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Date	May 2019	Marks available	3	Reference code	19M.2.SL.TZ2.T_2
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Find	Question number	T_2	Adapted from	N/A

Question

An archaeological site is to be made accessible for viewing by the public. To do this, archaeologists built two straight paths from point A to point B and from point B to point C as shown in the following diagram. The length of path AB is 185 m, the length of path BC is 250 m, and angle ${\text{A}}\mathop {\text{B}}\limits^ \wedge {\text{C}}$ is 125°.

The archaeologists plan to build two more straight paths, AD and DC. For the paths to go around the site, angle ${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{D}}$ is to be made equal to 85° and angle ${\text{B}}\mathop {\text{C}}\limits^ \wedge {\text{D}}$ is to be made equal to 70° as shown in the following diagram.

Find the distance from A to C.

[3]

Find the size of angle ${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{C}}$ .

[3]

b.i.

Find the size of angle ${\text{C}}\mathop {\text{A}}\limits^ \wedge {\text{D}}$ .

[1]

b.ii.

Find the size of angle ${\text{A}}\mathop {\text{C}}\limits^ \wedge {\text{D}}$ .

[2]

The length of path AD is 287 m.

Find the area of the region ABCD.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

AC² = 185² + 250² − 2 × 185 × 250 × cos(125°) (M1)(A1)

Note: Award (M1) for substitution in the cosine formula; (A1) for correct substitution.

387 (387.015…) (m) (A1)(G2)

Note: If radians are used the answer is 154 (154.471…), award at most (M1)(A1)(A0).

[3 marks]

$\frac{{250}}{{{\text{sin}}\,{\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{C}}}} = \frac{{387.015 \ldots }}{{{\text{sin}}\,\left( {125^\circ } \right)}}$ (M1)(A1)(ft)

${\text{co}}{{\text{s}}^{ - 1}}\left( {\frac{{{{185}^2} + 387.015{ \ldots ^2} - {{250}^2}}}{{2 \times 185 \times 387.015 \ldots }}} \right)$ (M1)(A1)(ft)

Note: Award (M1) for substitution in the sine or cosine formulas; (A1)(ft) for correct substitution.

${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{C}} = {\text{31}}{\text{.9}}^\circ$ (31.9478…°) (A1)(ft)(G2)

Note: Follow through from part (a).

[3 marks]

b.i.

(CAD =) 53.1° (53.0521…°) (A1)(ft)

Note: Follow through from their part (b)(i) only if working seen.

[1 mark]

b.ii.

(ACD = ) 70° − (180° − 125° − 31.9478°…) (M1)

Note: Award (M1) for subtracting their angle ${\text{A}}\mathop {\text{C}}\limits^ \wedge {\text{B}}$ from 70°.

(ADC =) 360 − (85 + 70 + 125) = 80

(ACD =) 180 − 80 − 53.0521... (M1)

46.9° (46.9478…°) (A1)(ft)(G2)

Note: Follow through from part (b)(i).

[2 marks]

$\frac{{185 \times 250 \times {\text{sin}}\,\left( {125^\circ } \right)}}{2} + \frac{{287 \times 387.015 \ldots \times {\text{sin}}\,\left( {53.0521 \ldots ^\circ } \right)}}{2}$ (M1)(M1)(M1)

Note: Award (M1) for substitution in the area formula for either triangle; (M1) for correct substitution for both areas; (M1) for adding their two areas;

18942.8… + 44383.9…

63300 (m²) (63326.8… (m²)) (A1)(ft)(G3)

Note: Follow through from parts (a) and (b)(ii).

${\text{DC}} = \frac{{287 \times {\text{sin}}\,\left( {53.0521 \ldots } \right)}}{{{\text{sin}}\,\left( {46.9478 \ldots } \right)}} = 313.884 \ldots$

$0.5 \times 287 \times 185 \times {\text{sin}}\,85^\circ + 0.5 \times 250 \times 313.884 \ldots \times {\text{sin}}\,70^\circ$ M1M1M1

Note: Award (M1) for substitution in the area formula for either triangle; (M1) for correct substitution for both areas; (M1) for adding their two areas;

26446.4… + 36869.3…

63300 (63315.8…) (m²) (A1)(ft)(G3)

[4 marks]