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Date	November 2019	Marks available	2	Reference code	19N.2.SL.TZ0.S_4
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	S_4	Adapted from	N/A

Question

The following diagram shows a right-angled triangle, $ABC$ ${\text{ABC}}$ , with $AC = 10 cm$ ${\text{AC}} = 10\,{\text{cm}}$ , $AB = 6 cm$ ${\text{AB}} = 6\,{\text{cm}}$ and $BC = 8 cm$ ${\text{BC}} = 8\,{\text{cm}}$ .

The points $D$ ${\text{D}}$ and $F$ ${\text{F}}$ lie on $[AC]$ $\left[ {{\text{AC}}} \right]$ .
$[BD]$ $\left[ {{\text{BD}}} \right]$ is perpendicular to $[AC]$ $\left[ {{\text{AC}}} \right]$ .
$BEF$ ${\text{BEF}}$ is the arc of a circle, centred at $A$ ${\text{A}}$ .
The region $R$ $R$ is bounded by $[BD]$ $\left[ {{\text{BD}}} \right]$ , $[DF]$ $\left[ {{\text{DF}}} \right]$ and arc $BEF$ ${\text{BEF}}$ .

Find $B ˆ A C$ ${\text{B}}\widehat {\text{A}}{\text{C}}$ .

[2]

Find the area of $R$ $R$ .

[5]

Markscheme

correct working (A1)

eg $sin α = \frac{8}{10}$ ${\text{sin}}\,\alpha = \frac{8}{{10}}$ , $cos θ = \frac{6}{10}$ ${\text{cos}}\,\theta = \frac{6}{{10}}$ , $cos B ˆ A C = \frac{6^{2} + 10^{2} - 8^{2}}{2 \times 6 \times 10}$ ${\text{cos}}\,{\text{B}}\widehat {\text{A}}{\text{C}} = \frac{{{6^2} + {{10}^2} - {8^2}}}{{2 \times 6 \times 10}}$

$0.927295$ $0.927295$

$B ˆ A C = 0.927$ ${\text{B}}\widehat {\text{A}}{\text{C}} = 0.927$ $(= {53.1}^{\circ})$ $\left( { = 53.1^\circ } \right)$ (A1) N2

[2 marks]

Note: There may be slight differences in the final answer, depending on the approach the candidate uses in part (b). Accept a final answer that is consistent with their working.

correct area of sector $ABF$ ${\text{ABF}}$ (seen anywhere) (A1)

eg $\frac{1}{2} \times 6^{2} \times 0.927$ $\frac{1}{2} \times {6^2} \times 0.927$ , $\frac{{53.1301}^{\circ}}{360^{\circ}} \times π \times 6^{2}$ $\frac{{53.1301^\circ }}{{360^\circ }} \times \pi \times {6^2}$ , $16 .6913$ ${\text{16}}{\text{.6913}}$

correct expression (or value) for either $[AD]$ $\left[ {{\text{AD}}} \right]$ or $[BD]$ $\left[ {{\text{BD}}} \right]$ (seen anywhere) (A1)

eg $AD = 6 cos (B ˆ A C) (= 3.6)$ ${\text{AD}} = 6\,{\text{cos}}\left( {{\text{B}}\widehat {\text{A}}{\text{C}}} \right)\,\,\,\left( { = 3.6} \right)$

$BD = 6 sin ({53.1}^{\circ}) (= 4.8)$ ${\text{BD}} = 6\,{\text{sin}}\left( {53.1^\circ } \right)\,\,\,\left( { = 4.8} \right)$

correct area of triangle $ABD$ ${\text{ABD}}$ (seen anywhere) (A1)

eg $\frac{1}{2} \times 6 cos B ˆ A D \times 6 sin B ˆ A D$ $\frac{1}{2} \times 6\,{\text{cos}}\,{\text{B}}\widehat {\text{A}}{\text{D}} \times 6\,{\text{sin}}\,{\text{B}}\widehat {\text{A}}{\text{D}}$ , $9 sin (2 B ˆ A C)$ $9\,{\text{sin}}\left( {{\text{2}}\,{\text{B}}\widehat {\text{A}}{\text{C}}} \right)$ , $8.64$ $8.64$ (exact)

appropriate approach (seen anywhere) (M1)

eg $A_{triangle ABD} - A_{sector}$ ${{\text{A}}_{{\text{triangle ABD}}}} - {{\text{A}}_{{\text{sector}}}}$ , their sector − their triangle $ABD$ ${\text{ABD}}$

$8.05131$ $8.05131$

area of shaded region $= 8.05$ $= 8.05$ (cm²) A1 N2

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.4—Circle: radians, arcs, sectors

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17M.2.AHL.TZ1.H_8b:
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21M.2.SL.TZ1.8e.i:
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16N.2.AHL.TZ0.H_9a:
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18M.1.SL.TZ2.S_4:
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