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