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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.

[3]
a.

The area of triangle ABD is 18.5 cm2. Find the possible values of θ.

[4]
b.

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    c 2 = a 2 + b 2 2 a b cos C

correct substitution into RHS       (A1)

eg    4.83 2 + 3.80 2 2 × 4.83 × 3.80 × cos 78.2 30.2622 ,

4.83 2 + 3.80 2 2 ( 4.83 ) ( 3.80 ) × cos 1.36

5.50111

5.50 (cm)     A1 N2

[3 marks]

a.

correct substitution for area of triangle ABD       (A1)

eg    1 2 × 6.73 × 5.50111 sin θ

correct equation      A1

eg    1 2 × 6.73 × 5.50111 sin θ = 18.5 ,   sin θ = 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]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.2—2d and 3d trig, sine rule, cosine rule, area
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Topic 3— Geometry and trigonometry » SL 3.5—Unit circle definitions of sin, cos, tan. Exact trig ratios, ambiguous case of sine rule
Topic 3— Geometry and trigonometry

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