# IB Math Analysis & Approaches Questionbank-Topic: SL 3.2 The cosine rule SL Paper 1

## Question

Note: In this question, distance is in metres and time is in seconds.

Two particles $${P_1}$$ and $${P_2}$$ start moving from a point A at the same time, along different straight lines.

After $$t$$ seconds, the position of $${P_1}$$ is given by r = $$\left( {\begin{array}{*{20}{c}} 4 \\ { – 1} \\ 3 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { – 1} \end{array}} \right)$$.

Two seconds after leaving A, $${P_1}$$ is at point B.

Two seconds after leaving A, $${P_2}$$ is at point C, where $$\overrightarrow {{\text{AC}}}Â = \left( {\begin{array}{*{20}{c}} 3 \\ 0 \\ 4 \end{array}} \right)$$.

Find the coordinates of A.

[2]
a.

FindÂ $$\overrightarrow {{\text{AB}}}$$;

[3]
b.i.

FindÂ $$\left| {\overrightarrow {{\text{AB}}} } \right|$$.

[2]
b.ii.

Find $$\cos {\rm{B\hat AC}}$$.

[5]
c.

Hence or otherwise, find the distance between $${P_1}$$ and $${P_2}$$ two seconds after they leave A.

[4]
d.

## Markscheme

recognizing $$t = 0$$ at A Â  Â  (M1)

A is $$(4,{\text{ }} – 1,{\text{ }}3)$$ Â  Â  A1 Â  Â  N2

[2 marks]

a.

METHOD 1

valid approach Â  Â  (M1)

eg$$\,\,\,\,\,$$$$\left( {\begin{array}{*{20}{c}} 4 \\ { – 1} \\ 3 \end{array}} \right) + 2\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { – 2} \end{array}} \right),{\text{ }}(6,{\text{ }}3,{\text{ }} – 1)$$

correct approach to find $$\overrightarrow {{\text{AB}}}$$ Â  Â  (A1)

eg$$\,\,\,\,\,$$$${\text{AO}} + {\text{OB}},{\text{ B}} – {\text{A, }}\left( {\begin{array}{*{20}{c}} 6 \\ 3 \\ { – 1} \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 4 \\ { – 1} \\ 3 \end{array}} \right)$$

$$\overrightarrow {{\text{AB}}}Â = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 4} \end{array}} \right)$$ Â  Â  A1 Â  Â  N2

METHOD 2

recognizing $$\overrightarrow {{\text{AB}}}$$ is two times the direction vector Â  Â  (M1)

correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$\overrightarrow {{\text{AB}}}Â = 2\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { – 2} \end{array}} \right)$$

$$\overrightarrow {{\text{AB}}}Â = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 4} \end{array}} \right)$$ Â  Â  A1 Â  Â  N2

[3 marks]

b.i.

correct substitution Â  Â  (A1)

eg$$\,\,\,\,\,$$$$\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {{2^2} + {4^2} + {4^2}} ,{\text{ }}\sqrt {4 + 16 + 16} ,{\text{ }}\sqrt {36}$$

$$\left| {\overrightarrow {{\text{AB}}} } \right| = 6$$ Â  Â  A1 Â  Â  N2

[2 marks]

b.ii.

METHOD 1 (vector approach)

valid approach involving $$\overrightarrow {{\text{AB}}}$$ and $$\overrightarrow {{\text{AC}}}$$ Â  Â  (M1)

eg$$\,\,\,\,\,$$$$\overrightarrow {{\text{AB}}}Â \bullet \overrightarrow {{\text{AC}}} ,{\text{ }}\frac{{\overrightarrow {{\text{BA}}}Â \bullet \overrightarrow {{\text{AC}}} }}{{{\text{AB}} \times {\text{AC}}}}$$

finding scalar product and $$\left| {\overrightarrow {{\text{AC}}} } \right|$$ Â  Â  (A1)(A1)

scalar product $$2(3) + 4(0) – 4(4){\text{ }}( =Â – 10)$$

$$\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {{3^2} + {0^2} + {4^2}} {\text{ }}( = 5)$$

substitution of their scalar product and magnitudes into cosine formula Â  Â  (M1)

eg$$\,\,\,\,\,$$$$\cos {\rm{B\hat AC = }}\frac{{6 + 0 – 16}}{{6\sqrt {{3^2} + {4^2}} }}$$

$${\text{cos}}\,B\hat AC = Â – \frac{{10}}{{30}}\left( { = Â – \frac{1}{3}} \right)$$ Â  Â Â A1 Â  Â  N2

Â

METHOD 2 (triangle approach)

valid approach involving cosine rule Â  Â  (M1)

eg$$\,\,\,\,\,$$$$\cos {\rm{B\hat AC = }}\frac{{{\text{A}}{{\text{B}}^2} + {\text{A}}{{\text{C}}^2} – {\text{B}}{{\text{C}}^2}}}{{2 \times {\text{AB}} \times {\text{AC}}}}$$

finding lengths AC and BC Â  Â  (A1)(A1)

$${\text{AC}} = 5,{\text{ BC}} = 9$$

substitution of their lengths into cosine formula Â  Â  (M1)

eg$$\,\,\,\,\,$$$$\cos {\rm{B\hat AC}} = \frac{{{5^2} + {6^2} – {9^2}}}{{2 \times 5 \times 6}}$$

$$\cos {\rm{B\hat AC}} =Â – \frac{{20}}{{60}}{\text{ }}\left( { =Â – \frac{1}{3}} \right)$$ Â  Â  A1 Â  Â  N2

[5 marks]

c.

Note: Â  Â  Award relevant marks for working seen to find BC in part (c) (if cosine rule used in part (c)).

Â

METHOD 1 (using cosine rule)

recognizing need to find BC Â  Â  (M1)

choosing cosine rule Â  Â  (M1)

eg$$\,\,\,\,\,$$$${c^2} = {a^2} + {b^2} – 2ab\cos {\text{C}}$$

correct substitution into RHS Â  Â  A1

eg$$\,\,\,\,\,$$$${\text{B}}{{\text{C}}^2} = {(6)^2} + {(5)^2} – 2(6)(5)\left( { – \frac{1}{3}} \right),{\text{ }}36 + 25 + 20$$

distance is 9Â  Â  Â A1 Â  Â  N2

Â

METHOD 2 (finding magnitude of $$\overrightarrow {BC}$$)Â

recognizing need to find BC Â  Â  (M1)

valid approach Â  Â  (M1)

eg$$\,\,\,\,\,$$attempt to find $$\overrightarrow {{\rm{OB}}}$$ or $$\overrightarrow {{\rm{OC}}}$$, $$\overrightarrow {{\rm{OB}}}Â = \left( {\begin{array}{*{20}{c}} 6 \\ 3 \\ { – 1} \end{array}} \right)$$ or $$\overrightarrow {{\rm{OC}}}Â = \left( {\begin{array}{*{20}{c}} 7 \\ { – 1} \\ 7 \end{array}} \right),{\text{ }}\overrightarrow {{\rm{BA}}}Â + \overrightarrow {{\rm{AC}}}$$

correct working Â  Â  A1

eg$$\,\,\,\,\,$$$$\overrightarrow {{\rm{BC}}}Â = \left( {\begin{array}{*{20}{c}} 1 \\ { – 4} \\ 8 \end{array}} \right),{\text{ }}\overrightarrow {{\rm{CB}}}Â = \left( {\begin{array}{*{20}{c}} { – 1} \\ 4 \\ { – 8} \end{array}} \right),{\text{ }}\sqrt {{1^2} + {4^2} + {8^2}}Â = \sqrt {81}$$

distance is 9 Â  Â Â A1 Â  Â  N2

Â

METHOD 3 (finding coordinates and using distance formula)

recognizing need to find BC Â  Â  (M1)

valid approach Â  Â  (M1)

eg$$\,\,\,\,\,$$attempt to find coordinates of B or C, $${\text{B}}(6,{\text{ }}3,{\text{ }} – 1)$$ or $${\text{C}}(7,{\text{ }} – 1,{\text{ }}7)$$

correct substitution into distance formula Â  Â  A1

eg$$\,\,\,\,\,$$$${\text{BC}} = \sqrt {{{(6 – 7)}^2} + {{\left( {3 – ( – 1)} \right)}^2} + {{( – 1 – 7)}^2}} ,{\text{ }}\sqrt {{1^2} + {4^2} + {8^2}}Â = \sqrt {81}$$

distance is 9 Â  Â  A1 Â  Â  N2

[4 marks]

d.

## Question

The following diagram shows triangle ABC, with $${\text{AB}} = 3{\text{ cm}}$$, $${\text{BC}} = 8{\text{ cm}}$$, and $${\rm{A\hat BC = }}\frac{\pi }{3}$$.

Show that $${\text{AC}} = 7{\text{ cm}}$$.

[4]
a.

The shape in the following diagram is formed by adding a semicircle with diameter [AC] to the triangle.

Find the exact perimeter of this shape.

[3]
b.

## Markscheme

evidence of choosing the cosine rule Â  Â  (M1)

eg$$\,\,\,\,\,$$$${c^2} = {a^2} + {b^2} – ab\cos C$$

correct substitution into RHS of cosine rule Â  Â  (A1)

eg$$\,\,\,\,\,$$$${3^2} + {8^2} – 2 \times 3 \times 8 \times \cos \frac{\pi }{3}$$

evidence of correct value for $$\cos \frac{\pi }{3}$$ (may be seen anywhere, including in cosine rule) Â  Â  A1

eg$$\,\,\,\,\,$$$$\cos \frac{\pi }{3} = \frac{1}{2},{\text{ A}}{{\text{C}}^2} = 9 + 64 – \left( {48 \times \frac{1}{2}} \right),{\text{ }}9 + 64 – 24$$

eg$$\,\,\,\,\,$$$${\text{A}}{{\text{C}}^2} = 49,{\text{ }}b = \sqrt {49}$$

$${\text{AC}} = 7{\text{ (cm)}}$$ Â  Â  AG Â  Â  N0

Note: Â  Â  Award no marks if the only working seen is $${\text{A}}{{\text{C}}^2} = 49$$ or $${\text{AC}} = \sqrt {49}$$ (or similar).

[4 marks]

a.

correct substitution for semicircle Â  Â  (A1)

eg$$\,\,\,\,\,$$$${\text{semicircle}} = \frac{1}{2}(2\piÂ \times 3.5),{\text{ }}\frac{1}{2} \times \piÂ \times 7,{\text{ }}3.5\pi$$

valid approach (seen anywhere) Â  Â  (M1)

eg$$\,\,\,\,\,$$$${\text{perimeter}} = {\text{AB}} + {\text{BC}} + {\text{semicircle, }}3 + 8 + \left( {\frac{1}{2} \times 2 \times \piÂ \times \frac{7}{2}} \right),{\text{ }}8 + 3 + 3.5\pi$$

$$11 + \frac{7}{2}\pi {\text{ }}( = 3.5\piÂ + 11){\text{ (cm)}}$$ Â  Â  A1 Â  Â  N2

[3 marks]

b.