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Date November 2015 Marks available 5 Reference code 15N.2.sl.TZ0.8
Level SL only Paper 2 Time zone TZ0
Command term Find Question number 8 Adapted from N/A

Question

The following diagram shows the quadrilateral \(ABCD\).

\[{\text{AD}} = 6{\text{ cm}},{\text{ AB}} = 15{\text{ cm}},{\rm{ A\hat BC}} = 44^\circ ,{\rm{ A\hat CB}} = 83^\circ {\rm{ and D\hat AC}} = \theta \]

Find \(AC\).

[3]
a.

Find the area of triangle \(ABC\).

[3]
b.

The area of triangle \(ACD\) is half the area of triangle \(ABC\).

Find the possible values of \(\theta \).

[5]
c.

Given that \(\theta \) is obtuse, find \(CD\).

[3]
d.

Markscheme

evidence of choosing sine rule     (M1)

eg\(\;\;\;\frac{{{\text{AC}}}}{{\sin {\rm{C\hat BA}}}} = \frac{{{\text{AB}}}}{{\sin {\rm{A\hat CB}}}}\)

correct substitution     (A1)

eg\(\;\;\;\frac{{{\text{AC}}}}{{\sin 44^\circ }} = \frac{{15}}{{\sin 83^\circ }}\)

\(10.4981\)

\({\text{AC}} = 10.5{\text{ }}{\text{ (cm)}}\)     A1     N2

[3 marks]

a.

finding \({\rm{C\hat AB}}\) (seen anywhere)     (A1)

eg\(\;\;\;180^\circ  - 44^\circ  - 83^\circ ,{\rm{ C\hat AB}} = 53^\circ \)

correct substitution for area of triangle \(ABC\)     A1

eg\(\;\;\;\frac{1}{2} \times 15 \times 10.4981 \times \sin 53^\circ \)

62.8813

\({\text{area}} = 62.9{\text{ }}{\text{ (c}}{{\text{m}}^2}{\text{)}}\)     A1     N2

[3 marks]

b.

correct substitution for area of triangle \(DAC\)    (A1)

eg\(\;\;\;\frac{1}{2} \times 6 \times 10.4981 \times \sin \theta \)

attempt to equate area of triangle \(ACD\) to half the area of triangle \(ABC\)     (M1)

eg\(\;\;\;{\text{area ACD}} = \frac{1}{2} \times {\text{ area ABC; 2ACD}} = {\text{ABC}}\)

correct equation     A1

eg\(\;\;\;\frac{1}{2} \times 6 \times 10.4981 \times \sin \theta  = \frac{1}{2}(62.9),{\text{ }}62.9887\sin \theta  = 62.8813,{\text{ }}\sin \theta  = 0.998294\)

\(86.6531\), \(93.3468\)

\(\theta  = 86.7^\circ {\text{ }},{\text{ }}\theta  = 93.3^\circ {\text{ }}\)     A1A1     N2

[5 marks]

 

c.

Note:     Note: If candidates use an acute angle from part (c) in the cosine rule, award M1A0A0 in part (d).

 

evidence of choosing cosine rule     (M1)

eg\(\;\;\;{\text{C}}{{\text{D}}^2} = {\text{A}}{{\text{D}}^2} + {\text{A}}{{\text{C}}^2} - 2 \times {\text{AD}} \times {\text{AC}} \times \cos \theta \)

correct substitution into rhs     (A1)

eg\(\;\;\;{\text{C}}{{\text{D}}^2} = {6^2} + {10.498^2} - 2(6)(10.498)\cos 93.336^\circ \)

\(12.3921\)

\(12.4{\text{ }}{\text{ (cm)}}\)     A1     N2

[3 marks]

Total [14 marks]

d.

Examiners report

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

Syllabus sections

Topic 3 - Circular functions and trigonometry » 3.6 » Area of a triangle, \(\frac{1}{2}ab\sin C\) .
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