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Date May 2016 Marks available 4 Reference code 16M.1.hl.TZ1.5
Level HL only Paper 1 Time zone TZ1
Command term Find Question number 5 Adapted from N/A

Question

The following diagram shows the triangle ABC where \({\text{AB}} = 2,{\text{ AC}} = \sqrt 2 \) and \({\rm{B\hat AC}} = 15^\circ \).

M16/5/MATHL/HP1/ENG/TZ1/05.c

Expand and simplify \({\left( {1 - \sqrt 3 } \right)^2}\).

[1]
a.

By writing \(15^\circ \) as \(60^\circ  - 45^\circ \) find the value of \(\cos (15^\circ )\).

[3]
b.

Find BC in the form \(a + \sqrt b \) where \(a,{\text{ }}b \in \mathbb{Z}\).

[4]
c.

Markscheme

\({\left( {1 - \sqrt 3 } \right)^2} = 4 - 2\sqrt 3 \)    A1

Note:     Award A0 for \(1 - 2\sqrt 3  + 3\).

[1 mark]

a.

\(\cos (60^\circ  - 45^\circ ) = \cos (60^\circ )\cos (45^\circ ) + \sin (60^\circ )\sin (45^\circ )\)    M1

\( = \frac{1}{2} \times \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 3 }}{2} \times \frac{{\sqrt 2 }}{2}{\text{ }}\left( {{\text{or }}\frac{1}{2} \times \frac{1}{{\sqrt 2 }} + \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }}} \right)\)    (A1)

\( = \frac{{\sqrt 2  + \sqrt 6 }}{4}{\text{ }}\left( {{\text{or }}\frac{{1 + \sqrt 3 }}{{2\sqrt 2 }}} \right)\)    A1

[3 marks]

b.

\(B{C^2} = 2 + 4 - 2 \times \sqrt 2  \times 2\cos (15^\circ )\)    M1

\( = 6 - \sqrt 2 \left( {\sqrt 2  + \sqrt 6 } \right)\)

\( = 4 - \sqrt {12} {\text{ }}\left( { = 4 - 2\sqrt 3 } \right)\)    A1

\(BC =  \pm \left( {1 - \sqrt 3 } \right)\)    (M1)

\(BC =  - 1 + \sqrt 3 \)    A1

Note:     Accept \(BC = \sqrt 3  - 1\).

Note:     Award M1A0 for \(1 - \sqrt 3 \).

Note:     Valid geometrical methods may be seen.

[4 marks]

c.

Examiners report

The main error here was to fail to note the word ‘simplify’ in the question and some candidates wrote \(1 + 3\) in their final answer rather than 4.

a.

This was well done by the majority of candidates, though a few wrote \(\cos (60 - 45) = \cos 60 - \cos 45\).

b.

Candidates were able to use the cosine rule correctly but then failed to notice the result obtained was the same as that obtained in part (a).

c.

Syllabus sections

Topic 3 - Core: Circular functions and trigonometry » 3.7 » The cosine rule.
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