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Date May 2010 Marks available 2 Reference code 10M.1.sl.TZ2.14
Level SL only Paper 1 Time zone TZ2
Command term Calculate Question number 14 Adapted from N/A

Question

A rectangular cuboid has the following dimensions.

Length     0.80 metres     (AD)

Width      0.50 metres     (DG)

Height     1.80 metres     (DC)

Calculate the length of AG.

[2]
a.

Calculate the length of AF.

[2]
b.

Find the size of the angle between AF and AG.

[2]
c.

Markscheme

\({\text{AG}} = \sqrt {{{0.8}^2} + {{0.5}^2}} \)     (M1)

AG = 0.943 m     (A1)     (C2)

[2 marks]

a.

\({\text{AF}} = \sqrt {{\text{A}}{{\text{G}}^2} + {{1.80}^2}} \)     (M1)

= 2.03 m     (A1)(ft)     (C2)

 

Note: Follow through from their answer to part (a).

 

[2 marks]

b.

\(\cos {\rm{G\hat AF}} = \frac{{0.943(39 \ldots )}}{{2.03(22 \ldots )}}\)     (M1)

\(\operatorname{G\hat AF} = 62.3^\circ \)     (A1)(ft)     (C2)

 

Notes: Award (M1) for substitution into correct trig ratio.

Accept alternative ratios which give 62.4° or 62.5°.

Follow through from their answers to parts (a) and (b).

 

[2 marks]

c.

Examiners report

This question was well answered. Surprisingly few candidates used the basic trigonometric ratios (for right angle triangles), opting instead to use the sine or cosine laws.

a.

This question was well answered. Surprisingly few candidates used the basic trigonometric ratios (for right angle triangles), opting instead to use the sine or cosine laws.

b.

This question was well answered. Surprisingly few candidates used the basic trigonometric ratios (for right angle triangles), opting instead to use the sine or cosine laws.

c.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.0 » Pythagoras’ theorem
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