| Date | May 2012 | Marks available | 3 | Reference code | 12M.1.hl.TZ1.10 |
| Level | HL only | Paper | 1 | Time zone | TZ1 |
| Command term | Show that | Question number | 10 | Adapted from | N/A |
Question
In the triangle ABC, \({\rm{A\hat BC}} = 90^\circ\) , \({\text{AC}} = \sqrt {\text{2}}\) and AB = BC + 1.
Show that cos \(\hat A - \sin \hat A = \frac{1}{{\sqrt 2 }}\).
By squaring both sides of the equation in part (a), solve the equation to find the angles in the triangle.
Apply Pythagoras’ theorem in the triangle ABC to find BC, and hence show that \(\sin \hat A = \frac{{\sqrt 6 - \sqrt 2 }}{4}\).
Hence, or otherwise, calculate the length of the perpendicular from B to [AC].
Markscheme
\(\cos \hat A = \frac{{{\text{BA}}}}{{\sqrt 2 }}\) A1
\(\sin \hat A = \frac{{{\text{BC}}}}{{\sqrt 2 }}\) A1
\(\cos \hat A - \sin \hat A = \frac{{{\text{BA}} - {\text{BC}}}}{{\sqrt 2 }}\) R1
\( = \frac{1}{{\sqrt 2 }}\) AG
[3 marks]
\({\cos ^2}\hat A - 2\cos \hat A\sin \hat A + {\sin ^2}\hat A = \frac{1}{2}\) M1A1
\(1 - 2\sin \hat A\cos \hat A = \frac{1}{2}\) M1A1
\(\sin 2\hat A = \frac{1}{2}\) M1
\(2\hat A = 30^\circ \) A1
angles in the triangle are 15° and 75° A1A1
Note: Accept answers in radians.
[8 marks]
\({\text{B}}{{\text{C}}^2} + {({\text{BC}} + 1)^2} = 2\) M1A1
\(2{\text{B}}{{\text{C}}^2} + 2{\text{BC}} - 1 = 0\) A1
\({\text{BC}} = \frac{{ -2 + \sqrt {12} }}{4}\left( { = \frac{{\sqrt 3 - 1}}{2}} \right)\) M1A1
\(\sin \hat A = \frac{{{\text{BC}}}}{{\sqrt 2 }} = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\) A1
\( = \frac{{\sqrt 6 - \sqrt 2 }}{4}\) AG
[6 marks]
EITHER
\(h = {\text{ABsin}}\hat A\) M1
\( = ({\text{BC}} + 1)\sin \hat A\) A1
\( = \frac{{\sqrt 3 + 1}}{2} \times \frac{{\sqrt 6 - \sqrt 2 }}{4} = \frac{{\sqrt 2 }}{4}\) M1A1
OR
\(\tfrac{1}{2}AB.BC = \tfrac{1}{2}AC.h\) M1
\(\frac{{\sqrt 3 - 1}}{2} \cdot \frac{{\sqrt {3 + 1} }}{2} = \sqrt {2h} \) A1
\(\frac{2}{4} = \sqrt 2 h\) M1
\(h = \frac{1}{{2\sqrt 2 }}\) A1
[4 marks]
Examiners report
Many good solutions to this question, although some students incorrectly stated the value of \(\arcsin \left( {\frac{1}{2}} \right)\). A surprising number of students had greater difficulties with part (d).
Many good solutions to this question, although some students incorrectly stated the value of \(\arcsin \left( {\frac{1}{2}} \right)\). A surprising number of students had greater difficulties with part (d).
Many good solutions to this question, although some students incorrectly stated the value of \(\arcsin \left( {\frac{1}{2}} \right)\). A surprising number of students had greater difficulties with part (d).
Many good solutions to this question, although some students incorrectly stated the value of \(\arcsin \left( {\frac{1}{2}} \right)\). A surprising number of students had greater difficulties with part (d).
