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Date	May 2018	Marks available	2	Reference code	18M.2.SL.TZ2.T_5
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Calculate	Question number	T_5	Adapted from	N/A

Question

The Tower of Pisa is well known worldwide for how it leans.

Giovanni visits the Tower and wants to investigate how much it is leaning. He draws a diagram showing a non-right triangle, ABC.

On Giovanni’s diagram the length of AB is 56 m, the length of BC is 37 m, and angle ACB is 60°. AX is the perpendicular height from A to BC.

Giovanni’s tourist guidebook says that the actual horizontal displacement of the Tower, BX, is 3.9 metres.

Use Giovanni’s diagram to show that angle ABC, the angle at which the Tower is leaning relative to the
horizontal, is 85° to the nearest degree.

[5]

a.i.

Use Giovanni's diagram to calculate the length of AX.

[2]

a.ii.

Use Giovanni's diagram to find the length of BX, the horizontal displacement of the Tower.

[2]

a.iii.

Find the percentage error on Giovanni’s diagram.

[2]

Giovanni adds a point D to his diagram, such that BD = 45 m, and another triangle is formed.

Find the angle of elevation of A from D.

[3]

Markscheme

$\frac{{{\text{sin BAC}}}}{{37}} = \frac{{{\text{sin 60}}}}{{56}}$ (M1)(A1)

Note: Award (M1) for substituting the sine rule formula, (A1) for correct substitution.

angle ${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{C}}$ = 34.9034…° (A1)

Note: Award (A0) if unrounded answer does not round to 35. Award (G2) if 34.9034… seen without working.

angle ${\text{A}}\mathop {\text{B}}\limits^ \wedge {\text{C}}$ = 180 − (34.9034… + 60) (M1)

Note: Award (M1) for subtracting their angle BAC + 60 from 180.

85.0965…° (A1)

85° (AG)

Note: Both the unrounded and rounded value must be seen for the final (A1) to be awarded. If the candidate rounds 34.9034...° to 35° while substituting to find angle ${\text{A}}\mathop {\text{B}}\limits^ \wedge {\text{C}}$ , the final (A1) can be awarded but only if both 34.9034...° and 35° are seen.
If 85 is used as part of the workings, award at most (M1)(A0)(A0)(M0)(A0)(AG). This is the reverse process and not accepted.

a.i.

sin 85… × 56 (M1)

= 55.8 (55.7869…) (m) (A1)(G2)

Note: Award (M1) for correct substitution in trigonometric ratio.

a.ii.

$\sqrt {{{56}^2} - 55.7869{ \ldots ^2}}$ (M1)

Note: Award (M1) for correct substitution in the Pythagoras theorem formula. Follow through from part (a)(ii).

cos(85) × 56 (M1)

Note: Award (M1) for correct substitution in trigonometric ratio.

= 4.88 (4.88072…) (m) (A1)(ft)(G2)

Note: Accept 4.73 (4.72863…) (m) from using their 3 s.f answer. Accept equivalent methods.

[2 marks]

a.iii.

$\left| {\frac{{4.88 - 3.9}}{{3.9}}} \right| \times 100$ (M1)

Note: Award (M1) for correct substitution into the percentage error formula.

= 25.1 (25.1282) (%) (A1)(ft)(G2)

Note: Follow through from part (a)(iii).

[2 marks]

${\text{ta}}{{\text{n}}^{ - 1}}\left( {\frac{{55.7869 \ldots }}{{40.11927 \ldots }}} \right)$ (A1)(ft)(M1)

Note: Award (A1)(ft) for their 40.11927… seen. Award (M1) for correct substitution into trigonometric ratio.

(37 − 4.88072…)² + 55.7869…²

(AC =) 64.3725…

64.3726…² + 8² − 2 × 8 × 64.3726… × cos120

(AD =) 68.7226…

$\frac{{{\text{sin 120}}}}{{68.7226 \ldots }} = \frac{{{\text{sin A}}\mathop {\text{D}}\limits^ \wedge {\text{C}}}}{{64.3725 \ldots }}$ (A1)(ft)(M1)