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Date	November 2019	Marks available	1	Reference code	19N.1.AHL.TZ0.H_9
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Show that	Question number	H_9	Adapted from	N/A

Question

In the following diagram, the points ${\text{A}}$ , ${\text{B}}$ , ${\text{C}}$ and ${\text{D}}$ are on the circumference of a circle with centre ${\text{O}}$ and radius $r$ . $\left[ {{\text{AC}}} \right]$ is a diameter of the circle. ${\text{BC}} = r$ , ${\text{AD}} = {\text{CD}}$ and ${\text{A}}\mathop {\text{B}}\limits^ \wedge {\text{C}} = {\text{A}}\mathop {\text{D}}\limits^ \wedge {\text{C}} = 90^\circ$ .

Given that ${\text{cos}}\,75^\circ = q$ , show that ${\text{cos}}\,105^\circ = - q$ .

[1]

a.

Show that ${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{D}} = 75^\circ$ .

[3]

b.

By considering triangle ${\text{ABD}}$ , show that ${\text{B}}{{\text{D}}^2} = 5{r^2} - 2{r^2}q\sqrt 6$ .

[4]

c.i.

By considering triangle ${\text{CBD}}$ , find another expression for ${\text{B}}{{\text{D}}^2}$ in terms of $r$ and $q$ .

[3]

c.ii.

Use your answers to part (c) to show that ${\text{cos}}\,75^\circ = \frac{1}{{\sqrt 6 + \sqrt 2 }}$ .

[3]

d.

Markscheme

${\text{cos}}\,105^\circ = {\text{cos}}\left( {180^\circ - 75^\circ } \right) = - {\text{cos}}\,75^\circ$ R1

$= - q$ AG

Note: Accept arguments using the unit circle or graphical/diagrammatical considerations.

[1 mark]

a.

${\text{AD}} = {\text{CD}} \Rightarrow {\text{C}}\mathop {\text{A}}\limits^ \wedge {\text{D}} = 45^\circ$ A1

valid method to find ${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{C}}$ (M1)

for example: ${\text{BC}} = r \Rightarrow {\text{B}}\mathop {\text{C}}\limits^ \wedge {\text{A}} = 60^\circ$

$\Rightarrow {\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{C}} = 30^\circ$ A1

hence ${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{D}} = 45^\circ + 30^\circ = 75^\circ$ AG

[3 marks]

b.

${\text{AB}} = r\sqrt 3$ , ${\text{AD}} = \left( {{\text{CD}}} \right) = r\sqrt 2$ A1A1

applying cosine rule (M1)

${\text{B}}{{\text{D}}^2} = {\left( {r\sqrt 3 } \right)^2} + {\left( {r\sqrt 2 } \right)^2} - 2\left( {r\sqrt 3 } \right)\left( {r\sqrt 2 } \right){\text{cos}}\,75^\circ$ A1

$= 3{r^2} + 2{r^2} - 2{r^2}\sqrt 6 \,{\text{cos}}\,75^\circ$

$5{r^2} - 2{r^2}q\sqrt 6$ AG

[4 marks]

c.i.

${\text{B}}\mathop {\text{C}}\limits^ \wedge {\text{D}} = 105^\circ$ (A1)

attempt to use cosine rule on $\Delta {\text{BCD}}$ (M1)

${\text{B}}{{\text{D}}^2} = {r^2} + {\left( {r\sqrt 2 } \right)^2} - 2r\left( {r\sqrt 2 } \right){\text{cos}}\,105^\circ$

$= 3{r^2} + 2{r^2}q\sqrt 2$ A1

[3 marks]

c.ii.

$5{r^2} - 2{r^2}q\sqrt 6 = 3{r^2} + 2{r^2}q\sqrt 2$ (M1)(A1)

$2{r^2} = 2{r^2}q\left( {\sqrt 6 + \sqrt 2 } \right)$ A1

Note: Award A1 for any correct intermediate step seen using only two terms.

$q = \frac{1}{{\sqrt 6 + \sqrt 2 }}$ AG

Note: Do not award the final A1 if follow through is being applied.

[3 marks]

d.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

d.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.2—2d and 3d trig, sine rule, cosine rule, area

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Find the size of angle ${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{C}}$ .
19M.2.SL.TZ2.T_2b.ii:
Find the size of angle ${\text{C}}\mathop {\text{A}}\limits^ \wedge {\text{D}}$ .
19M.2.SL.TZ2.T_2c:
Find the size of angle ${\text{A}}\mathop {\text{C}}\limits^ \wedge {\text{D}}$ .
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16N.2.SL.TZ0.T_5c:
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16N.2.SL.TZ0.T_5e:
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