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Date November 2011 Marks available 2 Reference code 11N.2.sl.TZ0.4
Level SL only Paper 2 Time zone TZ0
Command term Find and Hence Question number 4 Adapted from N/A

Question

Consider the triangle ABC, where AB =10 , BC = 7 and \({\rm{C}}\widehat {\rm{A}}{\rm{B}}\) = \({30^ \circ }\) .

Find the two possible values of \({\rm{A}}\widehat {\rm{C}}{\rm{B}}\) .

[4]
a.

Hence, find \({\rm{A}}\widehat {\rm{B}}{\rm{C}}\) , given that it is acute.

[2]
b.

Markscheme

Note: accept answers given in degrees, and minutes.

evidence of choosing sine rule     (M1)

e.g. \(\frac{{\sin A}}{a} = \frac{{\sin B}}{b}\)

correct substitution     A1

e.g. \(\frac{{\sin \theta }}{{10}} = \frac{{\sin {{30}^ \circ }}}{7}\) , \(\sin \theta  = \frac{5}{7}\)

\({\rm{A}}\widehat {\rm{C}}{\rm{B}} = {45.6^ \circ }\) , \({\rm{A}}\widehat {\rm{C}}{\rm{B}} = {134^ \circ }\)     A1A1     N1N1

Note: If candidates only find the acute angle in part (a), award no marks for (b).

[4 marks]

a.

attempt to substitute their larger value into angle sum of triangle     (M1)

e.g. \({180^ \circ } - (134.415{ \ldots ^ \circ } + {30^ \circ })\)

\({\rm{A}}\widehat {\rm{B}}{\rm{C}} = {15.6^ \circ }\)     A1     N2

[2 marks]

b.

Examiners report

Most candidates were comfortable applying the sine rule, although many were then unable to find the obtuse angle, demonstrating a lack of understanding of the ambiguous case. This precluded them from earning marks in part (b). Those who found the obtuse angle generally had no difficulty with part (b).

a.

Most candidates were comfortable applying the sine rule, although many were then unable to find the obtuse angle, demonstrating a lack of understanding of the ambiguous case. This precluded them from earning marks in part (b). Those who found the obtuse angle generally had no difficulty with part (b).

b.

Syllabus sections

Topic 3 - Circular functions and trigonometry » 3.6 » The sine rule, including the ambiguous case.
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