangle DCE will be used for a charity race. A track will be marked along the sides of this section.

Find the length of the path BD.

[3]
a.

Show that angle DBC is 48.7°, correct to three significant figures.

[3]
b.

Find the area of the park.

[4]
c.

Find the length of the path CE.

[2]
d.

Calculate the total length of the track.

[3]
e.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

( B D 2 = )   95 2 + 120 2 2 × 95 × 120 × cos 70     (M1)(A1)

 

Note:     Award (M1) for substituted cosine rule, (A1) for correct substitution.

 

( BD = )   125  (m)  ( 125.007  (m) )     (A1)(G2)

 

[3 marks]

a.

sin DBC 100 = sin 110 125.007     (M1)(A1)(ft)

 

Note:     Award (M1) for substituted sine rule, (A1)(ft) for correct substitution.

Follow through from their answer to part (a).

 

( DBC = )   48.7384     (A1)(ft)

( DBC = )   48.7     (AG)

 

Notes:     Award the final (A1)(ft) only if both their unrounded answer and 48.7° is seen. Follow through from their answer to part (a), only if their unrounded answer rounds to 48.7°.

 

[3 marks]

b.

1 2 × 125.007 × 100 × sin 21.3 + 1 2 × 95 × 120 × sin 70     (A1)(M1)(M1)

 

Note:     Award (A1) for 21.3° (21.2615…) seen, (M1) for substitution into (at least) one area of triangle formula in the form 1 2 a b sin c , (M1) for their correct substitutions and adding the two areas.

 

7630   m 2   ( 7626.70 m 2 )     (A1)(ft)(G3)

 

Notes:     Follow through from their answers to part (a). Accept 7620   m 2   ( 7622.79 m 2 ) from use of 48.7384…

 

[4 marks]

c.

( CE = )   100 × sin 21.3     (M1)

( CE = )   36.3  (m)  ( 36.3251  (m) )     (A1)(ft)(G2)

 

Note:     Follow through from their angle 21.3° in part (c). Award (M0)(A0) for halving 110° and/or assuming E is the midpoint of BD in any method seen.

 

OR

area of BCD = 1 2 BD × CE     (M1)

( CE = )   36.3  (m)  ( 36.3251  (m) )     (A1)(ft)(G2)

 

Note:     Follow through from parts (a) and (c). Award (M0)(A0) for halving 110° and/or assuming E is the midpoint of BD in any method seen.

 

[2 marks]

d.

100 2 36.3251 2 + 100 + 36.3251     (M1)(M1)

 

Note:     Award (M1) for correct use of Pythagoras to find DE (or correct trigonometric equation, 100 × cos 21.3 , to find DE), (M1) for the sum of 100, their DE and their CE.

 

229  (m)  ( 229.494  (m) )     (A1)(ft)(G2)

 

Note:     Follow through from part (d). Use of 3 sf values gives an answer of 230  (m)  ( 229.5  (m) ) .

 

[3 marks]

e.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.2—2d and 3d trig, sine rule, cosine rule, area
Show 133 related questions
Topic 3— Geometry and trigonometry

View options