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Date May 2013 Marks available 3 Reference code 13M.2.sl.TZ2.3
Level SL only Paper 2 Time zone TZ2
Command term Find Question number 3 Adapted from N/A

Question

The following diagram shows a triangle ABC.


The area of triangle ABC is \(80\) cm2 , AB \( = 18\) cm , AC \( = x\) cm and \({\rm{B}}\hat {\rm{A}}{\rm{C}} = {50^ \circ }\) .

Find \(x\) .

[3]
a.

Find BC.

[3]
b.

Markscheme

correct substitution into area formula     (A1)

eg   \(\frac{1}{2}(18x)\sin 50\)

setting their area expression equal to \(80\)     (M1)

eg   \(9x\sin 50 = 80\)

\(x = 11.6\)     A1     N2

[3 marks]

a.

evidence of choosing cosine rule     (M1)

eg   \({c^2} = {a^2} + {b^2} + 2ab\sin C\)

correct substitution into right hand side (may be in terms of \(x\))     (A1)

eg   \({11.6^2} + {18^2} - 2(11.6)(18)\cos 50\)

BC \( = 13.8\)     A1     N2

[3 marks]

b.

Examiners report

The vast majority of candidates were very successful with this question. A small minority drew an altitude from C and used right triangle trigonometry. Errors included working in radian mode, assuming that the angle at C was \(90^\circ \), and incorrectly applying the order of operations when evaluating the cosine rule.

a.

The vast majority of candidates were very successful with this question. A small minority drew an altitude from C and used right triangle trigonometry. Errors included working in radian mode, assuming that the angle at C was \(90^\circ \), and incorrectly applying the order of operations when evaluating the cosine rule.

b.

Syllabus sections

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